CHAPTER 11
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consent of McGraw-Hill Education.
1865
PROBLEM 11.1
A snowboarder starts from rest at the top of a double black diamond hill. As he rides down the slope, GPS
3
2
coordinates are used to determine his displacement as a function of time: x = 0.5t + t + 2t where x and t are
expressed in ft and seconds, respectively. Determine the position, velocity, and acceleration of the boarder
when t = 5 seconds.
SOLUTION
Position:
x = 0.5t 3 + t 2 + 2t
Velocity:
v=
dx
= 1.5t 2 + 2t + 2
dt
Acceleration:
a=
dv
= 3t + 2
dt
At t = 5 s,
x = 0.5(5) + 52 + 2 (5)
3
v = 1.5(5) + 2 (5) + 2
2
a = 3(5) + 2
x = 9 7 .5 ft
◀
v = 4 9 .5 f t / s
◀
a = 17 ft/s2 ◀
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written consent of McGraw-Hill Education.
1866
PROBLEM 11.2
The motion of a particle is defined by the relation x = t 3 -12t 2 + 36t + 30, where x and t are expressed in
feet and seconds, respectively. Determine the time, the position, and the acceleration of the particle when v
= 0.
SOLUTION
x = t 3 - 12t 2 + 36t + 30
Differentiating,
v=
dx
= 3t 2 - 24t + 36 = 3(t 2 - 8t + 12)
dt
= 3(t - 6)(t - 2)
a=
dv
= 6t - 24
dt
So v = 0 at t = 2 s and t = 6 s.
At t = 2 s,
x1 = (2)3 - 12(2)2 + 36(2) + 30 = 62
a1 = 6(2) - 24 = -12
t = 2.00 s ◀
x1 = 62.00 ft ◀
a1 = -12.00 ft/s2 ◀
At t = 6 s,
x2 = (6)3 - 12(6)2 + 36(6) + 30 = 30
t = 6.00 s ◀
x2 = 30.00 ft ◀
a2 = (6)(6) - 24 = 12
a2 = 12.00 ft/s2 ◀
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written consent of McGraw-Hill Education.
1867
PROBLEM 11.3
The vertical motion of mass A is defined by the relation x = cos (10 t ) - 0.1sin(10 t ), where x
and t are expressed in mm and seconds, respectively. Determine (a) the position, velocity
and acceleration of A when t = 0.4 s, (b) the maximum velocity and acceleration of A.
A
SOLUTION
x = cos(10t ) - 0.1sin(10t )
v=
dx
= -10 sin10t - 0.1(10)cos(10t )
dt
a=
dv
= -100 cos10t + 0.1(100)sin10t
dt
For trigonometric functions set calculator to radians:
(a) At t = 0.4 s
x1 = cos 4 - 0.1sin(4) = 0.578
x1 = 0.578 mm ◀
v1 = -10 sin(4) - cos(4) = 8.222
v1 = 8.22 mm/s ◀
a1 = -100 cos(4) + 10 sin(4) = 57.80
a1 = 57.80 mm/s2 ◀
(b) Maximum velocity occurs when a = 0.
-100 cos(10t ) + 10sin(10t ) = 0
tan10t = 10 , 10t = tan-1 (10) , t = 0.147 s
So
t = 0.147s and 0.147+p s for vmax
vmax = -10 sin(10 * 0.147) - cos(10 * 0.147)
= 10.05
vmax = 10.05 mm/s ◀
Note that we could have also used vmax = 10 2 + 12 = 10.05
by combining the sine and cosine terms. For amax we can take the derivative and set equal to
zero or just combine the sine and cosine terms.
amax = 1002 + 102 = 100.5 mm/s2
amax = 100.5 mm/s2 ◀
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1868
PROBLEM 11.4
v0
A loaded railroad car is rolling at a constant velocity
when it couples with a spring and dashpot bumper
system. After the coupling, the motion of the car is
defined by the relation x = 60 e-4.8t sin16 t where x and
t are expressed in mm and seconds, respectively.
Determine the position, the velocity and the acceleration
of the railroad car when (a) t = 0, (b) t = 0.3 s.
k
c
SOLUTION
x = 60 e-4.8t sin16 t
dx
= 60(-4.8)e-4.8t sin16t + 60(16)e-4.8t cos16t
dt
v = -288e-4.8t sin16t + 960e-4.8t cos16t
v=
a=
dv
= 1382.4e-4.8t sin16t - 4608e-4.8t cos16t
dt
- 4608e-4.8t cos16t -15360e-4.8t sin16t
a = -13977.6e-4.8t sin16t - 9216e-4.8 cos16t
(a) At t = 0,
x0 = 0
x0 = 0 mm ◀
v0 = 960 mm/s
a0 = -9216 mm/s2
(b) At t = 0.3 s,
v0 = 960 mm/s
◀
a0 = 9220 mm/s2
◀
x0.3 = 14.16 mm
◀
v0.3 = 87.9 mm/s
◀
e-4.8t = e-1.44 = 0.23692
sin16t = sin 4.8 = -0.99616
cos16t = cos4.8 = 0.08750
x0.3 = (60)(0.23692)(-0.99616) = -14.16
v0.3 = -(288)(0.23692)(-0.99616)
+ (960)(0.23692)(0.08750) = 87.9
a0.3 = -(13977.6)(0.23692)(-0.99616)
- (9216)(0.23692)(0.08750) = 3108
a0.3 = 3110 mm/s2 ◀
or 3.11 m/s2
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written consent of McGraw-Hill Education.
1869
◀
PROBLEM 11.5
A group of hikers uses a GPS while doing a 40 mile trek in Colorado. A curve fit to the data shows that
their altitude can be approximated by the function, y(t ) = 0.12t 5 - 6.75t 4 +135t 3 -1120t 2 + 3200t + 9070
where y and t are expressed in feet and hours, respectively. During the 18 hour hike, determine (a) the
maximum altitude that the hikers reach, (b) the total feet they ascend, (c) the total feet they descend. Hint:
You will need to use a calculator or computer to solve for the roots of a fourth order polynomial.
SOLUTION
You can graph the function to get the elevation profile.
Differentiate y(t) and set to zero to find when the hikers are ascending and descending.
x = 0.12(5)t 4 -6.75(4)t 3 + 135(3)t 2 -1200(2)t + 3200 = 0
Use a your calculator to find the roots of this polynomial to get
t = 2.151, 8.606 and 14.884 hours
Next, evaluate y(t) at start, end and calculated times to find the elevations
x(0) = 9070 ft (begin to ascend)
y = 11,976 ft at t = 2.151 hours ◀
x(2.151) = 11,976 ft (begin to descend)
x(8.606) = 8,344 ft (begin to ascend)
x(14.884) = 10,103 ft (begin to descend)
x(18) = 9,270 ft
Add the accents to get the total ascent = 4,664 ft
Add the descents to get the total descent = 4,464 ft
total ascent = 4,660 ft ◀
total descent = -4,460 ft ◀
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1870
PROBLEM 11.6
The motion of a particle is defined by the relation x = t 3 - 6t 2 + 9t + 5 , where x is expressed in feet and t
in seconds. Determine (a) when the velocity is zero, (b) the position, acceleration, and total distance
traveled when t = 5 s.
SOLUTION
x = t 3 - 6 t 2 + 9t + 5
Given:
Differentiate twice.
v=
dx
dv
= 3t 2 - 12t + 9 and a =
= 6t - 12
dt
dt
(a) When velocity is zero. v = 0 , 3t 2 - 12t + 9 = 3(t - 1)(t - 3) = 0 ,
(b) Position at t = 5 s.
x5 = (5) - (6)(5) + (9)(5) + 5
Acceleration at t = 5 s.
a5 = (6)(5) - 12
3
2
Position at t = 0.
x0 = 5 ft
Over 0 £ t < 1 s
x is increasing.
Over 1 s < t < 3 s
x is decreasing.
Over 3 s < t £ 5 s
x is increasing.
Position at t = 1 s.
x1 = (1) - (6)(1) + (9)(1) + 5 = 9 ft
Position at t = 3 s.
x3 = (3) - (6)(3) + (9)(3) + 5 = 5 ft
3
3
t = 1 s and t = 3 s ◀
x5 = 25 ft ◀
a5 = 18 ft/s2 ◀
2
2
Distance traveled.
At t = 1 s
d1 = x1 - x0 = 9 - 5 = 4 ft
At t = 3 s
d3 = d1 + x3 - x1 = 4 + 5 - 9 = 8 ft
At t = 5 s
d5 = d3 + x5 - x3 = 8 + 25 - 5 = 28 ft
d5 = 28 ft ◀
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written consent of McGraw-Hill Education.
1871
PROBLEM 11.7
y (m)
A girl operates a radio-controlled model car in a vacant parking
lot. The girl’s position is at the origin of the xy coordinate axes,
and the surface of the parking lot lies in the x-y plane. She
drives the car in a straight line so that the x coordinate is
3
2
defined by the relation x(t) = 0.5t - 3t + 3t + 2, where x and t
are expressed in meters and seconds, respectively. Determine
(a) when the velocity is zero, (b) the position and total distance
travelled when the acceleration is zero.
6
0
x (m)
2
SOLUTION
Position:
x (t ) = 0.5t 3 - 3t 2 + 3t + 2
Velocity:
v (t ) =
dx
dt
v (t ) = 1.5t 2 - 6t + 3
(a) Time when v = 0
0 = 1.5t 2 - 6t + 3
t=
6  62 - 4 (1.5)(3)
a (t ) =
Acceleration:
t = 0.586 s and t = 3.414 s ◀
2 * 1.5
dv
dt
a ( t ) = 3t - 6
Time when a = 0
0 = 3t - 6
(b) Position at t = 2 s
x (2) = 0.5(2)3 -3(2)2 + 3 * 2 + 2
so
t = 2 s
x (2 ) = 0 m ◀
To find total distance note that car changes direction at t = 0.586 s
x (0) = 0.5(0)3 -3(0)2 + 3 * 0 + 2
Position at t = 0 s
x (0 ) = 2
x (0.586) = 0.5(0.586) - 3(0.586) + 3 * 0.586 + 2
3
Position at t = 0.586 s
2
x (0.586) = 2.828 m
Distances traveled:
From t = 0 to t = 0.586 s:
x (0.586) - x (0) = 0.828 m
From t = 0.586 to t = 2 s:
x (2) - x (0.586) = 2.828 m
Total distance traveled =
Total distance =
0 .8 2 8 m + 2 .8 2 8 m
3 .6 5 6 m
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1872
◀
PROBLEM 11.8
The motion of a particle is defined by the relation x = t -(t - 2) , where x and t are expressed in feet and
seconds, respectively. Determine (a) the two positions at which the velocity is zero, (b) the total distance
traveled by the particle from t = 0 to t = 4 s.
2
3
SOLUTION
(
)
Position:
x (t ) = t 2 - t - 2
Velocity:
v (t ) =
3
dx
dt
v(t ) = 2t - 3(t - 2)2
v (t ) = 2t - 3(t 2 - 4t + 4)
= -3t 2 + 14t -12
Time when v(t) = 0
0 = -3t 2 + 14 t - 12
t=
-14  142 - 4 (-3)(-12)
2 (-3)
t = 1.131 s and t = 3.535 s
(a) Position at t = 1.131 s
Position at t = 3.535 s
(
)
3
2
x (3.535) = (3.535) - (3.535 - 2)
x (1.131) = (1.131) - 1.1.31 - 2
2
3
x (1.131) = 1.935 ft. ◀
x (3.531) = 8.879 ft. ◀
To find total distance traveled note that the particle changes direction at t = 1.131 s and again at t = 3.535 s.
(
)
(
)
x ( 0) = ( 0) - 0 - 2
2
Position at t = 0 s
3
x (0 ) = 8 ft
x ( 4) = ( 4 ) - 4 - 2
2
Position at t = 4 s
3
x (4 ) = 8 ft
(b) Distances traveled:
From t = 0 to t = 1.131 s:
x (1.131) - x (0) = 6.065 ft.
From t = 1.131 to t = 3.531 s:
x (3.535) - x (1.131) = 6.944 ft.
From t = 3.531 to t = 4 s:
x ( 4) - x (3.535) = 0.879 ft.
Total distance traveled =
6 . 0 6 5 f t + 6 .9 4 4 f t + 0 .8 7 9 f t
Total distance = 1 3 .8 8 8
ft.
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written consent of McGraw-Hill Education.
1873
◀
v0
PROBLEM 11.9
v=0
A
The brakes of a car are applied, causing it to slow down at a rate
2
of 10 ft/s . Knowing that the car stops in 100 ft, determine
(a) how fast the car was traveling immediately before the brakes
were applied, (b) the time required for the car to stop.
300 ft
x
SOLUTION
a = - 10 ft/s2
(a)
Velocity at
x = 0.
dv
= a = -10
dx
v
ò
0
ò
v0
0-
(b)
xf
vdv = -
(-10)dx
0
v02
= -10 x f = -(10)(300)
2
v02 = 6000
v0 = 77.5 ft/s ◀
Time to stop.
dv
= a = -10
dx
ò
0
ò
tf
dv = -
v0
-10 dt
0
0 - v0 = -10t f
tf =
v0
77.5
=
10
10
t f = 7.75 s ◀
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written consent of McGraw-Hill Education.
1874
PROBLEM 11.10
-0.2 t
, where a and t are expressed in ft/s2 and
The acceleration of a particle is defined by the relation a = 3e
seconds, respectively. Knowing that x = 0 and v = 0 at t = 0, determine the velocity and position of the
particle when t = 0.5 s.
SOLUTION
a =3e-0.2t ft/s2
Acceleration:
Given :
Velocity:
v0 = 0 ft/s, x0 = 0 ft
a=
dv
 dv = adt
dt
v
t
ò dv = ò adt
0
v0
t
v - vo =
ò
3e-0.2 t dt  v - 0= -15 e-0.2 t
t
0
0
(
)
v = 15 1 - e-0.2 t ft/s
Position:
v=
dx
 dx = vdt
dt
x
t
ò dx = ò vdt
0
x0
t
x - xo =
ò (
)
(
15 1 - e-0.2 t dt  x - 0=15 t + 5 e-0.2 t
)0
t
0
(
)
(
)
x = 15 t + 5 e-0.2 t - 75 ft
Velocity at t = 0.5 s
v = 15 1 - e-0.2*0.5 ft/s
v (0.5) = 1.427 ft/s ◀
Position at t = 0.5 s
(
)
x = 15 0.5 + 5 e-0.2*0.5 - 75 ft
x (0.5) = 0.363 ft. ◀
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written consent of McGraw-Hill Education.
1875
PROBLEM 11.11
2
2
The acceleration of a particle is defined by the relation a = 9 – 3t , where a and t are expressed in ft/s and
seconds, respectively. The particle starts at t = 0 with v = 0 and x = 5 ft. Determine (a) the time when the
velocity is again zero, (b) the position and velocity when t = 4 s, (c) the total distance traveled by the particle
from t = 0 to t = 4 s.
SOLUTION
a = 9 - 3t 2
Separate variables and integrate. ò dv = ò a dt = ò (9 - 3t 2 ) dt = 9
v
t
0
0
(
v - 0 = 9 t - t3 , v = t 9 - t2
(a)When v is zero.
)
t (9 - t 2 ) = 0 so t = 0 and t = 3 s (2 roots)
t =3s ◀
(b)Position and velocity at t = 4 s.
ò
x
dx =
5
x-5=
At t = 4 s,
ò
t
v dt =
0
ò (9t - t ) dt
t
3
0
9 2 1 4
9
1
t - t , x = 5 + t2 - t4
2
4
2
4
æ9ö 2 æ1 ö 4
x4 = 5 + çç ÷÷÷(4) - çç ÷÷÷(4)
çè 2 ÷ø
çè 4 ÷ø
(
v4 = (4) 9 - 42
)
x4 = 13 ft ◀
v4 = -28 ft/s ◀
(c)Distance traveled.
Over 0 < t < 3 s,
v is positive, so x is increasing.
Over 3 s < t £ 4 s,
v is negative, so x is decreasing.
At t = 3 s,
æ9ö 2 æ1 ö 4
x3 = 5 + çç ÷÷÷(3) - çç ÷÷÷(3) = 25.25 ft
çè 4 ÷ø
èç 2 ÷ø
At t = 3 s
d3 = x3 - x0 = 25.25 - 5 = 20.25 ft
At t = 4 s
d4 = d3 + x4 - x3 = 20.25 + 13 - 25.25 = 32.5 ft
d4 = 32.5 ft ◀
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written consent of McGraw-Hill Education.
1876
PROBLEM 11.12
Many car companies are performing research on collision avoidance systems. A small prototype applies
engine braking that decelerates the vehicle according to the relationship a = -k t , where a and t are
2
expressed in m/s and seconds, respectively. The vehicle is travelling 20 m/s when its radar sensors detect a
stationary obstacle. Knowing that it takes the prototype vehicle 4 seconds to stop, determine (a) expressions
for its velocity and position as a function of time, (b) how far the vehicle travelled before it stopped.
SOLUTION
Starting with the given function for acceleration,
a = -k t
Differentiate with respect to time to get
v =-
2k 3 2
t + v0
3
Let t = 4 and v0 = 20 to find k when v = 0
2k 3 2
4 + 20
3
k = 3.75
0 =-
2(3.75) 3 2
t + 20
3
v = -2.5t 3/2 + 20
v=-
v = -2.5t 3/2 + 20 ◀
Integrate velocity with respect to time to get position
4
(3.75)t 5/2 + 20t
15
x (t ) = -t 5/2 + 20t
x (t ) = -
x (t ) = -t 5/2 + 20t ◀
Find the distance when t = 4 seconds to find when it stops,
x (t ) = -4 5/2 + 20(4) = 48
x = 48 m ◀
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior
written consent of McGraw-Hill Education.
1877
PROBLEM 11.13
x
A
C
A Scotch yoke is a mechanism that transforms the circular motion of a crank into the
reciprocating motion of a shaft (or vice versa). It has been used in a number of
different internal combustion engines and in control valves. In the Scotch yoke
shown, the acceleration of Point A is defined by the relation a = -1.8sin kt, where a
and t are expressed in m/s and seconds, respectively, and k = 3 rad/s. Knowing that
x = 0 and v = 0.6 m/s when t = 0, determine the velocity and position of Point A
when t = 0.5 s.
2
B
D
SOLUTION
Acceleration:
a = -1.8sin kt m/s2
Given :
v0 = 0.6 m/s, x0 = 0,
Velocity:
dv
a =  dv = adt 
dt
v - v0 =
ò
v - 0.6 =
t
k = 3 rad/s
a dt = -1.8
0
v
t
ò dv = ò adt
v0
ò
0
t
sin kt dt =
0
1.8
cos kt
k
t
0
1.8
(cos kt -1) = 0.6cos kt - 0.6
3
v = 0.6 cos kt m/s
Position:
dx
v=
 dx = vdt 
dt
x - x0 =
x -0 =
ò
t
v dt = 0.6
0
ò
x
t
ò dx = ò vdt
x0
t
0
cos kt dt =
0
0.6
sin kt
k
t
0
0.6
(sin kt - 0) = 0.2sin kt
3
x = 0 .2 s in k t m
When t = 0.5 s,
kt = (3)(0.5) = 1.5 rad
v = 0.6cos1.5 = 0.0424 m/s
x = 0 .2 s i n 1 .5 = 0 .1 9 9 5 m
v = 42.4 mm/s ◀
x = 1 9 9 .5 m m
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written consent of McGraw-Hill Education.
1878
◀
x
PROBLEM 11.14
A
For the scotch yoke mechanism shown, the acceleration of Point A is defined by the
2
relation a = - 1.08 sin kt - 1.44 cos kt , where a and t are expressed in m/s and
seconds, respectively, and k = 3 rad/s. Knowing that x = 0.16 m and v = 0.36 m/s
when t = 0, determine the velocity and position of Point A when t = 0.5 s.
C
B
D
SOLUTION
Acceleration:
a = -1.08sin kt -1.44 cos kt m/s2
Given :
v0 = 0.36 m/s, x0 = 0.16,
Velocity:
dv
a =  dv = adt 
dt
v - v0 = -1.08
Integrate:
v - 0.36 =
=
ò
t
k = 3 rad/s
v
t
ò dv = ò adt
v0
0
sin kt dt - 1.44
0
1.08
cos kt
k
t
-
0
1.44
sin kt
k
ò
t
cos kt dt
0
t
0
1.08
1.44
(cos3t - 1) (sin 3t - 0)
3
3
= 0.36 cos3t - 0.36 - 0.48sin 3t
v = 0.36 cos3t - 0.48sin 3t m/s
Evaluate at
t = 0 .5 s
dx
v=
 dx = vdt 
dt
Position:
v =-453 mm/s ◀
= 0 .3 6 c o s 1 .5 - 0 .3 6 - 0 .4 8 s i n 1 .5
x - x0 =
x - 0.16 =
ò
t
x
ò dx = ò vdt
v dt = 0.36
0
0.36
sin kt
k
t
0
t
+
x0
ò
0
t
cos kt dt - 0.48
0
0.48
cos kt
k
ò
t
sin kt dt
0
t
0
0.36
0.48
(sin 3t - 0) +
(cos3t - 1)
3
3
= 0.12 sin 3t + 0.16 cos3t - 0.16
=
x = 0.12 sin 3t + 0.16 cos3t m
Evaluate at
t = 0 .5 s
x = 0 .1 2 s i n 1 .5 + 0 .1 6 c o s 1 .5 = 0 .1 3 1 0 m
x = 1 3 1 .0 m m
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written consent of McGraw-Hill Education.
1879
◀
PROBLEM 11.15
A piece of electronic equipment that is surrounded by
packing material is dropped so that it hits the ground with a
speed of 4 m/s. After contact the equipment experiences an
acceleration of a = -kx , where k is a constant and x is the
compression of the packing material. If the packing material
experiences a maximum compression of 15 mm, determine
the maximum acceleration of the equipment.
SOLUTION
a=
vdv
= -k x
dx
Separate and integrate.
ò
vf
v0
ò
xf
vdv = -
k x dx
0
1 2 1 2
1
v f - v0 = - kx 2
2
2
2
xf
0
1
= - kx 2f
2
Use v0 = 4 m/s, x f = 0.015 m, and v f = 0. Solve for k.
1
1
0 - (4)2 = - k (0.015)2
2
2
k = 71,111 s-2
Maximum acceleration.
amax = -kxmax : (-71,111)(0.015) = -1,067 m/s2
a = 1,067 m/s2
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior
written consent of McGraw-Hill Education.
1880
◀
PROBLEM 11.16
A projectile enters a resisting medium at x = 0 with an initial velocity
v0 = 1000 ft/s and travels 3 in. before coming to rest. Assuming that the
velocity of the projectile is defined by the relation v = v0 - kx, where v is
expressed in ft/s and x is in feet, determine (a) the initial acceleration of the
projectile, (b) the time required for the projectile to penetrate 2.5 in. into the
resisting medium.
SOLUTION
When x =
æ3
0 = (1000 ft/s) - k çç
çè12
3
ft, v = 0:
12
k = 4000
Or
(a) We have
v = v0 - kx , a =
ö
ft ÷÷÷
ø÷
1
s
dv
d
= (v0 - kx ) = -kv
dt dt
a = -k (v0 - kx )
or
1
s
At t = 0: a = 4000 (1000 ft/s - 0)
a0 = -4.00 ´106 ft/s2 ◀
dx
= v = v0 - kx
dt
(b) We have
ò
At t = 0, x = 0:
x
0
dx
=
v0 - kx
ò
t
dt
0
1
- [ln(v0 - kx )]0x = t
k
or
t=
or
When x = 2.5 in.:
æ
ö
1 æç v0 ÷ö 1 çç 1 ÷÷
÷÷ = ln ç
÷÷
ln çç
k çè v0 - kx ÷ø k ççè1 - vk x ø÷÷
0
t=
é
1
1
ê
ln
ê
4000
1/s
4000 1s ê1 - 1000 ft/s
ë
ù
ú
ft úú
û
( 2.512 )
t = 4.48 ´10-4 s ◀
Or
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1881
PROBLEM 11.17
Point A oscillates with an acceleration a = 100(0.25 - x ), where a and
x are expressed in m /s 2 and meters, respectively. Knowing that the
system starts at time t = 0 with v = 0 and x = 0.2 m, determine the
position and the velocity of A when t = 0.2 s.
SOLUTION
(
)
a = 100 0.25 - x m/s2
a is a function of x:
(
)
Use v dv = a dx = 100 0.25 - x dx with limits v = 0 when x = 0.2 m
ò
v
v dv =
0
ò
x
(
)
(
)
100 0.25 - x dx
0.2
1 2
1
v - 0 = - (100) 0.25 - x
2
2
(
= - 50 0.25 - x
(
v 2 = 0.25 - 100 0.25 - x
So
dx = v dt
Use
or
ò
Integrate:
(
)
t
dt = 
0
2
2
0.2
+ 0.125
(
dx
=
v
)
2
dx
(
 0.5 1 - 400 0.25 - x
x
0.2
x
v =  0.5 1 - 400 0.25 - x
or
dt =
ò
)
2
)
2
dx
(
0.5 1 - 400 0.25 - x
)
2
)
Let u = 20 0.25 - x ; when x = 0.2 u = 1 and du = - 20dx
t=
So
Solve for u.
ò
u
1
du
10 1 - u2
=
u
1
1æ
pö
sin-1 u =  ççsin-1 u - ÷÷÷
10
10 çè
2 ÷ø
1
sin-1 u =
p
 10t
2
æp
ö
u = sin çç  10t ÷÷÷ = cos 10t = cos10t
çè 2
÷ø
(
)
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written consent of McGraw-Hill Education.
1882
PROBLEM 11.17 (CONTINUED)
(
u = cos 10t = 20 0.25 - x
)
Solve for x and v.
x = 0.25 v=
1
cos10t
20
1
sin10t
2
Evaluate at t = 0.2 s.
x = 0.25 v=
1
cos ((10)(0.2))
20
1
sin ((10)(0.2))
2
x = 0.271 m ◀
v = 0.455 m/s ◀
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior
written consent of McGraw-Hill Education.
1883
PROBLEM 11.18
A
A brass (nonmagnetic) block A and a steel magnet B are in equilibrium in a brass
tube under the magnetic repelling force of another steel magnet C located at a
distance x = 0.004 m from B. The force is inversely proportional to the square of
the distance between B and C. If block A is suddenly removed, the acceleration
2
of block B is a = -9.81 + k/x 2 , where a and x are expressed in m/s and m,
-4 3 2
respectively, and k = 4 ´10 m /s . Determine the maximum velocity and
acceleration of B.
B
x
C
SOLUTION
The maximum velocity occurs when
xm2 =
0 = -9.81 +
a = 0.
k
4 ´10-4
=
= 40.775 ´10-6 m 2
9.81
9.81
k
x m2
xm = 0.0063855 m
The acceleration is given as a function of x.
v
dv
k
= a = -9.81 + 2
dx
x
Separate variables and integrate:
vdv = -9.81dx +
ò
v
0
ò
vdv = -9.81
x
k dx
x2
dx + k
x0
ò
x
x0
dx
x2
æ1
1 2
1ö
v = -9.81( x - x0 ) - k ççç - ÷÷÷
çè x x0 ø÷
2
æ 1
1 2
1ö
vm = -9.81( xm - x0 ) - k ççç - ÷÷÷
çè xm x0 ÷ø
2
æ
1
1 ö÷
= -9.81(0.0063855 - 0.004) - (4 ´10-4 )çç
è 0.0063855 0.004 ÷÷ø
= -0.023402 + 0.037358 = 0.013956 m2 /s2
Maximum velocity:
vm = 0.1671 m/s
vm = 167.1 mm/s
◀
The maximum acceleration occurs when x is smallest, that is, x = 0.004 m.
am = - 9.81 +
4 ´10-4
(0.004)2
am =15.19 m/s2 ◀
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written consent of McGraw-Hill Education.
1884
PROBLEM 11.19
Based on experimental observations, the acceleration of a particle is defined by the relation a = -(0.1 + sin x/b),
2
where a and x are expressed in m/s and meters, respectively. Knowing that b = 0 .8 m and that v = 1 m / s when
x = 0, determine (a) the velocity of the particle when x = - 1 m, (b) the position where the velocity is
maximum, (c) the maximum velocity.
SOLUTION
We have
When x = 0, v = 1 m/s:
ò
v
æ
dv
x ö÷
= a = -çç0.1 + sin
÷
è
dx
0.8 ø÷
vdv =
1
ò
x
0
æ
x ö÷
-çç0.1 + sin
÷ dx
è
0.8 ÷ø
é
1 2
x ùx
(v - 1) = - ê0.1x - 0.8 cos
ú
2
0.8 ûú 0
ëê
or
1 2
x
v = -0.1x + 0.8 cos
- 0.3
2
0.8
or
(a)
v
When x = -1 m:
1 2
-1
v = -0.1(-1) + 0.8 cos
- 0.3
2
0.8
v = 0.323 m/s ◀
or
(b)
æ
x ö÷
When v = vmax, a = 0: -çç0.1 + sin
÷= 0
è
0.8 ø÷
or
(c)
x = -0.080 134 m
x = - 0 .0 8 0 1 m
◀
When x = -0.080 134 m:
-0.080 134
1 2
vmax = -0.1(-0.080 134) + 0.8 cos
- 0.3
2
0.8
= 0.504 m 2 /s2
vmax = 1.004 m/s ◀
or
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written consent of McGraw-Hill Education.
1885
PROBLEM 11.20
A
A spring AB is attached to a support at A and to a collar. The
unstretched length of the spring is l. Knowing that the collar is
l
B
C
released from rest at
x = x0
and has an acceleration defined by
the relation a = -100( x - lx/ l 2 + x 2 ), determine the velocity
x0
of the collar as it passes through Point C.
SOLUTION
a=v
Since a is function of x,
æ
dv
lx
= -100 ççç x 2
dx
l + x2
èç
ö÷
÷÷÷
ø÷
Separate variables and integrate:
ò
vf
v0
vdv = -100
ò
0æ
lx
çç x ç
2
x0 ç
l + x2
è
æ x2
ö
1 2 1 2
v f - v0 = -100 ççç - l l 2 + x 2 ÷÷÷
÷ø
2
2
è2
ö÷
÷÷÷ dx
÷ø
0
x0
æ x2
ö
1 2
v f - 0 = -100 ççç- 0 - l 2 + l l 2 + x02 ÷÷÷
÷ø
çè 2
2
1 2 100
vf =
(-l 2 + x02 - l 2 - 2l l 2 + x02 )
2
2
100 2
=
( l + x02 - l )2
2
v f = 10( l 2 + x02 - l ) ◀
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written consent of McGraw-Hill Education.
1886
PROBLEM 11.21
The acceleration of a particle is defined by the relation a = k (1- e ), where k is a constant. Knowing that
the velocity of the particle is v = + 9 m/s when x = -3 m and that the particle comes to rest at the origin,
determine (a) the value of k, (b) the velocity of the particle when x = -2 m.
-x
SOLUTION
Acceleration:
a = k (1 - e-x )
Given :
at x = -3 m,
at x = 0 m,
m/s
v = 9
v = 0
m/s
adx = vdv
k (1 - e- x ) dx = vdv
Integrate using x = -3 m and v = 9 m/s as the lower limits of the integrals
x
ò
v
k (1 - e-x ) dx =
-3
ò vdv
9
x
k ( x + e-x )
Velocity:
v
1
= v2
2 9
-3
(
)
1
1
k x + e-x -(-3 + e3 ) = v2 - (9)2
2
2
(1)
(a) Now substitute v = 0 m/s and x = 0 m into (1) and solve for k
(
)
1
1
k 0 + e-0 -(-3 + e3 ) = 02 - (9)2
2
2
k = 2.52 m 2 /s2 ◀
(b) Find velocity when x = -2 m using the equation (1) and the value of k
(
)
1
1
2.518 -2 + e2 - (-3 + e3 ) = v2 - (9)2
2
2
v = 4 .7 0 m /s
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written consent of McGraw-Hill Education.
1887
◀
PROBLEM 11.22
Starting from x = 0 with no initial velocity, a particle is given an acceleration, a = 0.8 v2 + 49 where a
and v are expressed in ft/s 2 and ft/s, respectively. Determine (a) the position of the particle when v = 24
ft/s, (b) the speed of the particle when x = 40 ft.
SOLUTION
a = 0.8 v 2 + 49
v dv = a dx
dx =
v dv
a
=
v dv
0.8 v2 + 49
Integrating using x = 0 when v = 0,
ò
x
dx =
0
1
0.8
ò
v dv
v
0
v2 + 49
=
1
ù
v2 + 49 ú
0.8
û
(
v2 + 49 - 7
(
24 2 + 49 - 7
x = 1.25
v
0
)
(1)
)
x = 22.5 ft ◀
(a) When v = 24 ft/s,
x = 1.25
(b) Solving equation (1) for v2 ,
v 2 + 49 = 7 + 0.8 x
v2 = (7 + 0.8 x ) - 49
2
When x = 40 ft,
2
v2 = éê 7 + (0.8)(40)ùú - 49 = 1472 ft 2 /s2
ë
û
v = 38.4 ft/s ◀
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written consent of McGraw-Hill Education.
1888
PROBLEM 11.23
A ball is dropped from a boat so that it strikes the surface of a
lake with a speed of 16.5 ft/s. While in the water the ball
experiences an acceleration of a = 10 - 0.8 v, where a and v
2
are expressed in ft/s and ft/s, respectively. Knowing the
ball takes 3 s to reach the bottom of the lake, determine (a) the
depth of the lake, (b) the speed of the ball when it hits the
bottom of the lake.
d
SOLUTION
a=
dv
= 10 - 0.8v
dt
Separate and integrate:
ò
v
dv
=
v0 10 - 0.8v
ò
t
dt
0
v
-
1
ln(10 - 0.8v) = t
0.8
v0
æ 10 - 0.8v ÷ö
ln ççç
÷÷ = -0.8t
èç10 - 0.8v0 ø÷
10 - 0.8v = (10 - 0.8v0 )e-0.8t
0.8v = 10 - (10 - 0.8v0 )e-0.8t
or
v = 12.5 - (12.5 - v0 )e-0.8t
v = 12.5 + 4 e-0.8t
With v0 = 16.5ft/s
Integrate to determine x as a function of t.
v=
ò
x
0
dx =
dx
= 12.5 + 4e-0.8t
dt
ò
t
(12.5 + 4 e-0.8t )dt
0
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written consent of McGraw-Hill Education.
1889
PROBLEM 11.23 (CONTINUED)
x = 12.5t - 5e- 0.8 t
(a)
t
0
At t = 35 s,
x = 12.5(3) - 5e-2.4 + 5 = 42.046 ft
(b)
= 12.5t - 5e- 0.8 t + 5
v = 12.5 + 4e-2.4 = 12.863 ft/s
x = 4 2 .0 ft
◀
v = 1 2 .8 6 f t/s
◀
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written consent of McGraw-Hill Education.
1890
PROBLEM 11.24
The acceleration of a particle is defined by the relation a = - k v , where k is a constant. Knowing that x = 0
and v = 81 m/s at t = 0 and that v = 36 m/s when x = 18 m, determine (a) the velocity of the particle when
x = 20 m, (b) the time required for the particle to come to rest.
SOLUTION
(a)
We have
v
dv
= a = -k v
dx
v dv = -kdx
so that
When x = 0, v = 81 m/s:
ò
v
v dv =
81
ò
x
or
2 3/2 v
[v ]81 = -kx
3
or
2 3/2
[v - 729] = -kx
3
When x = 18 m, v = 36 m/s:
-k dx
0
2 3/2
(36 - 729) = -k (18)
3
k = 19 m/s2
or
Finally
When x = 20 m:
2 3/2
(v - 729) = -19(20)
3
3/2
v
or
(b)
We have
At t = 0, v = 81 m/s:
or
or
When v = 0:
=159
v = 2 9 .3 m / s
◀
t = 0 .9 4 7 s
◀
dv
= a = -19 v
dt
ò
v
81
dv
=
v
ò
t
-19dt
0
v
2[ v ]81
=-19t
2( v - 9) = -19t
2(-9) = -19t
or
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written consent of McGraw-Hill Education.
1891
PROBLEM 11.25
The acceleration of a particle is defined by the relation a = -kv 2.5 , where k is a constant. The particle starts
at x = 0 with a velocity of 16 mm/s, and when x = 6 mm the velocity is observed to be 4 mm/s. Determine
(a) the velocity of the particle when x = 5 mm, (b) the time at which the velocity of the particle is 9 mm/s.
SOLUTION
Acceleration:
a = -kv2.5
Given :
at t = 0,
mm,
v = 16
v = 4
mm/s
x = 0
at x = 6 mm,
mm/s
adx = vdv
-kv 2.5 dx = vdv
Separate variables
-3 2
-kdx = v
dv
Integrate using x = -0 m and v = 16 mm/s as the lower limits of the integrals
x
v
-3 2
ò -kdx = ò v
0
dv
16
x
-1 2 v
-kx
=-2v
16
0
1
2
Now substitute v = 4 mm/s and x = 6 mm into (1) and solve for k
Velocity and position:
-1 2
kx = 2v
k (6 ) = 4
-
-1 2
-
1
2
k = 0.0833 mm
(a)
(1)
-3 2 1 2
s
Find velocity when x = 5 mm using the equation (1) and the value of k
-1 2
k (5) = 2v
-
1
2
v = 4 .7 6 m m / s
(b)
Separate variables
a=
dv
or adt = dv
dt
a=
dv
or adt = dv
dt
-kdt = v-2.5 dv
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written consent of McGraw-Hill Education.
1892
◀
PROBLEM 11.25 (CONTINUED)
Integrate using t = 0 and v = 16 mm/s as the lower limits of the integrals
t
v
-5 2
ò -kdt = ò v
0
dv
16
t
v
2 -3 2
-kt =- v
3
0
16
Velocity and time:
2 -3 2 1
kt = v
3
96
(2)
Find time when v = 9 mm/s using the equation (2) and the value of k
2
1
kt = (9)-3 2 3
96
t = 0 .1 7 1 s
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written consent of McGraw-Hill Education.
1893
◀
Problem 11.26
A human powered vehicle (HPV) team wants to model the
acceleration during the 260 m sprint race (the first 60 m is called a
2
2
flying start) using a = A - Cv , where a is acceleration in m/s and v
is the velocity in m/s. From wind tunnel testing, they found that
-1
C = 0.0012 m . Knowing that the cyclist is going 100 km/h at the
260 meter mark, what is the value of A?
SOLUTION
Acceleration:
a = A - Cv2 m/s2
Given :
C = 0.0012 m-1 , v f =100 km/hr when x f = 260 m
Note: 100 km/hr = 27.78 m/s
adx = vdv
Separate variables
(A -
Cv 2 ) dx = vdv
dx =
vdv
A - Cv2
Integrate starting from rest and traveling a distance xf with a final velocity vf.
xf
vf
ò dx = ò
0
0
vdv
A - Cv2
v
f
1
ln ( A - Cv 2 )
2C
0
1
1
2
xf =ln ( A - v f ) +
ln ( A)
2C
2C
1 æç A ÷÷ö
xf =
ln ç
÷
2C çèç A - Cv 2f ÷÷ø
x
x 0f = -
Next solve for A
A=-
Cv2f e
1- e
2Cx f
2Cx f
Now substitute values of C, xf and vf and solve for A
A = 1.995 m/s2 ◀
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written consent of McGraw-Hill Education.
1894
PROBLEM 11.27
v
Experimental data indicate that in a region downstream of a given louvered
supply vent the velocity of the emitted air is defined by v = 0.18v0 /x,
where v and x are expressed in m/s and meters, respectively, and v 0 is the
initial discharge velocity of the air. For v0 = 3.6 m/s, determine (a) the
acceleration of the air at x = 2 m, (b) the time required for the air to flow
from x = 1 to x = 3 m .
x
SOLUTION
(a)
dv
dx
0.18v0 d æ 0.18v0 ÷ö
çç
=
÷
x dx è x ÷ø
a=v
We have
=-
When x = 2 m:
a =-
0.0324 v02
x3
0.0324(3.6)2
(2)3
a = -0.0525 m/s2 ◀
or
(b)
0.18v0
dx
=v=
dt
x
We have
From x = 1 m to x = 3 m:
or
or
ò
3
xdx =
1
ò
t3
t1
0.18v0 dt
é 1 2 ù3
ê x ú = 0.18v0 (t3 - t1 )
ëê 2 úû1
(t3 - t1 ) =
(9 - 1)
0.18(3.6)
1
2
t3 - t1 = 6.17 s ◀
or
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written consent of McGraw-Hill Education.
1895
PROBLEM 11.28
Based on observations, the speed of a jogger can be approximated by
the relation v = 7.5(1 - 0.04 x )0.3 , where v and x are expressed in km/h
and kilometers, respectively. Knowing that x = 0 at t = 0, determine
(a) the distance the jogger has run when t = 1 h, (b) the jogger’s
2
acceleration in m/s at t = 0, (c) the time required for the jogger to run
6 km.
SOLUTION
Given: v = 7.5(1 - 0.04 x )
0.3
with units km and km/h
(a) Distance at t = 1 hr.
Using dx = v dt , we get dt =
dx
dx
=
v
7.5(1 - 0.04 x )0.3
Integrating, using t = 0 when x = 0,
ò
t
0
dt =
1
7.5
ò
x
0
dx
(1 - 0.04)
0.3
or [t ]t0 =
-1
1
1 - 0.04 x 0.7
⋅
(7.5) (0.7)(0.04)
{
ì
t = 4.7619 ïí1 - 1 - 0.04 x
ï
ï
î
(
ì
ï
x = 25 í1 - 1 - 0.210t
ï
ï
î
(
Solving for x,
x
0
0.7 ü
ï
)
ï
ý
ï
ï
þ
(1)
1/0.7 ï
ü
)
ï
ý
ï
ï
þ
1/0.7 ü
ì
ïý
x = 25 ï
í1 - éêë1 -(0.210)(1)ùúû
ï
ïþï
ï
î
When t = 1 h
}
x = 7.15 km ◀
(b) Acceleration when t = 0.
dv
- 0.7
= (7.5)(0.3)(- 0.04)(1 - 0.04 x )-0.7 = - 0.0900(1 - 0.04 x )
dx
When t = 0 and x = 0,
a=v
v = 7.5 km/h,
dv
- 0.0900 h-1
dx
dv
= (7.5)(-0.0900) = - 0.675 km/h 2
dx
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written consent of McGraw-Hill Education.
1896
PROBLEM 11.28 (CONTINUED)
=-
(0.675)(1000)
(3600)
2
m/s2
a = - 52.1 ´ 10-6 m/s2 ◀
(c) Time to run 6 km.
Using x = 6 km in equation (1),
0.7 ï
ü
ïì
t = 4.7619 í1 - éê1 - (0.04)(6)ùú ý = 0.8323 h
ë
û
îïï
þïï
t = 49.9 min ◀
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written consent of McGraw-Hill Education.
1897
PROBLEM 11.29
P
y
The acceleration due to gravity at an altitude y above the surface of the earth can be
expressed as
a=
- 32.2
[1 + ( y/20.9 ´10 6 )]2
2
where a and y are expressed in ft/s and feet, respectively. Using this expression, compute
the height reached by a projectile fired vertically upward from the surface of the earth if
its initial velocity is (a) 1800 ft/s, (b) 3000 ft/s, (c) 36,700 ft/s.
SOLUTION
v
We have
dv
32.2
= a =dy
æ
ö2
ç1 + y 6 ÷÷÷
çè
20.9 ´10 ø
y = 0,
When
y = ymax , v = 0
provided that v does reduce to zero,
ò
Then
0
v dv =
v0
ò
-32.2
ymax
0
æ
ö2
ç1 + y 6 ÷÷
çè
20.9 ´10 ø
dy
é
ê
1
1
- v02 = -32.2 ê- 20.9 ´10 6
y
ê
2
1 + 20.9 ´
106
êë
or
or
or
ymax =
v0 = 1800 ft/s:
ù ymax
ú
ú
ú
úû 0
æ
ö÷
ç
1
÷
= 1345.96 ´10 çç1 ÷÷÷
ymax
ç 1 + 20.9 ´106 ÷÷
èç
ø
v02
(a)
v = v0
ymax =
6ç
v02
v2
64.4 - 20.9 ´0 106
(1800)2
2
(1800)
64.4 - 20.9
´106
ymax = 50.4 ´103 ft ◀
or
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1898
PROBLEM 11.29 (CONTINUED)
(b)
v0 = 3000 ft/s:
ymax =
(3000)2
2
(3000)
64.4 - 20.9
´106
ymax = 140.7 ´103 ft ◀
or
(c)
v0 = 36, 700 ft/s:
ymax =
(36,700)2
(36,700)2
64.4 - 20.9
´106
= -3.03´1010 ft
This solution is invalid since the velocity does not reduce to zero. The velocity 36,700 ft/s is above the
escape velocity vR from the earth. For vR and above.
y m ax
¥ ◀
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written consent of McGraw-Hill Education.
1899
PROBLEM 11.30
P
The acceleration due to gravity of a particle falling toward the earth is a =-gR /r , where
r is the distance from the center of the earth to the particle, R is the radius of the earth, and g
is the acceleration due to gravity at the surface of the earth. If R = 3960 mi, calculate the
escape velocity, that is, the minimum velocity with which a particle must be projected
vertically upward from the surface of the earth if it is not to return to the earth. (Hint: v = 0
for r = ¥ .)
2
r
2
R
SOLUTION
We have
v
dv
gR 2
= a =- 2
dr
r
r = R, v = ve
When
r = ¥, v = 0
then
ò
0
ve
vdv =
ò
¥
-
R
gR 2
dr
r2
or
é 1 ù¥
1
- ve2 = gR 2 ê ú
2
ëê r ûú R
or
ve = 2 gR
1/2
æ
5280 ft ö÷
= çç2 ´ 32.2 ft/s2 ´ 3960 mi ´
÷÷
è
1 mi ø
ve = 36.7´103 ft/s ◀
or
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written consent of McGraw-Hill Education.
1900
PROBLEM 11.31
The velocity of a particle is v = v0 [1 - sin(pt/T )]. Knowing that the particle starts from the origin with an
initial velocity v 0 , determine (a) its position and its acceleration at t = 3T , (b) its average velocity during the
interval t = 0 to t = T .
SOLUTION
(a)
é
dx
æ pt öù
= v = v0 ê1 - sin çç ÷÷÷ú
è T øúû
êë
dt
We have
At t = 0, x = 0:
At t = 3T :
ò
x
dx =
0
(b)
t
0
é
æ p t öù
v0 ê1 - sin çç ÷÷÷ú dt
è T øúû
êë
t
é
é
æ pt öù
æ pt ö T ù
T
T
x = v0 êt + cos çç ÷÷÷ú = v0 êt + cos çç ÷÷÷ - ú
è T øûú 0
è T ø p ûú
p
p
ëê
ëê
(1)
é
æ p ´ 3T ÷ö T ù
æ
T
2T ö
- ú = v0 çç3T - ÷÷÷
x3T = v0 ê3T + cos çç
è T ÷÷ø p úû
è
êë
p
p ø
x3T = 2.36 v0T ◀
a=
At t = 3T :
ò
é
æ pt öùüï
dv d ìï
p
pt
= ïív0 ê1 - sin çç ÷÷÷úïý = -v0 cos
è T øûúþ
dt dt î
T
T
ïï ëê
ïï
a3T = -v0
p
p ´3T
cos
T
T
a3T =
pv0
◀
T
Using Eq. (1)
At t = 0:
At t = T :
Now
é
T
Tù
x 0 = v0 ê 0 + cos(0) - ú = 0
p
p ûú
ëê
é
æ pT ö T ù
æ
T
2T ö
xT = v0 êêT + cos ççç ÷÷÷ - úú = v0 ççT - ÷÷÷ = 0.363v0T
è
p
pø
è T ø÷ p ûú
ëê
vave =
xT - x0 0.363v0T - 0
=
Dt
T -0
vave = 0.363v0 ◀
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1901
Problem 11.32
e
O
θ
r
An eccentric circular cam, which serves a similar function as the
Scotch yoke mechanism in Problem 11.13, is used in conjunction
with a flat face follower to control motion in pumps and in steam
engine valves. Knowing that the eccentricity is denoted by e, the
maximum range of the displacement of the follower is dmax, and
the maximum velocity of the follower is vmax, determine the
displacement, velocity, and acceleration of the follower.
A
y
SOLUTION
Constraint:
y = r + e cos q
(1)
Differentiate:
y =-eq sin q
(2)
Differentiate again:
y =-eqsin q -eq2 cos q
(3)
ymax occurs when cosθ = 1 and ymin occurs when cosθ = -1
dmax = ymax - ymin
dmax = r + e - (r - e)
dmax = 2e
e=
dmax
2
(4)
y=r +
Substitute (4) into (1) to get Position
Max Velocity occurs when
dmax
cos q ◀
2
s in q =  1
vmax = eq
v
q = max
e
(5)
y = -vmax sin q
Substitute (5) into (2) to get velocity and assume cw rotation.
Substitute (4) and (5) into (3)
y = -
2 ö
dmax 
dmax æç 4vmax
÷÷
ç
q sin q ÷ cos q
ç
2
2
2 çè dmax ÷ø÷

y=-
Acceleration:
dmax 
2v2
q sin q - max cos q ◀
2
dmax
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1902
PROBLEM 11.33
An airplane begins its take-off run at A with zero
velocity and a constant acceleration a. Knowing that it
becomes airborne 30 s later at B with a take-off
velocity of 270 km/h, determine (a) the acceleration
a, (b) distance AB.
SOLUTION
Since it is constant acceleration you can find the acceleration from the distance over time
a = dv / dt
270 km = 75m/s
a=
75m/s
30 s
= 2.5m/s2
a = 2.50 m/s2 ◀
Next find the distance
1 2
at + vo t + x0
2
1
x = 2.5(30)2 + 0t + 0 = 1125
2
x=
x = 1125 m ◀
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written consent of McGraw-Hill Education.
1903
PROBLEM 11.34
A minivan is tested for acceleration and braking. In the street-start
acceleration test, elapsed time is 8.2 s for a velocity increase from 10
km/h to 100 km/h. In the braking test, the distance traveled is 44 m
during braking to a stop from 100 km/h. Assuming constant values of
acceleration and deceleration, determine (a) the acceleration during the
street-start test, (b) the deceleration during the braking test.
SOLUTION
10 km/h = 2.7778 m/s
100 km/h = 27.7778 m/s
(a) Acceleration during start test.
dv
dt
a=
ò
8.2
a dt =
0
ò
27.7778
v dt
2.7778
8.2 a = 27.7778 - 2.7778
a = 3.05 m/s2 ◀
(b) Deceleration during braking.
a=v
ò
44
a dx =
0
a ( x)
44
0
dv
=
dx
0
ò
( )
1
= v2
2
44 a = -
v dv =
27.7778
0
27.7778
2
1
(27.7778)
2
a = - 8.77 m/s2
deceleration = - a = 8.77 m/s2 ◀
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written consent of McGraw-Hill Education.
1904
Problem 11.35
v0
Steep safety ramps are built beside mountain highways to
enable vehicles with defective brakes to stop safely. A truck
enters a 750-ft ramp at a high speed v 0 and travels 540 ft in 6 s
at constant deceleration before its speed is reduced to v0 / 2.
Assuming the same constant deceleration, determine (a) the
additional time required for the truck to stop, (b) the additional
distance traveled by the truck.
S
TRY
VER
MO
N
OU
SS C
CRO
SOLUTION
Given :
xo = 0 , x A = 540 m , t A = 6 s , vA =
Uniform Acceleration:
v A = vo + at A
Substitute known values:
1
vo = vo + (a)(6)
2
a =-
Uniform Acceleration:
Substitute known values:
1
vo , Lramp = 750 ft
2
1
vo
12
1
x A = xo + vo t A + at A2
2
540 = 0 + vo (6) -
1
vo (6)2
24
vo = 120 ft/s and a = - 10 ft/s2
v B = vo + at B
(a)
Substitute known values:
0 = 120 - 10t B
Solve for tB
t B = 12 s
t B - t A = 6.0 s ◀
Additional time to stop
(b)
1
xB = xo + vo tB + atB2
2
Substitute known values:
1
xB = 0 +120(12) - 10(12)2
2
Solve for xB
x B = 720 ft
x B - x A = 180.0 ft ◀
Additional time to stop
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1905
v1
89.6 ft
PROBLEM 11.36
A group of students launches a model rocket in the vertical direction. Based on tracking
data, they determine that the altitude of the rocket was 89.6 ft at the end of the powered
portion of the flight and that the rocket landed 16 s later. Knowing that the descent
parachute failed to deploy so that the rocket fell freely to the ground after reaching its
2
maximum altitude and assuming that g = 32.2 ft/s , determine (a) the speed v 1 of the
rocket at the end of powered flight, (b) the maximum altitude reached by the rocket.
SOLUTION
(a)
We have
1
y = y1 + v1t + at 2
2
At
y=0
t la n d ,
0 = 89.6 ft + v1 (16 s)
Then
+
1
(-32.2 ft/s2 )(16 s)2
2
v1 = 252 ft/s ◀
or
(b)
v2 = v12 + 2a( y - y1 )
We have
At
y = ymax ,
v=0
Then
0 = (252 ft/s)2 + 2(-32.2 ft/s2 )( ymax - 89.6) ft
ymax = 1076 ft ◀
or
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1906
PROBLEM 11.37
A
3m
B
3m
C
d
A small package is released from rest at A and
moves along the skate wheel conveyor ABCD.
The package has a uniform acceleration of
4.8 m/s2 as it moves down sections AB and CD,
and its velocity is constant between B and C. If
the velocity of the package at D is 7.2 m/s,
determine (a) the distance d between C and D,
(b) the time required for the package to reach D.
SOLUTION
(a)
For A
B and C
D we have
v2 = v02 + 2a( x - x0 )
2
vBC
= 0 + 2(4.8 m/s2 )(3 - 0) m
Then, at B
= 28.8 m 2 /s2
2
vD2 = vBC
+ 2aCD ( xD - xC )
and at D
d = 2.40 m ◀
or
For A
d = x D - xC
(7.2 m/s)2 = (28.8 m 2 /s2 ) + 2(4.8 m/s2 )d
or
(b)
(vBC = 5.3666 m/s)
B and C
D we have
v = v0 + at
Then A
B
5.3666 m/s = 0 + (4.8 m/s2 )t AB
t AB = 1.118 04 s
or
and C
7.2 m/s = 5.3666 m/s + (4.8 m/s2 )tCD
D
tCD = 0.38196 s
or
Now, for B
C, we have
xC = x B + v BC t BC
or
3 m = (5.3666 m/s)t BC
or
t BC = 0.55901 s
Finally,
t D = t AB + t BC + tCD = (1.11804 + 0.55901 + 0.38196) s
t D = 2.06 s ◀
or
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written consent of McGraw-Hill Education.
1907
PROBLEM 11.38
v
A sprinter in a 100-m race accelerates uniformly for the first 35 m and then runs with
constant velocity. If the sprinter’s time for the first 35 m is 5.4 s, determine (a) his
acceleration, (b) his final velocity, (c) his time for the race.
SOLUTION
Given:
0 £ x £ 35 m, a = constant
35 m < x £ 100 m,
v = constant
At t = 0, v = 0 when
x = 35 m, t = 5.4 s
Find:
(a)
a
(b)
v when x = 100 m
(c)
t when x = 100 m
(a)
We have
At t = 5.4 s:
or
1
x = 0 + 0t + at 2
2
1
35 m = a(5.4 s)2
2
0 £ x £ 35 m
for
a = 2.4005 m/s2
a = 2.40 m/s2 ◀
(b)
First note that v = vmax for 35 m £ x £ 100 m.
Now
(c)
v2 = 0 + 2a( x - 0)
for
0 £ x £ 35 m
When x = 35 m :
2
vmax
= 2(2.4005 m/s2 )(35 m)
or
vmax = 12.9628 m/s
We have
When x = 100 m :
x = x1 + v0 (t - t1 )
vmax = 12.96 m/s ◀
for
35 m < x £ 100 m
100 m = 35 m + (12.9628 m/s)( t 2 - 5.4) s
t2 = 10.41 s ◀
or
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1908
B
A
O
PROBLEM 11.39
B
A
Automobile A starts from O and
accelerates at the constant rate of
2
0.75 m/s . A short time later it is
passed by bus B which is traveling
in the opposite direction at a
constant speed of 6 m/s. Knowing
that bus B passes point O 20 s after
automobile A started from there,
determine when and where the
vehicles passed each other.
x
SOLUTION
Place origin at 0.
Motion of auto.
( x A )0 = 0, (vA )0 = 0, aA = 0.75 m/s2
x A = ( x A )0 + (v A )0 t +
æ1ö
1
a A t 2 = 0 + 0 + çç ÷÷÷(0.75) t 2
è2ø
2
x A = 0.375t 2 m
Motion of bus.
( xB )0 = ?, (vB )0 = -6 m/s, aB = 0
xB = ( xB )0 - (vB )0 t = ( xB )0 - 6t m
At t = 20 s, x B = 0.
0 = ( xB )0 - (6)(20)
Hence,
( xB )0 = 120 m
x B = 120 - 6 t
When the vehicles pass each other, x B = x A .
120 - 6t = 0.375 t 2
0.375 t 2 + 6 t - 120 = 0
t=
t=
(
- 6  (6)2 - ( 4 )(0.375) -120
)
(2)(0.375)
- 6  14.697
0.75
= 11.596 s
and - 27.6 s
t = 11.60 s ◀
Reject the negative root.
Corresponding values of xA and xB.
x A = (0.375)(11.596) = 50.4 m
2
xB = 120 - (6)(11.596) = 50.4 m
x = 50.4 m ◀
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written consent of McGraw-Hill Education.
1909
PROBLEM 11.40
50 m
vB
vA
A
B
In a boat race, boat A is leading boat B by 50 m
and both boats are traveling at a constant speed
of 180 km/h. At t = 0, the boats accelerate at
constant rates. Knowing that when B passes A,
t = 8 s and v A = 225 km/h, determine (a) the
acceleration of A, (b) the acceleration of B.
SOLUTION
(a)
We have
v A = (v A )0 + a A t
(v A )0 = 180 km/h = 50 m/s
At t = 8 s:
Then
vA = 225 km/h = 62.5 m/s
62.5 m/s = 50 m/s + a A (8 s)
a A = 1.563 m/s2 ◀
or
(b)
We have
1
1
x A = ( x A )0 + (vA )0 t + a A t 2 = 50 m + (50 m/s)(8 s) + (1.5625 m/s2 )(8 s)2 = 500 m
2
2
and
1
x B = 0 + (vB )0 t + aB t 2
2
At t = 8 s:
x A = xB
(vB )0 = 50 m/s
1
500 m = (50 m/s)(8 s) + aB (8 s)2
2
a B = 3.13 m/s2 ◀
or
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written consent of McGraw-Hill Education.
1910
PROBLEM 11.41
(vA)0 = 30 ft/s
(vB )0 = 0
A
As relay runner A enters the 65-ft-long exchange zone with a
speed of 30 ft/s, he begins to slow down. He hands the baton to
runner B 2.5 s later as they leave the exchange zone with the
same velocity. Determine (a) the uniform acceleration of each
of the runners, (b) when runner B should begin to run.
B
65 ft
SOLUTION
Let x = 0 at the start of the exchange zone, and t = 0 as runner A enters the exchange zone.
Then,
(vA )0 = 30 ft/s.
Motion of runner A :
v A = (v A )0 + a A t
1
x A = (vA )0 t + aAt 2
2
(a) Solving for a A , and noting that x A = 65 ft when t = 2.5 s
2 éê x A - (vA )0 t ùú 2 é65 - (30)(2.5)ù
û= ë
û
aA = ë
2.52
t2
aA = -3.20 ft/s2 ◀
(vA ) f = 30 +(-3.20)(2.5) = 22.0 ft/s
At t = 2.5 s,
Motion of runner B: (v B )0 = 0 and x B = 0 at the starting time t B of runner B.
vB2 - (vB )0 = 2aB xB
2
Then,
Solving for a B , and noting that vB = (vA ) f = 22.0 ft/s
When x B = 65 ft,
aB =
2
2
(vB ) - (vB )0
2 xB
=
22.02 - 0
(2)(65)
aB = 3.723 ft/s2 ◀
v B = (v B )0 + a B (t - t B ) = a B (t - t B )
t - tB =
vB
aB
(b) Solving for t B , and noting that vB = 22.0 ft/s when t = 2.5 s
tB = t -
vB
22.0
= 2.5 = -3.41 s
aB
3.723
Runner B should begin 3.41 s before runner A reaches the exchange zone. ◀
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written consent of McGraw-Hill Education.
1911
PROBLEM 11.42
(vA)0 = 24 mi/h
(vB)0 = 36 mi/h
A
B
75 ft
x
Automobiles A and B are traveling in adjacent highway lanes
and at t = 0 have the positions and speeds shown. Knowing
that automobile A has a constant acceleration of 1.8 ft/s2 and
that B has a constant deceleration of 1.2 ft/s2 , determine
(a) when and where A will overtake B, (b) the speed of each
automobile at that time.
SOLUTION
a A = +1.8 ft/s2
aB = -1.2 ft/s2
(v A )0 = 24 mi/h = 35.2 ft/s
A overtakes B
(vB )0 = 36 mi/h = 52.8 ft/s
Motion of auto A:
v A = (v A )0 + a A t = 35.2 + 1.8t
(1)
1
1
x A = ( x A )0 + (vA )0 t + aAt 2 = 0 + 35.2t + (1.8)t 2
2
2
(2)
vB = (vB )0 + aB t = 52.8 - 1.2t
(3)
1
1
xB = ( xB )0 + (vB )0 t + aB t 2 = 75 + 52.8t + (-1.2)t 2
2
2
(4)
Motion of auto B:
(a)
A overtakes B at t = t1.
x A = xB : 35.2t + 0.9t12 = 75 + 52.8t1 - 0.6t12
1.5t12 -17.6t1 - 75 = 0
t1 = - 3.22 s
Eq. (2):
and
t1 = 15.0546
x A = 35.2(15.05) + 0.9(15.05)2
t1 = 15.05 s ◀
x A = 734 ft ◀
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written consent of McGraw-Hill Education.
1912
PROBLEM 11.42 (CONTINUED)
(b)
Velocities when t1 = 15.05 s
Eq. (1):
v A = 35.2 + 1.8(15.05)
v A = 62.29 ft/s
Eq. (3):
vA = 42.5 mi/h
◀
vB = 23.7 mi/h
◀
vB = 52.8 - 1.2(15.05)
vB = 34.74 ft/s
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written consent of McGraw-Hill Education.
1913
PROBLEM 11.43
Two automobiles A and B are approaching each other in adjacent highway lanes. At t = 0, A and B are 3200 ft
apart, their speeds are v A = 65 mi/h and vB = 40 mi/h, and they are at Points P and Q, respectively. Knowing
that A passes Point Q 40 s after B was there and that B passes Point P 42 s after A was there, determine (a) the
uniform accelerations of A and B, (b) when the vehicles pass each other, (c) the speed of B at that time.
vA = 65 mi/h
vB = 40 mi/h
A
B
P
3200 ft
Q
SOLUTION
(a)
1
x A = 0 + (vA )0 t + aAt 2 (vA )0 = 65 mi/h = 95.33 ft/s
2
We have
(x is positive
; origin at P.)
1
3200 m = (95.333 m/s)(40 s) + aA (40 s)2
2
At t = 40 s:
1
2
Also, x B = 0 + (vB )0 t + aB t
2
(xB is positive
At t = 42 s:
(vB )0 = 40 mi/h = 58.667 ft/s
; origin at Q.)
1
3200 ft = (58.667 ft/s)(42 s) + aB (42 s)2
2
aB = 0.83447 ft/s2
or
(b)
aA =-0.767 ft/s2 ◀
When the cars pass each other
aB = 0.834 ft/s2 ◀
x A + x B = 3200 ft
1
1
2
2 2
= 3200 ft
Then (95.333 ft/s)t AB + (-0.76667 ft/s)t AB + (58.667 ft/s)t AB + (0.83447 ft/s )t AB
2
2
or
Solving
(c)
2
0.03390t AB
+ 154t AB - 3200 = 0
t = 20.685 s
and
t = - 4563 s
We have
v B = ( v B )0 + a B t
At t = t AB :
vB = 58.667 ft/s + (0.83447 ft/s2 )(20.685 s)
t>0
t AB = 20.7 s ◀
vB = 51.8 mi/h ◀
= 75.927 ft/s
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written consent of McGraw-Hill Education.
1914
PROBLEM 11.44
An elevator is moving upward at a constant speed of 4 m/s. A man standing 10 m
above the top of the elevator throws a ball upward with a speed of 3 m/s. Determine
(a) when the ball will hit the elevator, (b) where the ball will hit the elevator with
respect to the location of the man.
10 m
SOLUTION
Place the origin of the position coordinate at the level of the standing man, the positive direction being up.
The ball undergoes uniformly accelerated motion.
1
yB = ( yB )0 + (vB )0 t - gt 2
2
with ( yB )0 = 0, (vB )0 = 3 m/s, and g = 9.81 m/s2 .
yB = 3t - 4.905t 2
The elevator undergoes uniform motion.
yE = ( yE )0 + vE t
with ( yE )0 = - 10 m and vE = 4 m/s.
(a)
Set y B = yE
Time of impact.
3t - 4.905t 2 = -10 + 4t
4.905t 2 + t -10 = 0
t = 1.3295 and - 1.5334
(b)
t = 1.330 s ◀
Location of impact.
yB = (3)(1.3295) - (4.905)(1.3295)2 = -4.68 m
yE = -10 + (4)(1.3295) = -4.68 m
(checks)
4.68 m below the man ◀
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1915
PROBLEM 11.45
A
v0
B
300 m
Two rockets are launched at a fireworks display. Rocket A is launched with
an initial velocity v0 = 100 m/s and rocket B is launched t1 seconds later
with the same initial velocity. The two rockets are timed to explode
simultaneously at a height of 300 m as A is falling and B is rising. Assuming
a constant acceleration g = 9.81 m/s2 , determine (a) the time t1, (b) the
velocity of B relative to A at the time of the explosion.
v0
SOLUTION
Place origin at ground level. The motion of rockets A and B is
Rocket A:
Rocket B:
v A = (v A )0 - gt = 100 - 9.81t
(1)
1
yA = ( yA )0 + (vA )0 t - gt 2 = 100t - 4.905t 2
2
(2)
vB = (vB )0 - g(t - t1 ) = 100 - 9.81(t - t1 )
(3)
1
yB = ( yB )0 + (vB )0 (t - t1 ) - g(t - t1 )2
2
= 100(t - t1 ) - 4.905(t - t1 )2
(4)
Time of explosion of rockets A and B. y A = y B = 300 ft
From (2),
300 = 100t - 4.905t 2
4.905t 2 -100t + 300 = 0
t = 16.732 s and
From (4),
300 = 100(t - t1 ) - 4.905(t - t12 )
t - t1 = 16.732 s
Since rocket A is falling,
Since rocket B is rising,
(a)
Time t1:
(b)
Relative velocity at explosion.
3.655 s
and
3.655 s
t = 16.732 s
t - t1 = 3.655 s
t1 = t - (t - t1 )
From (1),
v A = 100 - (9.81)(16.732) = - 64.15 m/s
From (3),
v B = 100 - (9.81)(16.732 - 13.08) = 64.15 m/s
Relative velocity:
vB/A = vB - vA
t1 = 13.08 s ◀
v B/A = 128.3 m/s ◀
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1916
PROBLEM 11.46
Car A is parked along the northbound lane of a highway, and car B is traveling in the southbound lane at a
constant speed of 60 mi/h. At t = 0, A starts and accelerates at a constant rate a A , while at t = 5 s, B begins to
slow down with a constant deceleration of magnitude a A /6. Knowing that when the cars pass each other
x = 294 ft and v A = v B , determine (a) the acceleration a A , (b) when the vehicles pass each other, (c) the
distance d between the vehicles at t = 0.
A
(vB)0 = 60 mi /h
(vA)0 = 0
B
x
d
SOLUTION
For t ³ 0 :
vA = 0 + aAt
1
x A = 0 + 0 + aA t 2
2
0 £ t < 5 s:
x B = 0 + ( v B )0 t
At t = 5 s :
x B = (88 ft/s)(5 s) = 440 ft
For t ³ 5 s :
( v B )0 = 60 mi/h = 88 ft/s
1
aB = - aA
6
1
xB = ( xB )S + (vB )0 (t - 5) + aB (t - 5)2
2
vB = (vB )0 + aB (t - 5)
Assume t > 5 s when the cars pass each other.
At that time (t AB ),
v A = vB :
aAt AB = (88 ft/s) -
x A = 294 ft :
1 2
294 ft = aAt AB
2
Then
or
a A( 67 t AB - 65 )
1
2
2
a A t AB
=
aA
(t AB - 5)
6
88
294
2
44t AB
- 343t AB + 245 = 0
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1917
PROBLEM 11.46 (CONTINUED)
Solving
(a)
With t AB > 5 s,
t AB = 0.795 s and
t AB = 7.00 s
1
294 ft = aA (7.00 s)2
2
aA = 12.00 ft/s2 ◀
or
(b)
t AB = 7.00 s ◀
From above
Note: An acceptable solution cannot be found if it is assumed that t AB £ 5 s.
(c)
We have
d = x + ( x B )t AB
= 294 ft + 440 ft + (88 ft/s)(2.00 s)
ö
1æ 1
+ çç- ´12.00 ft/s2 ÷÷÷ (2.00 s)2
ø
2è 6
d = 906 ft ◀
or
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1918
PROBLEM 11.47
W
C
The elevator shown in the figure moves downward with a constant velocity of
4 m/s. Determine (a) the velocity of the cable C, (b) the velocity of the
counterweight W, (c) the relative velocity of the cable C with respect to the
elevator, (d ) the relative velocity of the counterweight W with respect to the
elevator.
E
M
SOLUTION
Choose the positive direction downward.
(a)
Velocity of cable C.
yC + 2 yE = constant
vC + 2vE = 0
vE = 4 m/s
But,
or
(b)
vC = -2vE = -8 m/s
vC = 8.00 m/s ◀
Velocity of counterweight W.
yW + y E = constant
vW + v E = 0
(c)
vW = -v E = - 4 m/s
vW = 4.00 m/s ◀
Relative velocity of C with respect to E.
vC /E = vC - v E = (- 8 m/s) - (+ 4 m/s) = - 12 m/s
v C /E = 12.00 m/s ◀
(d )
Relative velocity of W with respect to E.
vW /E = vW - v E = (- 4 m/s) - (4 m/s) = - 8 m/s
v W /E = 8.00 m/s ◀
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1919
PROBLEM 11.48
The elevator shown starts from rest and moves upward with a constant
acceleration. If the counterweight W moves through 30 ft in 5 s, determine
(a) the acceleration of the elevator and the cable C, (b) the velocity of the
elevator after 5 s.
W
C
E
M
SOLUTION
We choose positive direction downward for motion of counterweight.
1
yW = aW t 2
2
At t = 5 s,
yW = 30 ft
1
30 ft = aW (5 s)2
2
aW = 2.4 ft/s2
(a)
Accelerations of E and C.
Since
Thus:
Also,
Thus:
(b)
aW = 2.4 ft/s2
yW + y E = constant
vW + v E = 0, and
aW + a E = 0
aE = -aW = -(2.4 ft/s2 ),
yC + 2 y E = constant,
vC + 2 v E = 0, and
a E = 2.40 ft/s2 ◀
aC + 2 a E = 0
aC = -2aE = -2(-2.4 ft/s2 ) = +4.8 ft/s2 ,
aC = 4.80 ft/s2 ◀
Velocity of elevator after 5 s.
vE = (vE )0 + aE t = 0 + (-2.4 ft/s2 )(5 s) = -12 ft/s
(v E )5 = 12.00 ft/s ◀
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1920
Problem 11.49
An athlete pulls handle A to the left with a
constant velocity of 0.5 m/s. Determine
(a) the velocity of the weight B, (b) the
relative velocity of weight B with respect
to the handle A.
A
B
SOLUTION
From the diagram:
x A + 4 yB = constant
(a)
v A + 4 vB = 0
Differentiate
Sketch:
(1)
x A = 0.5 m/s
vB = -0.125 m/s
Solve (1) for vB
vB = 0.125 m/s  ◀
(b) Finding the relative velocity:

vB = 0.125  m/s

v A = 0.5 ¬ m/s



vB / A = vB - vA
= 0.125  -(0.5 ¬) m/s
or
v B / A = 0.5  +0.125  m/s
v B / A = 0.5154 m/s
14 ◀
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1921
Problem 11.50
An athlete pulls handle A to the left with a constant
acceleration. Knowing that after the weight B has been lifted
4 in. its velocity is 2 ft/s, determine (a) the accelerations of
handle A and weight B (b) the velocity and change in
position of handle A after 0.5 sec.
A
B
SOLUTION
(a)
From the diagram:
Sketch:
x A + 3yB = constant
v A + 3vB = 0
(1)
a A + 3aB = 0
(2)
Uniform Acceleration of weight B
v B 2 = (v B )o +2 aB ( yB - ( yB )o )
2
Substitute in known values
æ 4
ö
ft - 0÷÷÷
(-2 ft/s)2 = 0 2 + 2 a B ççè 12
ø
aB = -6 ft/s2
Using (2)
aA + 3(-6 ft/s2 ) = 0
aA = 18 ft/s2 ¬ ◀
aB = 6 ft/s2  ◀
(b) Uniform Acceleration of Handle A
v A = (v A )o + a A t
v A = 0 + (18 ft/s2 )(0.5 s)
v A = 9 ft/s ¬ ◀
or
Also
1
yA = ( yA )o + (vA )o t + aAt 2
2
yA -( yA )o = 0 +
1
(18 ft/s2 )(0.5 s)2
2
DyA = 2.25 ft ¬ ◀
or
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1922
PROBLEM 11.51
In the position shown, collar B moves to the left with a constant
velocity of 300 mm/s. Determine (a) the velocity of collar A, (b) the
velocity of portion C of the cable, (c) the relative velocity of portion C
of the cable with respect to collar B.
SOLUTION
Let x be position relative to the right supports, increasing to the left.
xA
xB
(
)
2 x A + x B + x B - x A = constant
Constraint of entire cable:
2 vB + v A = 0
Constraint of point C of cable:
v A = - 2 vB
2 x A + xC = constant
2 v A + vC = 0
vC = - 2 v A
(a) Velocity of collar A.
v A = - 2 vB = -(2)(300) = - 600 mm/s
v A = 600 mm/s
◀
vC = 1200 mm/s
◀
vC /B = 900 mm/s
◀
(b) Velocity of point C of cable.
vC = - 2 v A = -(2)(-600) = 1200 mm/s
(c) Velocity of point C relative to collar B.
vC/B = vC - vB = 1200 - 300 = 900 mm/s
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1923
PROBLEM 11.52
Collar A starts from rest and moves to the right with a constant
acceleration. Knowing that after 8 s the relative velocity of collar B
with respect to collar A is 610 mm/s, determine (a) the accelerations of
A and B, (b) the velocity and the change in position of B after 6 s.
SOLUTION
Let x be position relative to the right supports, increasing to the left.
xA
xB
Constraint of entire cable:
(
)
2 x A + x B + x B - x A = constant,
2 vB + v A = 0,
1
vB = - v A ,
2
or
1
aB = - a A
2
and
(a) Accelerations of A and B.
1
v B /A = v B - v A = - v A - v A
2
vA = v A - (v A )0 = a A t ,
or
2
v A = - vB /A
3
2
(610) = - 406.67 mm/s
3
aA =
v A - (v A )0
t
=
- 406.67 - 0
1
1
aB = - a A = - (-50.8)
2
2
8
= - 50.8 mm/s2
a A = 50.8 mm/s2
◀
aB = 25.4 mm/s2
◀
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1924
PROBLEM 11.52 (CONTINUED)
(b) Velocity and change in position of B after 6 s.
vB = (vB )0 + aB t = 0 + (25.4)(6)
x B - ( x B )0 = (vB )0 t +
2
1
1
aB t 2 = (25.4)(6)
2
2
vB = 152.5 mm/s
◀
Dx B = 458 mm
◀
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1925
PROBLEM 11.53
A farmer wants to get his hay bales into the top loft of his
barn by walking his horse forward with a constant
velocity of 1 ft/s. Determine the velocity and acceleration
of the hay bale when the horse is 10 ft away from the
barn.
20 ft
SOLUTION
From the diagram, we can write an expression for the total length of rope:
L = 20 - yB + x H2 + (20)2
Differentiate
dL
=0
dt
x H x H
0 = -y B +
Velocities
vB = y B and vH = x H
vB =
Differentiate again:
x H2 + (20)2
aB =
aB =
or
aB =
at xH = 10
vB =
and
aB =
x H vH
x H2 + (20)2
dvB
dt
x H vH
x H2 + (20)2
v H2
x H2 + (20)2
x H v H
+
x H2 + (20)2
x H aH
+
x H2 + (20)2
+
-
x H vH (- 12 )(2 x H ) x H
(x
2
H
+ (20)2
)
3/2
x H2 v H2
( x H2 + (20)2 )
3/2
10(1)
vB = 0.447 ft/s ◀
(10)2 + (20)2
(1)2
(10)2 + (20)2
+
(10)(0)
(10)2 + (20)2
aB = 0.0447 + 0 - 0.000894
-
(10)2 (1)2
((10)2 + (20)2 )
3/ 2
aB = 0.0358 ft/s2 ◀
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1926
PROBLEM 11.54
The motor M reels in the cable at a constant rate of 100 mm/s. Determine
(a) the velocity of load L, (b) the velocity of pulley B with respect to load L.
B
100 mm/s
L
M
SOLUTION
Let xB and xL be the positions, respectively, of pulley B and load L measured downward from a fixed elevation
above both. Let xM be the position of a point on the cable about to enter the reel driven by the motor. Then,
considering the lengths of the two cables,
x M + 3x B = constant
vM + 3vB = 0
x L + ( x L - x B ) = constant
2 vL + vB = 0
vM = 100 mm/s
with
vB = vL =
vM
= -33.333 m/s
3
vB
= -16.667 mm/s
2
v L = 16.67 mm/s ◀
(a)
Velocity of load L.
(b)
Velocity of pulley B with respect to load L. v B/L = v B - v L = - 33.333 - (- 16.667) = - 16.667
v B /L = 16.67 mm/s ◀
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1927
PROBLEM 11.55
A
B
Collar A starts from rest at t = 0 and moves upward with a constant acceleration of
2
3.6 in./s . Knowing that collar B moves downward with a constant velocity of 18 in./s,
determine (a) the time at which the velocity of block C is zero, (b) the corresponding
position of block C.
C
SOLUTION
Define positions as positive downward from a fixed level.
Constraint of cable.
( yB - yA ) + ( yC - yA ) + 2( yC - yB ) = constant
3 yC - y B - 2 y A = constant
Differentiate
3vC - v B - 2 v A = 0
(1)
Differentiate again
3aC - a B - 2 a A = 0
(2)
Motion of block C:
(vA )0 = 0, aA = -3.6 in./s2 , vB = (vB )0 = 18 in./s, aB = 0
Using (1)
1
(vC )0 = éëê(vB )0 + 2(vA )0 ùûú = 6 in./s
Using (2)
aC =
Constant acceleration
Constant acceleration
3
(
)
(
)úùû = -2.4 in./s2
1
1
aB + 2aA = êé0 + (2) -3.6
3
3ë
vC = (vC )0 + aC t
(3)
= 6 - 1.2 t
xC - ( xC )0 = (vC )0 t +
1
aC t 2
2
(4)
= 6t - 0.6t 2
(a) Time at vC = 0 :
Using (3)
0 = 6 - 2.4t
t = 2.5 s ◀
(b) Corresponding position of block C:
Using (4)
æ1ö
2
xC - ( xC )0 = (6 )(2.5) + çç ÷÷÷(-2.4 )(2.5)
è2ø
xC - ( xC )0 = 7.5 in.  ◀
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1928
PROBLEM 11.56
Collars A and B start from rest, and collar A moves upward with an
acceleration of 3t 2 mm/s2 . Knowing that collar B moves downward
with a constant acceleration and that its velocity is 150 mm/s after
moving 700 mm, determine (a) the acceleration of block C, (b) the
distance through which block C will have moved after 3 s.
SOLUTION
Let x be position relative to upper support, positive downward.
d
xA
xB
xC
Let d = value of x at lower support.
Constraint of cable:
(d - x A ) + ( xC - x A ) + 2 xC + ( xC - x B ) = constant
4 vC - vB - 2 v A = 0
and
4 aC - aB - 2a A = 0
2
vB2 - (vB )0 = 2aB éê x B - ( x B )0 ùú
ë
û
Accelerations:
vB2 - (vB )0
2
aB =
2 éê x B - ( x B )0 ùú
ë
û
=
1502 - 0
= 16.07 mm/s2
2 (700 - 0)
a A = -3t 2
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1929
PROBLEM 11.56 (CONTINUED)
aC =
(a)
(
(b) Velocity and position:
vC = (vC )0 +
xC - ( xC )0 =
At t = 3 s,
)
1
1
(a + 2aA ) = éêë16.07 + (2) -3t 2 ùúû
4 B
4
ò
t
ò
t
0
0
aC = 4.02 - 1.5t 2 mm/s2 ◀
aC dt = 0 + 4.018t - 0.5t 3
vC dt = 2.009t 2 - 0.125t 4
xC - ( xC )0 = (2.009)(3) - (0.125)(3)
2
4
DxC = 7.96 mm
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1930
◀
PROBLEM 11.57
C
Block B starts from rest, block A moves with a constant
acceleration, and slider block C moves to the right with a constant
acceleration of 75 mm/s2 . Knowing that at t = 2 s the velocities
of B and C are 480 mm/s downward and 280 mm/s to the right,
respectively, determine (a) the accelerations of A and B, (b) the
initial velocities of A and C, (c) the change in position of slider
block C after 3 s.
B
A
SOLUTION
From the diagram
3 y A + 4 yB + xC = constant
Then
3v A + 4 vB + vC = 0
(1)
and
3a A + 4 aB + aC = 0
(2)
(vB ) = 0,
Given:
a A = constent
(aC ) = 75 mm/s2
v B = 480 mm/s
At t = 2 s,
v C = 280 mm/s
(a)
Eq. (2) and a A = constant and aC = constant  a B = constant
vB = 0 + aB t
Then
At t = 2 s:
480 mm/s = aB (2 s)
aB = 240 mm/s2
or
a B = 240 mm/s2 ◀
Substituting into Eq. (2)
3aA + 4(240 mm/s2 ) + (75 mm/s2 ) = 0
aA = -345 mm/s
or
a A = 345 mm/s2 ◀
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1931
PROBLEM 11.57 (CONTINUED)
(b)
vC = (vC )0 + aC t
We have
At t = 2 s:
280 mm/s = ( vC )0 + (75 mm/s)(2 s)
vC = 130 mm/s
or
(vC )0 = 130.0 mm/s
◀
Then, substituting into Eq. (1) at t = 0
3(v A )0 + 4(0) + (130 mm/s) = 0
vA = -43.3 mm/s
(c)
We have
At t = 3 s :
or
(v A )0 = 43.3 mm/s ◀
1
xC = ( xC )0 + (vC )0 t + aC t 2
2
1
xC - ( xC )0 = (130 mm/s)(3 s) + (75 mm/s2 )(3 s)2
2
= 728 mm
or xC - (x C )0 = 728 mm
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1932
◀
PROBLEM 11.58
C
Block B moves downward with a constant velocity of 20 mm/s. At
t = 0, block A is moving upward with a constant acceleration, and
its velocity is 30 mm/s. Knowing that at t = 3 s slider block C has
moved 57 mm to the right, determine (a) the velocity of slider
block C at t = 0, (b) the accelerations of A and C, (c) the change in
position of block A after 5 s.
B
A
SOLUTION
From the diagram
3 y A + 4 yB + xC = constant
Then
3v A + 4 vB + vC = 0
(1)
and
3a A + 4 aB + aC = 0
(2)
v B = 20 mm/s ;
Given:
(v A )0 = 30 mm/s
(a)
Substituting into Eq. (1) at t = 0
3(-30 mm/s) + 4(20 mm/s) + ( vC )0 = 0
(vC )0 = 10 mm/s
(b)
We have
At t = 3 s :
◀
1
xC = ( xC )0 + (vC )0 t + aC t 2
2
1
57 mm = (10 mm/s)(3 s) + aC (3 s)2
2
aC = 6 mm/s2
Now
( v C )0 = 10.00 mm/s
or
or
aC = 6.00 mm/s2
v B = constant  a B = 0
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1933
◀
PROBLEM 11.58 (CONTINUED)
Then, substituting into Eq. (2)
3aA + 4(0) + (6 mm/s2 ) = 0
a A = -2 mm/s2
(c)
We have
At t = 5 s :
or
a A = 2.00 mm/s2 ◀
1
yA = ( yA )0 + (vA )0 t + aA t 2
2
1
y A - ( y A )0 = (-30 mm/s)(5 s) + (-2 mm/s2 )(5 s)2
2
= -175 mm
y A - ( y A )0 = 175.0 mm ◀
or
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1934
PROBLEM 11.59
The system shown starts from rest, and each component moves with a constant
acceleration. If the relative acceleration of block C with respect to collar B is
60 mm/s2 upward and the relative acceleration of block D with respect to block A
is 110 mm/s2 downward, determine (a) the velocity of block C after 3 s, (b) the
change in position of block D after 5 s.
A
B
C
D
SOLUTION
From the diagram
Cable 1:
2 y A + 2 yB + yC = constant
Then
2 v A + 2 vB + vC = 0
(1)
and
2 a A + 2aB + aC = 0
(2)
( yD - y A ) + ( yD - yB ) = constant
Cable 2:
Then
-v A - vB + 2 vD = 0
(3)
and
-a A - aB + 2 aD = 0
(4)
At t = 0, v = 0; all accelerations constant;
aC/B = 60 mm/s2 , aD/A = 110 mm/s2
Given:
(a)
We have
aC /B = aC - a B = - 60
or
a B = aC + 60
and
a D /A = a D - a A = 110
or
a A = a D - 110
Substituting into Eqs. (2) and (4)
Eq. (2):
2(aD - 110) + 2(aC + 60) + aC = 0
3aC + 2aD = 100
or
Eq. (4):
-(aD - 110) - (aC + 60) + 2 aD = 0
-aC + aD = -50
or
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1935
(5)
(6)
PROBLEM 11.59 (CONTINUED)
Solving Eqs. (5) and (6) for aC and a D
aC = 40 mm/s2
aD =-10 mm/s2
Now
vC = 0 + aC t
At t = 3 s :
vC = (40 mm/s2 )(3 s)
vC = 120.0 mm/s ◀
or
(b)
We have
At t = 5 s :
1
yD = ( yD )0 + (0)t + aD t 2
2
1
yD - ( yD )0 = (-10 mm/s2 )(5 s)2
2
y D - ( y D )0 = 125.0 mm ◀
or
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1936
PROBLEM 11.60*
A
B
C
The system shown starts from rest, and the length of the upper cord is adjusted so
that A, B, and C are initially at the same level. Each component moves with a
constant acceleration, and after 2 s the relative change in position of block C with
respect to block A is 280 mm upward. Knowing that when the relative velocity of
collar B with respect to block A is 80 mm/s downward, the displacements of A
and B are 160 mm downward and 320 mm downward, respectively, determine
(a) the accelerations of A and B if aB > 10 mm/s2 , (b) the change in position of
block D when the velocity of block C is 600 mm/s upward.
D
SOLUTION
From the diagram
Cable 1:
2 yA + 2 yB + yC = constant
Then
2v A + 2vB + vC = 0
(1)
and
2 a A + 2 aB + aC = 0
(2)
( yD - yA ) + ( yD - yB ) = constant
Cable 2:
Then
-vA - vB - 2 vD = 0
(3)
and
-a A - aB + 2 aD = 0
(4)
t =0
Given: At
v=0
( yA )0 = ( yB )0 = ( yC )0
All accelerations constant.
At t = 2 s
yC/A = 280 mm
vB/A = 80 mm/s
When
y A - ( yA )0 = 160 mm
yB - ( yB )0 = 320 mm
aB >10 mm/s2
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1937
PROBLEM 11.60* (CONTINUED)
(a)
We have
1
yA = ( yA )0 + (0)t + a A t 2
2
and
1
yC = ( yC )0 + (0)t + aC t 2
2
Then
1
yC/A = yC - yA = (aC - aA )t 2
2
At t = 2 s, yC/A = -280 mm:
1
-280 mm = (aC - aA )(2 s)2
2
or
aC = a A -140
(5)
Substituting into Eq. (2)
2 a A + 2 aB + ( a A - 140) = 0
or
1
aA = (140 - 2aB )
3
Now
vB = 0 + a B t
(6)
vA = 0 + aA t
vB/A = vB - vA = (aB - a A )t
Also
When
1
yB = ( yB )0 + (0)t + aB t 2
2
v B/A = 80 mm/s :
80 = ( a B - a A )t
DyA = 160 mm :
1
160 = aAt 2
2
DyB = 320 mm :
1
320 = aB t 2
2
(7)
Then
1
160 = (aB - aA )t 2
2
Using Eq. (7)
320 = (80)t
Then
1
160 = a A (4)2
2
or
a A = 20.0 mm/s2 ◀
and
1
320 = aB (4)2
2
or
aB = 40.0 mm/s2 ◀
or
t=4 s
Note that Eq. (6) is not used; thus, the problem is over-determined.
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1938
PROBLEM 11.60* (CONTINUED)
(b)
Substituting into Eq. (5)
aC = 20 -140 =-120 mm/s2
and into Eq. (4)
-(20 mm/s2 ) - (40 mm/s2 ) + 2aD = 0
or
aD = 30 mm/s2
Now
vC = 0 + aC t
When vC = - 600 mm/s:
or
Also
At t = 5 s :
-600 mm/s = (-120 mm/s2 )t
t=5s
1
yD = ( yD )0 + (0)t + aD t 2
2
1
yD - ( yD )0 = (30 mm/s2 )(5 s)2
2
y D - (y D )0 = 375 mm ◀
or
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1939
PROBLEM 11.61
a (ft/s2)
4
6
0
10
14
t(s)
–4
A particle moves in a straight line with a constant acceleration of
-4 ft/s2 for 6s, zero acceleration for the next 4s, and a constant
acceleration of +4 ft/s2 for the next 4s. Knowing that the
particle starts from the origin and that its velocity is -8 ft/s
during the zero acceleration time interval, (a) construct the v - t
and x - t curves for 0 £ t £ 14s, (b) determine the position
and the velocity of the particle and the total distance traveled
when t = 14s.
SOLUTION
From the a - t curve
A1 = - 24 ft/s, A2 = 16 ft/s
(a) Construct the v - t curve :
v6 = - 8 ft/s
v0 = v6 - A1
= -8 - (-24 )
= 16 ft/s
v10 = - 8 ft/s
v14 = v10 + A2
= -8 +16
◀
= 8 ft/s
From the v - t curve
A3 = 32 ft, A4 = -8 ft
A5 = -32 ft, A6 = -8 ft , A7 = 8 ft
(b) Construct the x - t curve
x0 = 0
x4 = x0 + A3 = 32 ft
x6 = x4 + A4 = 24 ft
x10 = x6 + A5 = -8 ft
x12 = x10 + A6 = -16 ft
x14 = x12 + A7 = -8 ft
Total Distance Traveled
◀
0 £ t £ 4 s,
d1 = 32 - 0 = 32 ft
4 s £ t £ 12 s,
d2 = -16 - 32 = 48 ft
12 s £ t £ 14 s,
d3 = - 8 - (- 16 ) = 8 ft
d = 32 + 48 + 8
d = 88 ft ◀
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1940
Problem 11.62
a (ft/s2)
4
0
6
10
t(s)
14
–4
A particle moves in a straight line with a constant acceleration of
-4 ft/s2 for 6s, zero acceleration for the next 4 s, and a constant
acceleration of +4 ft/s2 for the next 4s. Knowing that the
particle starts from the origin with v0 = 16 ft/s, (a) construct the
v - t and x - t curves for 0 £ t £ 14s, (b) determine the
amount of time during which the particle is further than 16 ft
from the origin.
SOLUTION
From the a - t curve
A1 = - 24 ft/s, A2 = 16 ft/s
(a) Construct the v - t curve :
v0 = 16 ft/s
v6 = v0 + A1
= 16 + (-24)
= -8 ft/s
v10 = v6 = - 8 ft/s
v14 = v10 + A2
= -8 +16
◀
= 8 ft/s
From the v - t curve
A3 = 32 ft, A4 = -8 ft
A5 = -32 ft, A6 = -8 ft , A7 = 8 ft
Construct the x - t curve
x0 = 0
x4 = x0 + A3 = 32 ft
x6 = x4 + A4 = 24 ft
x10 = x6 + A5 = -8 ft
x12 = x10 + A6 = -16 ft
x14 = x12 + A7 = -8 ft
◀
Time for x > 16 ft.
From the x - t diagram, this is time interval t1 to t 2 .
dx
= v = 16 - 4t
From 0 < t < 6 s,
dt
(b)
Integrating, using limits x = 0 when t = 0 and x = 16 ft when t = t1
[ x ]16
0
t
= éëê16 t - 2 t 2 ùûú
0
or
16 = 16 t1 - 2 t12
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1941
PROBLEM 11.62 (CONTINUED)
t12 -8t1 + 8 = 0
or
8  (8) - (4)(1)(8)
2
Using the quadratic formula:
t1 =
(2)(1)
= 4  2.828 = 1.172 s
and
6.828 s
The larger root is out of range, thus t1 = 1.172 s
From 6 < t < 10,
x = 24 - 8(t - 6) = 72 - 8t
Setting x = 16,
16 = 72 - 8t2
Time Interval:
Dt = t2 - t1
or
t2 = 7 s
D t = 5.83 s ◀
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1942
PROBLEM 11.63
v (m/s)
60
26
–5
–20
41
46
10
t (s)
A particle moves in a straight line with the
velocity shown in the figure. Knowing that
x = - 540 m at t = 0, (a) construct the a-t
and x-t curves for 0 < t < 50 s, and
determine (b) the total distance traveled by
the particle when t = 50 s, (c) the two times
at which x = 0.
SOLUTION
(a)
at = slope of v-t curve at time t
From t = 0 to t = 10 s:
v = constant  a = 0
-20 - 60
= -5 m/s2
26 -10
t = 10 s to t = 26 s:
a=
t = 26 s to t = 41 s:
v = constant  a = 0
t = 41 s to t = 46 s:
a=
t = 46 s:
-5 - (-20)
= 3 m/s2
46 - 41
v = constant  a = 0
x 2 = x1 + (area under v-t curve from t1 to t2 )
At t = 10 s:
x10 = -540 + 10(60) + 60 m
Next, find time at which v = 0. Using similar triangles
tv = 0 - 10
60
At
t = 22 s:
t = 26 s:
t = 41 s:
t = 46 s:
t = 50 s:
=
26 - 10
80
or
tv = 0 = 22 s
1
x22 = 60 + (12)(60) = 420 m
2
1
x26 = 420 - (4)(20) = 380 m
2
x41 = 380 -15(20) = 80 m
æ 20 + 5 ö÷
x46 = 80 - 5çç
÷ = 17.5 m
è 2 ÷ø
x50 = 17.5 - 4(5) = -2.5 m
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1943
PROBLEM 11.63 (CONTINUED)
(b)
From t = 0 to t = 22 s: Distance traveled = 420 - (- 540)
= 960 m
t = 22 s to t = 50 s: Distance traveled = |- 2.5 - 420|
= 422.5 m
Total distance traveled = (960 + 422.5) ft = 1382.5 m
Total distance traveled = 1383 m ◀
(c)
Using similar triangles
Between 0 and 10 s:
(t x=0 )1 - 0
10
=
540
600
( t x = 0 )1 = 9.00 s ◀
Between 46 s and 50 s :
(t x=0 )2 - 46
4
=
17.5
20
( t x = 0 )2 = 49.5 s ◀
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1944
PROBLEM 11.64
v (m/s)
A particle moves in a straight line with the
velocity shown in the figure. Knowing that
x = -540 m at t = 0, (a) construct the
a - t and x - t curves for 0 < t < 50 s,
and determine (b) the maximum value of
the position coordinate of the particle,
(c) the values of t for which the particle is
at x = 100 m.
60
26
–5
–20
41
46
10
t (s)
SOLUTION
(a)
at = slope of v-t curve at time t
From t = 0 to t = 10 s:
v = constant  a = 0
-20 - 60
= -5 m/s2
26 -10
t = 10 s to t = 26 s:
a=
t = 26 s to t = 41 s:
v = constant  a = 0
t = 41 s to t = 46 s:
a=
t = 46 s:
-5 - (-20)
= 3 m/s2
46 - 41
v = constant  a = 0
x 2 = x1 + (area under v-t curve from t1 to t2 )
At t = 10 s:
x10 = -540 + 10(60) = 60 m
Next, find time at which v = 0. Using similar triangles
tv = 0 - 10
60
At
t = 22 s:
t = 26 s:
t = 41 s:
t = 46 s:
t = 50 s:
=
26 - 10
80
or
tv = 0 = 22 s
1
x22 = 60 + (12)(60) = 420 m
2
1
x26 = 420 - (4)(20) = 380 m
2
x41 = 380 -15(20) = 80 m
æ 20 + 5 ö÷
x46 = 80 - 5çç
÷ = 17.5 m
è 2 ÷ø
x50 = 17.5 - 4(5) = -2.5 m
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1945
PROBLEM 11.64 (CONTINUED)
(b)
Reading from the x-t curve
(c)
Between 10 s and 22 s
xmax = 420 m ◀
100 m = 420 m - (area under v - t curve from t , to 22 s) m
1
100 = 420 - (22 - t1 )(v1 )
2
(22 - t1 )(v1 ) = 640
Using similar triangles
v1
60
=
22 - t1 12
Then
v1 = 5(22 - t1 )
or
(22 - t1 )[5(22 - t1 )] = 640
t1 = 10.69 s
and t1 = 33.3 s
10 s < t1 < 22 s 
We have
t1 = 10.69 s ◀
Between 26 s and 41 s:
Using similar triangles
41 - t2
15
=
20
300
t 2 = 40.0 s ◀
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1946
Problem 11.65
v (ft/s)
18
6
0
24 30
10
18
t (s)
–18
A particle moves in a straight line with the velocity shown in the
figure. Knowing that x = - 48 ft at t = 0, draw the a – t and x – t
curves for 0 < t < 40 s and determine (a) the maximum value of
the position coordinate of the particle, (b) the values of t for
which the particle is at a distance of 108 ft from the origin.
SOLUTION
The a - t curve is the slope of the v - t curve
0 < t < 10 s,
a=0
18 - 6
= 1.5 ft/s2
18 - 10
-18 - 18
a=
= - 3 ft/s2
30 - 18
a=
10 s < t < 18 s,
18 s < t < 30 s,
30 s < t < 40 s
a=0
◀
Points on the x–t curve may be calculated using areas of the v–t curve.
A1 = (10)(6) = 60 ft
A2 =
1
(6 + 18)(18 - 10) = 96 ft
2
A3 =
1
(18)(24 - 18) = 54 ft
2
A4 =
1
(-18)(30 - 24) = - 54 ft
2
A5 = (- 18)(40 - 30) = -180 ft
Value of x at specific times:
x 0 = - 48 ft
x10 = x 0 + A1 = 12 ft
x18 = x10 + A2 = 108 ft
x 24 = x18 + A3 = 162 ft
x 30 = x 24 + A4 = 108 ft
x 40 = x 30 + A5 = - 72 ft
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1947
◀
PROBLEM 11.65 (CONTINUED)
(a) Maximum value of x occurs when: v = 0, i.e. t = 24 s.
x max = 162 ft ◀
(b) Times when x = 108 ft.
From the x - t curve
t = 18 s and t = 30 s ◀
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1948
PROBLEM 11.66
v
A parachutist is in free fall at a rate of 200 km/h when he opens his parachute at
an altitude of 600 m. Following a rapid and constant deceleration, he then
descends at a constant rate of 50 km/h from 586 m to 30 m, where he maneuvers
the parachute into the wind to further slow his descent. Knowing that the
parachutist lands with a negligible downward velocity, determine (a) the time
required for the parachutist to land after opening his parachute, (b) the initial
deceleration.
SOLUTION
Assume second deceleration is constant. Also, note that
200 km/h = 55.555 m/s,
50 km/h = 13.888 m/s
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1949
PROBLEM 11.66 (CONTINUED)
(a)
Now D x = area under v- t curve for given time interval
Then
æ 55.555 + 13.888 ö÷
(586 - 600) m = -t1 çç
÷÷ø m/s
è
2
t1 = 0.4032 s
(30 - 586) m = -t2 (13.888 m/s)
t2 = 40.0346 s
1
(0 - 30) m = - (t3 )(13.888 m/s)
2
t3 = 4.3203 s
t total = (0.4032 + 40.0346 + 4.3203) s
(b)
We have
t total = 44.8 s ◀
Dvinitial
t1
[-13.888 - (-55.555)] m/s
=
0.4032 s
ainitial =
= 103.3 m/s2
a initial = 103.3 m/s2 ◀
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1950
PROBLEM 11.67
40 mi /h
A commuter train traveling at
40 mi/h is 3 mi from a station. The
train then decelerates so that its
speed is 20 mi/h when it is 0.5 mi
from the station. Knowing that the
train arrives at the station 7.5 min
after beginning to decelerate and
assuming constant decelerations,
determine (a) the time required for
the train to travel the first 2.5 mi, (b)
the speed of the train as it arrives at
the station, (c) the final constant
deceleration of the train.
3 mi
SOLUTION
Given: At t = 0, v = 40 mi/h, x = 0; when x = 2.5 mi, v = 20 mi/h;
at t = 7.5 min, x = 3 mi; constant decelerations.
The v-t curve is first drawn as shown.
(a)
A1 = 2.5 mi
We have
æ 40 + 20 ö÷
1h
= 2.5 mi
(t1 min) çç
mi/h ´
è 2 ÷÷ø
60 min
t1 = 5.00 min ◀
(b)
A2 = 0.5 mi
We have
æ 20 + v2 ÷ö
1h
(7.5 - 5) min ´ çç
= 0.5 mi
÷ mi/h ´
è 2 ÷ø
60 min
v2 = 4.00 mi/h ◀
(c)
We have
afinal = a12
=
(4 - 20) mi/h 5280 ft 1 min
1h
´
´
´
(7.5 - 5)min
mi
60 s 3600 s
afinal = -0.1564 ft/s2 ◀
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1951
PROBLEM 11.68
A temperature sensor is attached to slider AB which moves
back and forth through 60 in. The maximum velocities of the
slider are 12 in./s to the right and 30 in./s to the left. When the
slider is moving to the right, it accelerates and decelerates at a
2
constant rate of 6 in./s ; when moving to the left, the slider
2
accelerates and decelerates at a constant rate of 20 in./s .
Determine the time required for the slider to complete a full
cycle, and construct the v - t and x-t curves of its motion.
x
60 in.
A
B
SOLUTION
The v-t curve is first drawn as shown. Then
ta =
vright
aright
=
12 in./s
=2s
6 in./s2
vleft 30 in./s
=
aleft 20 in./s
= 1.5 s
td =
A1 = 60 in.
Now
[( t1 - 2) s](12 in./s) = 60 in.
or
t1 = 7 s
or
A2 = 60 in.
and
or
{[( t2 - 7) - 1.5] s}(30 in./s) = 60 in.
or
t 2 = 10.5 s
tcycle = t2
Now
t cycle = 10.5 s ◀
We have x ii = x i + (area under v- t curve from ti to tii )
At
t = 5 s:
1
x2 = (2)(12) = 12 in.
2
x5 = 12 + (5 - 2)(12)
t = 7 s:
= 48 in.
x 7 = 60 in.
t = 2 s:
t = 8.5 s:
t = 9 s:
t = 10.5 s:
1
x8.5 = 60 - (1.5)(30)
2
= 37.5 in.
x9 = 37.5 - (0.5)(30)
= 22.5 in.
x10.5 = 0
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written consent of McGraw-Hill Education.
1952
PROBLEM 11.69
x
In a water-tank test involving the launching of a small model
boat, the model’s initial horizontal velocity is 6 m/s, and its
horizontal acceleration varies linearly from -12 m/s2 at t = 0
to -2 m/s2 at t = t1 and then remains equal to -2 m/s2 until
t = 1.4 s. Knowing that v = 1.8 m /s when t = t1 , determine
(a) the value of t1 , (b) the velocity and the position of the
model at t = 1.4 s.
v0 = 6 m/s
SOLUTION
Given:
v0 = 6 m/s; for 0 < t < t1 ,
for
t1 < t < 1.4 s a = -2 m/s2 ;
at
t = 0 a = -12 m/s2 ;
at t = t1
a = -2 m/s2 , v = 1.8 m/s2
The a-t and v-t curves are first drawn as shown. The time axis is
not drawn to scale.
(a)
We have
vt1 = v0 + A1
æ12 + 2 ö÷
1.8 m/s = 6 m/s - (t1 s) çç
m/s 2
è 2 ÷÷ø
t1 = 0.6 s ◀
(b)
We have
v1.4 = vt1 + A2
v1.4 = 1.8 m/s - (1.4 - 0.6) s ´ 2 m/s2
v1.4 = 0.20 m/s ◀
Now x1.4 = A3 + A4 , where A3 is most easily determined using
integration. Thus,
for 0 < t < t1 :
a=
-2 - (-12)
50
t -12 = t -12
0.6
3
dv
50
= a = t -12
dt
3
Now
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written consent of McGraw-Hill Education.
1953
PROBLEM 11.69 (CONTINUED)
At t = 0, v = 6 m/s:
ò
v
dv =
6
ò
t æ 50
0
ö
çç t - 12÷÷ dt
÷ø
è3
v=6+
or
25 2
t -12t
3
We have
dx
25
= v = 6 -12t + t 2
dt
3
Then
A3 =
ò
xt1
0
dx =
ò
0.6
(6 - 12t +
0
25 2
t )dt
3
0.6
é
25 ù
= ê6t - 6t 2 + t 3 ú = 2.04 m
9 ûú 0
ëê
Also
æ 1.8 + 0.2 ö÷
A4 = (1.4 - 0.6) çç
÷ø÷ = 0.8 m
è
2
Then
x1.4 = (2.04 + 0.8) m
x1.4 = 2.84 m ◀
or
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1954
PROBLEM 11.70
a (m/s2)
The acceleration record shown was obtained for a small
airplane traveling along a straight course. Knowing that x = 0
and v = 60 m/s when t = 0, determine (a) the velocity and
position of the plane at t = 20 s, (b) its average velocity
during the interval 6 s < t < 14 s.
0.75
6
8 10
0
12 14
20 t(s)
–0.75
SOLUTION
Geometry of “bell-shaped” portion of v-t curve
The parabolic spandrels marked by * are of equal area. Thus, total area of shaded portion of v-t diagram is:
= Dx = 6 m
(a)
v20 = 60 m/s ◀
W hen t = 2 0 s:
x 20 = (60 m/s) (20 s) - (shaded area)
x20 = 1194 m ◀
= 1200 m - 6 m
(b)
From t = 6 s to t = 14 s:
Dt = 8 s
D x = (60 m/s)(14 s - 6 s) - (shaded area)
= (60 m/s)(8 s) - 6 m = 480 m - 6 m = 474 m
vaverage =
Dx
Dt
=
474 m
8s
vaverage = 59.25 m/s ◀
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written consent of McGraw-Hill Education.
1955
B
PROBLEM 11.71
A
200 m
200 m
In a 400-m race, runner A reaches her maximum velocity vA in
4 s with constant acceleration and maintains that velocity until
she reaches the half-way point with a split time of 25 s. Runner
B reaches her maximum velocity vB in 5 s with constant
acceleration and maintains that velocity until she reaches the
half-way point with a split time of 25.2 s. Both runners then run
the second half of the race with the same constant deceleration
of 0.1 m/s 2 . Determine (a) the race times for both runners,
(b) the position of the winner relative to the loser when the
winner reaches the finish line.
SOLUTION
Sketch v-t curves for first 200 m.
t1 = 4 s, t2 = 25 - 4 = 21 s
Runner A:
1
A1 = (4)(vA )max = 2(vA )max
2
A2 = 21(vA )max
A1 + A2 = D x = 200 m
23( v A ) max = 200
or
t1 = 5 s,
Runner B:
(v A ) max = 8.6957 m/s
t2 = 25.2 - 5 = 20.2 s
1
A1 = (5)(vB )max = 2.5(vB )max
2
A2 = 20.2(vB )max
A1 + A2 = D x = 200 m
22.7( v B )max = 200
Sketch v-t curve for second 200 m.
D v = | a |t3 = 0.1t3
1
A3 = vmax t3 - Dvt3 = 200
2
t3 =
Runner A:
0.05t32 - vmax t3 + 200 = 0
or
vmax  (vmax )2 - (4)(0.05)(200)
(vmax ) A = 8.6957,
Reject the larger root. Then total time
(v B )max = 8.8106 m/s
or
(2)(0.05)
(t3 ) A = 146.64 s
(
= 10 vmax  (vmax )2 - 40
and
)
27.279 s
t A = 25 + 27.279 = 52.279 s
t A = 52.2 s ◀
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1956
PROBLEM 11.71 (CONTINUED)
Runner B: (vmax )B = 8.8106, (t3 ) B = 149.45 s
and
26.765 s
t B = 25.2 + 26.765 = 51.965 s
Reject the larger root. Then total time
t B = 52.0 s ◀
Velocity of A at t = 51.965 s:
v1 = 8.6957 - (0.1)(51.965 - 25) = 5.999 m/s
Velocity of A at t = 51.279 s:
v2 = 8.6957 - (0.1)(52.279 - 25) = 5.968 m/s
Over 51.965 s £ t £ 52.965 s, runner A covers a distance D x
Dx = vave (Dt ) =
1
(5.999 + 5.968)(52.279 - 51.965)
2
D x = 1.879 m ◀
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1957
PROBLEM 11.72
A
16 ft
B
40 ft
50 ft
40 ft
A car and a truck are both traveling at the
constant speed of 35 mi/h; the car is 40 ft
behind the truck. The driver of the car
wants to pass the truck, i.e., he wishes to
place his car at B, 40 ft in front of the truck,
and then resume the speed of 35 mi/h; The
maximum acceleration of the car is 5ft/s2
and the maximum deceleration obtained by
applying the brakes is 20 ft/s2 What is the
shortest time in which the driver of the car
can complete the passing operation if he
does not at any time exceed a speed of
50 mi/h? Draw the v - t curve.
SOLUTION
Relative to truck, car must move a distance:
Allowable increase in speed:
D x = 16 + 40 + 50 + 40 = 146 ft
D vm = 50 - 35 = 15 mi/h = 22 ft/s
Acceleration Phase:
t1 = 22 / 5 = 4.4 s
Deceleration Phase:
t3 = 22 / 20 = 1.1 s
But: D x = A1 + A2 + A3 :
1
A1 = (22)(4.4) = 48.4 ft
2
1
A3 = (22)(1.1) = 12.1ft
2
146 ft = 48.4 + (22)t 2 + 12.1
t total = t1 + t2 + t3 = 4.4 s + 3.89 s + 1.1s = 9.39 s
t2 = 3.89 s
t B = 9.39 s ◀
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1958
PROBLEM 11.73
Solve Problem 11.72, assuming that the driver of the car does not pay any attention to the speed limit while
passing and concentrates on reaching position B and resuming a speed of 35 mi/h in the shortest possible time.
What is the maximum speed reached? Draw the v - t curve.
PROBLEM 11.72 A car and a truck are both traveling at the constant speed of 35 m i/h; the car is 40 ft
behind the truck. The driver of the car wants to pass the truck, i.e., he wishes to place his car at B, 40 ft in
front of the truck, and then resume the speed of 35 m i/h; The maximum acceleration of the car is 5 ft/s 2 and
the maximum deceleration obtained by applying the brakes is 2 0 ft/s 2 What is the shortest time in which the
driver of the car can complete the passing operation if he does not at any time exceed a speed of 50 mi/h?
Draw the v - t curve.
A
B
16 ft
40 ft
50 ft
40 ft
SOLUTION
Relative to truck, car must move a distance:
D x = 16 + 40 + 50 + 40 = 146 ft
Dx = A1 + A2 :
1
Dvm = 5t1 = 20t2 ; t2 = t1
4
1
146 ft = (Dvm )(t1 + t2 )
2
146 ft =
æ
1
1 ö
(5t1 ) ççt1 + t1 ÷÷÷
è
2
4 ø
t12 = 46.72
t1 = 6.835 s
1
t2 = t1 = 1.709
4
t total = t1 + t 2 = 6.835 + 1.709
t B = 8.54 s ◀
D vm = 5t1 = 5(6.835) = 34.18 ft/s = 23.3 mi/h
Speed vtotal = 35 mi/h, vm = 35 mi/h + 23.3 mi/h
vm = 58.3 mi/h ◀
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1959
PROBLEM 11.74
Car A is traveling on a highway at a
constant speed ( v A )0 = 60 mi/h,
and is 380 ft from the entrance of
an access ramp when car B enters
the acceleration lane at that point
at a speed ( v B )0 = 15 mi/h, Car B
accelerates uniformly and enters
the main traffic lane after traveling
200 ft in 5 s. It then continues to
accelerate at the same rate until it
reaches a speed of 60 mi/h, which
it then maintains. Determine the
final distance between the two
cars.
380 ft
A
(vA)0
(vB)0
B
SOLUTION
Given:
(v A )0 = 60 mi/h, (vB )0 = 1.5 mi/h, at t = 0,
( x A )0 = -380 ft, ( x B )0 = 0; at t = 5 s,
x B = 200 ft; for 15 mi/h < vB £ 60 mi/h,
aB = constant; for vB = 60 mi/h,
aB = 0
First note
60 mi/h = 88 ft/s
15 mi/h = 22 ft/s
The v - t curves of the two cars are then drawn as shown.
Using the coordinate system shown, we have
at t = 5 s, x B = 200 ft:
é 22 + (v B )5 ù
(5 s) ê
ú ft/s = 200 ft
2
ëê
ûú
(vB )5 = 58 ft/s
or
Then, using similar triangles, we have
(88 - 22) ft/s
t1
=
(58 - 22) ft/s
5s
(= aB )
t1 = 9.1667 s
or
Finally, at t = t1
é
æ 22 + 88 ö÷ ù
ft/sú
x B / A = x B - x A = ê(9.1667 s)çç
è 2 ÷÷ø úû
êë
- [-380 ft + (9.1667 s)(88 ft/s)]
x B / A = 77.5 ft ◀
or
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1960
PROBLEM 11.75
2
12 m
An elevator starts from rest and moves upward, accelerating at a rate of 1.2 m/s
until it reaches a speed of 7.8 m/s, which it then maintains. Two seconds after the
elevator begins to move, a man standing 12 m above the initial position of the top of
the elevator throws a ball upward with an initial velocity of 20 m/s. Determine when
the ball will hit the elevator.
SOLUTION
Given:
At t = 0 vE = 0; For 0 < v E £ 7.8 m/s, aE = 1.2 m/s2 ;
For v E = 7.8 m/s, a E = 0;
At t = 2 s, v B = 20m/s 
The v - t curves of the ball and the elevator are first drawn as shown. Note that the initial slope of the curve
for the elevator is 1.2 m/s2 , while the slope of the curve for the ball is -g(-9.81 m / s2 ).
The time t1 is the time when vE reaches 7.8 m/s
Thus,
or
or
vE = (0) + aE t
7.8 m/s = (1.2 m/s2 )t1
t1 = 6.5 s
The time ttop is the time at which the ball reaches the top of its trajectory.
Thus,
or
or
vB = (vB )0 - g(t - 2)
0 = 20 m/s - (9.81 m/s2 )(ttop - 2) s
t top = 4.0387 s
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written consent of McGraw-Hill Education.
1961
PROBLEM 11.75 (CONTINUED)
Using the coordinate system shown, we have
0 < t < t1 :
At t = t top :
æ1
ö
yE = - 12 m + çç aE t 2 ÷÷÷ m
è2
ø
1
(4.0387 - 2) s ´ (20 m/s)
2
= 20.387 m
yB =
1
yE = -12 m + (1.2 m/s2 )(4.0387s)2
2
= -2.213 m
and
t = [2+2(4.0387 - 2)]s = 6.0774 s, y B = 0
At
and at t = t1 ,
1
yE = -12 m + (6.5 s)(7.8 m/s) = 13.35 m
2
The ball hits the elevator (yB = yE ) when ttop £ t £ t1 .
For t ³ ttop :
é1
ù
yB = 20.387 m - ê g(t - t top )2 ú m
ëê 2
ûú
Then,
yB = yB
when
1
20.387m - (9.81m/s2 )(t - 4.0387)2
2
1
= -12m + (1.2m/s2 )(t s)2
2
5.505t 2 - 39.6196t + 47.619 = 0
or
Solving
t = 1.525 s
and
t = 5.67 s
t = 5.67 s ◀
Since 1.525 s is less than 2 s,
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1962
PROBLEM 11.76
(vB)0 = 45 mi/h
(vA)0 = 40 mi/h
B
A
60 ft
Car A is traveling at 40 mi/h when it
enters a 30 mi/h speed zone. The driver
2
of car A decelerates at a rate of 16 ft/s
until reaching a speed of 30 mi/h, which
she then maintains. When car B, which
was initially 60 ft behind car A and
traveling at a constant speed of 45 mi/h,
enters the speed zone, its driver
2
decelerates at a rate of 20 ft/s until
reaching a speed of 28 mi/h. Knowing
that the driver of car B maintains a speed
of 28 mi/h, determine (a) the closest that
car B comes to car A, (b) the time at
which car A is 70 ft in front of car B.
SOLUTION
(vA )0 = 40 mi/h; For 30 mi/h < vA £ 40 mi/h, a A = -16ft/s2 ; For v A = 30 mi/h, a A = 0;
Given:
( x A / B )0 = 60 ft; (vB )0 = 45 mi/h;
When x B = 0, aB = -20 ft/s2 ;
For vB = 28 mi/h, aB = 0
First note
40 mi/h = 58.667 ft/s
30 mi/h = 44 ft/s
45 mi/h = 66 ft/s
28 mi/h = 41.067 ft/s
At t = 0
The v - t curves of the two cars are as shown.
At
t = 0:
Car A enters the speed zone.
t = ( t B )1 :
Car B enters the speed zone.
t = tA :
Car A reaches its final speed.
t = tmin :
v A = vB
t = ( t B ) 2 : Car B reaches its final speed.
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1963
PROBLEM 11.76 (CONTINUED)
(a)
aA =
We have
- 16 ft/s 2 =
or
( v A )final - ( v A )0
tA
(44 - 58.667) ft/s
tA
t A = 0.91669 s
or
Also
60 ft = (t B )1 (v B )0
or
60 ft = ( t B )1 (66 ft/s) or ( t B )1 = 0.90909 s
aB =
and
- 20 ft/s 2 =
or
( vB )final - ( vB )0
(t B )2 - (t B )1
(41.067 - 66) ft/s
[(t B )2 - 0.90909]s
Car B will continue to overtake car A while v B > v A . Therefore, ( x A / B )min will occur when v A = v B ,
which occurs for
( t B )1 < t min < ( t B ) 2
For this time interval
vA = 44 ft/s
vB = (vB )0 + aB [t - (t B )1 ]
Then
at t = t
min:
44 ft/s = 66 ft/s + (-20 ft/s2 )(tmin - 0.90909)s
tmin = 2.00909 s
or
Finally ( x A / B )min = ( x A )t min - ( x B )t min
ïì é (v ) + (v A )final ù
ïü
= ïít A ê A 0
ú + (tmin - t A )(v A )final ïý
ïîï ëê
ï
2
ûú
ï
þ
ìïï
é (vB )0 + (v A )final ùüïï
-í( x B )0 + (t B )1 (vB )0 + [tmin - (t B )1 ] ê
úý
ïîï
2
ëê
ûúï
ï
þ
é
ù
æ 58.667 + 44 ö÷
= ê(0.91669 s) çç
÷ ft/s + (2.00909 - 0.91669)s´(44 ft/s)ú
÷
êë
úû
è
ø
2
é
æ 66 + 44 ö÷ ù
- ê- 60 ft + (0.90909 s)(66 ft/s) + (2.00909 - 0.90909) s ´çç
ft/sú
è 2 ø÷÷ úû
êë
= (47.057 + 48.066) ft - (-60 + 60.000 + 60.500) ft
or ( x A / B )min = 34.6 ft ◀
= 34.623 ft
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written consent of McGraw-Hill Education.
1964
PROBLEM 11.76 (CONTINUED)
(b)
Since ( x A / B ) £ 60 ft for t £tmin , it follows that x A / B = 70 ft for t > (t B )2
[Note ( t B )2  t min ]. Then, for t > ( t B ) 2
x A / B = ( x A / B )min +[(t - tmin )(v A )final ]
é (v A )final + (vB )final ù
ïì
ïü
-ï
ú + [t - (t B )2 ](vB )final ïý
í[(t B )2 - (tmin )] ê
ïî
ïþï
êë
úû
2
ï
70ft = 34.623ft + [(t - 2.00909)s ´ (44ft/s)]
or
é
ù
æ 44 + 41.06 ÷ö
- ê(2.15574 - 2.00909)s ´çç
ft/s + (t - 2.15574)s ´ (41.067)ft/sú
÷
÷
è
ø
êë
úû
2
t = 14.14 s ◀
or
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written consent of McGraw-Hill Education.
1965
PROBLEM 11.77
a (ft/s2)
a = 24 – 200t2
24
16
0
a = 32 – 80t
0
0.2
0.4 t (s)
An accelerometer record for the motion of a given part of a
mechanism is approximated by an arc of a parabola for 0.2 s and a
straight line for the next 0.2 s as shown in the figure. Knowing that
v = 0 when t = 0 and x = 0.8 ft when t = 0.4 s, (a) construct the
v - t curve for 0 £ t £ 0.4 s, (b) determine the position of the part
at t = 0.3 s and t = 0.2 s.
SOLUTION
Divide the area of the a - t curve into the four areas A1 , A 2 , A3 an d A 4 .
2
A1 = (8)(0.2) = 1.0667 ft/s
3
A2 = (16)(0.2) = 3.2 ft/s
1
A3 = (16 + 8)(0.1) = 1.2 ft/s
2
1
A4 = (8)(0.1) = 0.4 ft/s
2
Velocities: v0 = 0
v0.2 = v0 + A1 + A2
v0.2 = 4.27 ft/s ◀
v0.3 = v0.2 + A3
v0.3 = 5.47 ft/s ◀
v0.4 = v0.3 + A4
v0.4 = 5.87 ft/s ◀
Sketch the v - t curve and divide its area into A 5 , A 6 , an d A 7 as shown.
ò
0.8
dx = 0.8 - x =
x
ò
0.4
vdt
or
ò
x = 0.8 -
t
0.4
vdt
t
x0.3 = 0.8 - A5 - (5.47)(0.1)
At t = 0.3 s,
2
With A5 = (0.4)(0.1) = 0.0267ft,
3
At t = 0.2 s,
x 0.3 = 0.227 ft ◀
x0.2 = 0.8 - ( A5 + A6 ) - A7
2
With A5 + A6 = (1.6)(0.2) = 0.2133ft,
3
and
A 7 = (4.27)(0.2) = 0.8533 ft
x0.2 = 0.8 - 0.2133 - 0.8533
x 0.2 = - 0.267 ft ◀
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written consent of McGraw-Hill Education.
1966
PROBLEM 11.78
a (m/s2)
2
1
0
2
t1 t(s)
4.5
–6
A car is traveling at a constant speed of
54 km/h when its driver sees a child run into
the road. The driver applies her brakes until
the child returns to the sidewalk and then
accelerates to resume her original speed of
54 km/h; the acceleration record of the car is
shown in the figure. Assuming x = 0 when
t = 0, determine (a) the time t1 at which the
velocity is again 54 km/h, (b) the position of
the car at that time, (c) the average velocity of
the car during the interval 1 s £ t £ t1 .
SOLUTION
Given:
At
t = 0, x = 0, v = 54 km/h;
For t = t1 ,
v = 54 km/h
First note
(a)
54 km/h = 15 m/s
vb = v a + (area under a - t curve from t a to tb )
We have
Then
at
t = 2s :
v = 15 - (1)(6) = 9 m/s
t = 4.5 s:
1
v = 9 - (2.5)(6) = 1.5 m/s
2
t = t1 :
1
15 = 1.5 + (t1 - 4.5)(2)
2
t1 = 18.00 s ◀
or
(b)
Using the above values of the velocities, the v - t curve is drawn as shown.
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written consent of McGraw-Hill Education.
1967
PROBLEM 11.78 (CONTINUED)
Now
x at t = 18 s
x18 = 0 + å (area under the v - t curve from t = 0 to t = 18 s)
æ15 + 9 ö÷
= (1s)(15m/s) + (1s) çç
÷ m/s
è 2 ø÷
é
ù
1
+ ê(2.5 s)(1.5 m/s) + (2.5s)(7.5m/s)ú
êë
úû
3
é
ù
2
+ ê(13.5 s)(1.5 m/s) + (13.5s)(13.5m/s)ú
3
ëê
ûú
= [15 + 12 + (3.75 + 6.25) + (20.25 + 121.50)]m
= 178.75 m
(c)
First note
Now
or
x18 = 178.8 m ◀
x1 = 15m
x18 = 178.75m
vave =
Dx (178.75 - 15)m
=
= 9.6324 m/s
(18 - 1)s
Dt
vave = 34.7 km/h ◀
or
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written consent of McGraw-Hill Education.
1968
PROBLEM 11.79
An airport shuttle train travels between two terminals that are 1.6 mi apart. To maintain passenger comfort,
2
the acceleration of the train is limited to 4 ft/s , and the jerk, or rate of change of acceleration, is limited to
2
0.8 ft/s per second. If the shuttle has a maximum speed of 20 mi/h, determine (a) the shortest time for the
shuttle to travel between the two terminals, (b) the corresponding average velocity of the shuttle.
SOLUTION
xmax = 1.6 mi; amax = 4 ft/s2
Given:
æ da ö÷
çç ÷
= 0.8ft/s2 /s; vmax = 20 mi/h
è dt ÷ømax
First note
20 mi/h = 29.333 ft/s
1.6 mi = 8448 ft
(a)
To obtain tmin , the train must accelerate and decelerate at the maximum rate to maximize the
2
time for which v = vmax . The time D t required for the train to have an acceleration of 4 ft/s is found
from
æ da ö÷
a
çç ÷ = max
è dt ø÷max
Dt
4 ft/s2
0.8 ft/s2 /s
or
Dt =
or
Dt = 5 s
Now,
æ
ö
da
çsince
= constant ÷÷÷
çè
ø
dt
after 5 s, the speed of the train is
1
v5 = (Dt )(amax )
2
or
1
v5 = (5 s)(4 ft/s2 ) = 10 ft/s
2
Then, since v5 < vmax , the train will continue to accelerate at 4 ft/s until v = vmax . The a - t curve
must then have the shape shown. Note that the magnitude of the slope of each inclined portion of the
2
curve is 0.8 ft/s /s.
2
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written consent of McGraw-Hill Education.
1969
PROBLEM 11.79 (CONTINUED)
Now
at t = (10 + D t1 )s, v = vmax :
é1
ù
2 ê (5 s)(4 ft/s2 )ú + (D t1 )(4 ft/s2 ) = 29.333 ft/s
úû
ëê 2
or
D t1 = 2.3333 s
Then
at t = 5 s :
1
v = 0 + (5)(4) = 10 ft/s
2
t = 7.3333 s : v = 10 + (2.3333)(4) = 19.3332 ft/s
1
t = 12.3333 s : v = 19.3332 + (5)(4) = 29.3332 ft/s
2
Using symmetry, the v - t curve is then drawn as shown.
Noting that A1 = A2 = A3 = A4 and that the area under the v - t curve is equal to x max , we have
é
æ10 + 19.3332 ÷ö ù
2 ê(2.3333 s) çç
ft/sú
è
ø÷÷ úû
êë
2
+ (10 + Dt2 )s ´ (29.3332 ft/s) = 8448 ft
or
D t2 = 275.67 s
Then
t min = 4(5 s) + 2(2.3333 s) + 275.67 s
= 300.34 s
tmin = 5.01min ◀
or
(b)
We have
vave =
1.6 mi
3600 s
Dx
=
´
Dt 300.34 s
1h
vave = 19.18 mi/h ◀
or
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1970
PROBLEM 11.80
During a manufacturing process, a conveyor belt starts from rest and travels a total of 1.2 ft before
2
temporarily coming to rest. Knowing that the jerk, or rate of change of acceleration, is limited to 4.8 ft/s per
second, determine (a) the shortest time required for the belt to move 1.2 ft, (b) the maximum and average
values of the velocity of the belt during that time.
SOLUTION
Given:
At
t = 0, x = 0, v = 0; x max = 1.2 ft;
when
æ da ö
x = xmax , v = 0; çç ÷÷÷
è dt ø
= 4.8 ft/s2
max
(a)
1
Observing that vmax must occur at t = tmin , the a - t curve must have the shape shown. Note that the
2
2
magnitude of the slope of each portion of the curve is 4.8 ft/s /s.
We have
at
t = Dt :
1
1
v = 0 + (Dt )(amax ) = amax Dt
2
2
t = 2D t :
1
1
vmax = amax Dt + (Dt )(amax ) = amax Dt
2
2
Using symmetry, the v - t is then drawn as shown.
Noting that A1 = A2 = A3 = A4 and that the area under the v - t curve is equal to x max , we have
(2Dt )(vmax ) = xmax
vmax = amax Dt  2 amax Dt 2 = xmax
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1971
PROBLEM 11.80 (CONTINUED)
Now
or
Then
amax
= 4.8 ft/s2 /s so that
Dt
2(4.8Dt ft/s3 )Dt 2 = 1.2 ft
Dt = 0.5 s
tmin = 4Dt
t min = 2.00 s ◀
or
(b)
We have
vmax = amax Dt
= (4.8ft/s2 /s ´Dt)Dt
= 4.8ft/s2 /s ´ (0.5 s)2
vmax = 1.2 ft/s ◀
or
Also
vave =
1.2 ft
Dx
=
Dttotal
2.00 s
vave = 0.6 ft/s ◀
or
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1972
PROBLEM 11.81
–a (m/s2)
Two seconds are required to bring the piston
rod of an air cylinder to rest; the acceleration
record of the piston rod during the 2 s is as
shown. Determine by approximate means
(a) the initial velocity of the piston rod, (b)
the distance traveled by the piston rod as it is
brought to rest.
4.0
3.0
2.0
1.0
0
0
0.25
0.5
0.75
1.0
1.25
1.5
1.75
2.0
t (s)
SOLUTION
a - t curve; at t = 2 s, v = 0
Given:
1.
The a - t curve is first approximated with a series of rectangles, each of width D t = 0.25 s. The
area ( D t ) (aave) of each rectangle is approximately equal to the change in velocity D v for the
specified interval of time. Thus,
D v @ aave D t
where the values of a ave and D v are given in columns 1 and 2, respectively, of the following table.
2.
v(2) = v0 +
Now
and approximating the area
ò
2
0
ò
2
a dt = 0
0
a dt under the a - t curve by å aave Dt » å Dv, the initial velocity is
then equal to
v0 = -å Dv
Finally, using
v2 = v1 + Dv12
where Dv12 is the change in velocity between times t1 and t2, the velocity at the end of each
0.25 interval can be computed; see column 3 of the table and the v - t curve.
3.
The v - t curve is then approximated with a series of rectangles, each of width 0.25 s. The area
(Dt )(vave ) of each rectangle is approximately equal to the change in position D x for the specified
interval of time. Thus
Dx » vave Dt
where vave and D x are given in columns 4 and 5, respectively, of the table.
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1973
PROBLEM 11.81 (CONTINUED)
4.
With x0 = 0 and noting that
x2 = x1 + Dx12
where Dx12 is the change in position between times t1 and t2, the position at the end of each 0.25 s
interval can be computed; see column 6 of the table and the x -t curve.
(a)
We had found
(b)
At t = 2 s
v0 = 1.914 m/s ◀
x = 0.840 m ◀
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1974
PROBLEM 11.82
a (m/s2)
7.0
The acceleration record shown was obtained
during the speed trials of a sports car.
Knowing that the car starts from rest,
determine by approximate means (a) the
velocity of the car at t = 8 s, (b) the distance
the car has traveled at t = 20 s.
6.0
5.0
4.0
3.0
2.0
1.0
0
0
2
4
6
8
10
12
14
16
18
20
22
t (s)
SOLUTION
Given: a -t curve; at
1.
t = 0, x = 0, v = 0
The a - t curve is first approximated with a series of rectangles, each of width D t = 2s. The
area (Dt)(aave) of each rectangle is approximately equal to the change in velocity D v for the
specified interval of time. Thus,
Dv @ aave Dt
where the values of a ave and D v are given in columns 1 and 2, respectively, of the following table.
2.
Noting that v0 = 0 and that
v2 = v1 + Dv12
where Dv12 is the change in velocity between times t1 and t2, the velocity at the end of each 2 s
interval can be computed; see column 3 of the table and the v - t curve.
3.
The v - t curve is next approximated with a series of rectangles, each of width D t = 2 s. The
area (Dt )(vave ) of each rectangle is approximately equal to the change in position D x for the
specified interval of time.
Thus,
Dx @ vave Dt
where vave and D x are given in columns 4 and 5, respectively, of the table.
4.
With x0 = 0 and noting that
x2 = x1 + Dx12
where Dx12 is the change in position between times t1 and t2, the position at the end of each 2 s
interval can be computed; see column 6 of the table and the x - t curve.
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1975
PROBLEM 11.82 (CONTINUED)
(a)
At t = 8 s, v = 32.58 m/s
(b)
At t = 20 s
v = 117.3 km/h ◀
or
x = 660 m ◀
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written consent of McGraw-Hill Education.
1976
PROBLEM 11.83
–a (ft/s2)
60
A training airplane has a velocity of 126 ft/s when
it lands on an aircraft carrier. As the arresting gear
of the carrier brings the airplane to rest, the
velocity and the acceleration of the airplane are
recorded; the results are shown (solid curve) in the
figure. Determine by approximate means (a) the
time required for the airplane to come to rest,
(b) the distance traveled in that time.
50
40
30
20
10
0
0
20
40
60
80
100
120
140
v (ft/s)
SOLUTION
Given: a - v curve:
v0 = 126 ft/s
The given curve is approximated by a series of uniformly accelerated motions (the horizontal dashed lines on
the figure).
For uniformly accelerated motion
v22 = v12 + 2a( x2 - x1 )
v2 = v1 + a(t2 - t1 )
v22 - v12
2a
v2 - v1
Dt =
a
or
Dx =
For the five regions shown above, we have
Region
v1 , ft/s
v2 , ft/s
a, ft/s2
Dx , ft
Dt, s
1
126
120
-12.5
59.0
0.480
2
120
100
-33
66.7
0.606
3
100
80
-45.5
39.6
0.440
4
80
40
-54
44.4
0.741
5
40
0
-58
13.8
0.690
223.5
2.957
å
(a)
From the table, when v = 0
t = 2.96 s ◀
(b)
From the table and assuming x 0 = 0, when v = 0
x = 224 ft ◀
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1977
PROBLEM 11.84
v (in./s)
100
Shown in the figure is a portion of the experimentally
determined v - x curve for a shuttle cart. Determine by
approximate means the acceleration of the cart (a) when
x = 10 in., (b) when v = 80 in./s.
80
60
40
20
0
0
10
20
30
40
50
x (in.)
SOLUTION
Given: v - x curve
First note that the slope of the above curve is
dv
dx . Now
a=v
(a)
dv
dx
When
x = 10 in., v = 55 in./s
Then
æ 40 in./s ö÷
a = 55 in./s çç
÷
è 13.5 in. ø÷
a = 163.0 in./s2 ◀
or
(b)
When v = 80 in./s, we have
æ 40 in./s ö÷
a = 80 in./s çç
÷
è 28 in. ø÷
a = 114.3 in./s2 ◀
or
Note: To use the method of measuring the subnormal outlined at the end of Section 11.8, it is necessary
that the same scale be used for the x and v axes (e.g., 1 in. = 50 in., 1 in. = 50 in./s). In the above
solution, D v and D x were measured directly, so different scales could be used.
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written consent of McGraw-Hill Education.
1978
PROBLEM 11.85
a (m/s2)
0.6
0
T/3
T
t(s)
An elevator starts from rest and rises 40 m to its maximum
velocity in T s with the acceleration record shown in the figure.
Determine (a) the required time T, (b) the maximum velocity,
(c) the velocity and position of the elevator at t = T/2.
SOLUTION
Find the position of the elevator from the a -t curve using the moment-area method.
First, find the areas A1 and A2 on the a - t curve
(a)
A1 =
1
T
(0.6) = 0.1T m/s
2
3
A2 =
1
2T
(0.6)
= 0.2T m/s
2
3
Apply the moment-area formula: x - x0 = v0 t +
ò
T
(T - t ) adt
0
æ
æ
æ T 1 æ 2T ööö÷
2 æ T öö
x = v0 t + ( A1 ) çççT - ççç ÷÷÷÷÷÷ + A2 çççT - ççç + ççç ÷÷÷÷÷÷÷÷
è
è 3 3 è 3 øø÷÷ø
3 è 3 øø÷
èç
7
8
40 = 0 + T 2 + T 2
90
90
15 2
1 2
= T
= T
90
6
T 2 = (40)(6)
= 240 s2
T = 15.49 s ◀
(b)
vmax = v0 + A1 + A2 = 0 + 0.1T + 0.2T = 0.3T
vmax = 4.65 m/s ◀
(c)
To find the position at t = T/2, first find the areas A1 and A2 on the a - t curve
T
1
A3 = (0.6) = 0.05T
2
6
T
1
A4 = (0.45) = 0.0375T
2
6
A1 = 0.1T
Apply the moment-area formula: x - x 0 = v0 t +
ò
T /2æT
0
ö
çç - t ÷÷÷ adt
è2
ø
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1979
PROBLEM 11.85 (CONTINUED)
æ T 2T ö
æ2 T ö
æ1 T ö
T
+ A1 çç - ÷÷÷ + A3 çç ⋅ ÷÷÷ + A4 çç ⋅ ÷÷÷
è2
è3 6 ø
è3 6 ø
2
9 ø
æ 5T ö
T
æT ö
= 0 + (0.1T ) çç ÷÷÷ + (0.05T )çç ÷÷÷ + (0.0375T ) = 0.035417T 2
è 18 ø
è9ø
18
x = v0
= (0.035417)(15.49)2
x = 8.50 m ◀
v = v0 + A1 + A3 + A4 = 0.1875T
v = 2.90 m/s ◀
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1980
PROBLEM 11.86
5 mi
A
3 mi
C
Two road rally checkpoints A and B are located
on the same highway and are 8 mi apart. The
speed limits for the first 5 mi and the last 3 mi are
60 mi/h and 35 mi/h, respectively. Drivers must
stop at each checkpoint, and the specified time
between points A and B is 10 min 20 s. Knowing
that a driver accelerates and decelerates at the
same constant rate, determine the magnitude of
her acceleration if she travels at the speed limit as
much as possible.
B
SOLUTION
Given
10
20
+
60 3600
= 0.1722 h
10 min 20 s =
Sketch the v - t curve and note ta =
60
25
35
, t b = , tc =
a
a
a
Find Areas A1 and A2
1
1
A1 = 60t1 - (60)(ta ) - (25)tb
2
2
1
1
= 60 t1 -1800 - 312.5
a
a
1
A2 = 35(0.1722 - t1 ) - (35)tc
2
1
= 6.0278 - 35t1 - 612.5
a
(1)
(2)
Also given
A1 = 5 mi and A2 = 3 mi
Setting (1) equal to 5 mi:
1
60t1 - 2112.5 = 5
a
(3)
Setting (2) equal to 3 mi:
1
35t1 + 612.5 = 3 0278
a
(4)
Solving equations (3) and (4) for t1 and
1
: t1 = 85.45 ´10-3 h = 5.13min
a
1
= 60.23´10-6 h 2 / mi
a
a = 16.616 ´103 mi/ h2 =
(16.616 ´103 )(5280)
(3600)2
a = 6.77 ft / s2 ◀
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1981
PROBLEM 11.87
a (m/s2)
2
4
As shown in the figure, from t = 0 to t = 4 s, the acceleration of a
given particle is represented by a parabola. Knowing that x = 0 and
v = 8 m/s when t = 0, (a) construct the v - t and x – t curves for
0 < t < 4 s, (b) determine the position of the particle at
t =3s. (Hint: Use table inside the front cover.)
t(s)
a = – 3 (t – 2)2 m/s2
–12
SOLUTION
t = 0, x = 0, v = 8 m/s
Given
At
(a) We have
v 2 = v1
(area under a - t curve from t1 to t2)
and
x 2 = x1
(area under v - t curve from t1 to t2)
Then, using the formula for the area of a parabolic spandrel, we have
1
at t = 2 s: v = 8 - (2)(12) = 0
3
1
t = 4 s: v = 0 - (2)(12) = -8 m/s
3
The v - t curve is then drawn as shown.
Note: The area under each portion of the curve is a spandrel of Order no. 3.
(2)(8)
Now at t = 2 s: x = 0 +
=4 m
3 +1
(2)(8)
t = 4 s: x = 4 =0
3 +1
The x - t curve is then drawn as shown.
(b)
We have
At t = 3s: a = -3(3 - 2) = -3 m/s
2
◀
2
1
v = 0 - (1)(3) = -1 m/s
3
(1)(1)
x = 43 +1
x 3 = 3.75 m ◀
or
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1982
PROBLEM 11.88
a (m/s2)
6
2
8
12
–0.75
t(s)
A particle moves in a straight line with the acceleration
shown in the figure. Knowing that the particle starts from
the origin with v0 = -2 m/s, (a) construct the v - t and
x - t curves for 0 < t < 18 s, (b) determine the position
and the velocity of the particle and the total distance
traveled when t = 18 s.
SOLUTION
Find the indicated areas under the a - t curve:
A 1 = (-0.75)(8) = -6 m/s
A 2 = (2)(4) = 8 m/s
A 3 = (6)(6) = 36 m/s
Find the values of velocity at times indicated:
v0 = -2 m/s
v8 = v0 + A1 = -8 m/s
v12 = v8 + A2 = 0
v18 = v12 + A3 = 36 m/s
(a) Sketch v - t curve using straight line portions over the constant acceleration periods.
Find the indicated areas under the v - t curve:
1
A4 = (-2 - 8)(8) = -40 m
2
1
A5 = (-8)(4) = -16 m
2
1
A6 = (36)(6) = 108 m
2
Find the values of position at times indicated:
x0 = 0
x8 = x0 + A4 = -40 m
x12 = x8 + A5 = -56 m
x18 = x12 + A6 = 52 m
Sketch the x - t curve using the positions and times calculated:
(b) Position at t = 18s
x18 = 52 m ◀
Velocity at t - 18s
v18 = 36 m/s ◀
d = 164 m ◀
Total Distance Traveled: = 56 + 108
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1983
PROBLEM 11.89
A ball is thrown so that the motion is defined by the
2
equations x = 5 t and y = 2 + 6t - 4.9t , where
x and y are expressed in meters and t is expressed
in seconds. Determine (a) the velocity at t = 1s,
(b) the horizontal distance the ball travels before
hitting the ground.
y
x
SOLUTION
Units are meters and seconds.
Horizontal motion:
vx =
dx
=5
dt
Vertical motion:
vy =
dy
= 6 - 9.8t
dt
(a)
Velocity at t = 1 s.
vx = 5
v y = 6 - 9.8 = -3.8
v = v x2 + v y2 = 52 + 3.82 = 6.28 m/s
tanq =
(b)
vy
vx
=
-3.8
5
Horizontal distance:
v = 6.28 m/s
q = -37.2
37.2 ◀
( y = 0)
y = 2 + 6t - 4.9t 2
x = 7.49 m ◀
t = 1.4971 s
x = (5)(1.4971) = 7.4856 m
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written consent of McGraw-Hill Education.
1984
PROBLEM 11.90
y
3
The motion of a vibrating particle is defined by the position vector
r = 10(1 - e-3t )i + (4e-2 t sin15t ) j, where r and t are expressed in
millimeters and seconds, respectively. Determine the velocity and
acceleration when (a) t = 0, (b) t = 0.5 s.
2
1
0
2
4
6
8
10 x
−1
−2
SOLUTION
r = 10(1 - e-3t )i + (4e-2 t sin15t ) j
Then
and
(a)
v=
dr
= 30e-3t i + [60e-2 t cos15t - 8e-2 t sin15t ]j
dt
dv
= -90e-3t i + [-120e-2 t cos15t - 900e-2 t sin15t -120e-2 t cos15t + 16e-2 t sin15t ]j
dt
= -90e-3t i + [-240e-2 t cos15t - 884e-2 t sin15t ]j
a=
When t = 0 :
v = 30i + 60 j mm/s
v = 67.1 mm/s
63.4 ◀
a =-90i - 240 j mm/s2
a = 256 mm/s2
69.4 ◀
v = 8.29 mm/s
36.2 ◀
a = 336 mm/s2
86.6 ◀
When t = 0.5 s:
v = 30e-1.5 i + [60e-1 cos 7.5 - 8e-1 sin 7.5]
= 6.694 i + 4.8906 j mm/s
a = 90e-1.5 i + [-240e-1 cos 7.5 - 884e-1 sin 7.5 j]
= -20.08i - 335.65 j mm/s2
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written consent of McGraw-Hill Education.
1985
PROBLEM 11.91
The
motion
of
a
particle
is
defined
by
the
equations
x = (4 cos pt - 2) / (2 - cos pt ) and y = 3sin pt / (2 - cos pt ), where x
and y are expressed in feet and t is expressed in seconds. Show that the
path of the particle is part of the ellipse shown, and determine the
velocity when (a) t = 0, (b) t = 1/3 s, (c) t = 1 s.
SOLUTION
Substitute the given expressions for x and y into the given equation of the ellipse, and note that
the equation is satisfied.
(
)
16 cos2 pt - 16 cos pt + 4
x2
y2
9sin 2 pt
+
=
+
2
2
4
3
4 (2 - cos pt )
3(2 - cos pt )
=
4 cos2 pt - 4 cos pt + 1 + 3sin 2 pt
=
(2 - cos pt )
2
4 - 4 cos pt + cos2 pt
(2 - cos pt )
2
=1
Calculate x and y by differentiation.
x =
y =
-4 p sin pt (4 cos pt - 2)(p sin pt )
-6p sin pt
=
2
2
(2 - cos pt )
(2 - cos pt )
(2 - cos pt )
3sin pt (p sin pt ) 3p (2 cos pt - 1)
3p cos pt
=
2
(2 - cos pt ) (2 - cos pt )2
(2 - cos pt )
(a) When t = 0 s,
1
(b) When t = s,
3
(c) When t = 1 s,
x = 0
x =
-6p
( ) =-4 p3
2
(2 - )
3
and
y =
1
2
x = 0
y = 3p,
and
2
3, y = 0
3p (-3)
(3)
2
= -p,
v = 9.42 ft/s
v = 7.26 ft/s
v = 3.14 ft/s
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written consent of McGraw-Hill Education.
1986
◀
◀
◀
PROBLEM 11.92
The motion of a particle is defined by the equations x = 1 0 t - 5 s in t and y =10 - 5cos t, where x and y are
expressed in feet and t is expressed in seconds. Sketch the path of the particle for the time interval
0 £ t £ 2 p , and determine (a) the magnitudes of the smallest and largest velocities reached by the particle,
(b) the corresponding times, positions, and directions of the velocities.
SOLUTION
Sketch the path of the particle, i.e., plot of y versus x.
Using x = 10 t - 5 sin t , and y = 10 - 5 cos t obtain the values in the table below. Plot as shown.
t(s)
x(ft)
y(ft)
0
0.00
5
p
2
10.71
10
p
31.41
15
p
2
52.12
10
2p
62.83
5
3
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written consent of McGraw-Hill Education.
1987
PROBLEM 11.92 (Continued)
(a)
Differentiate with respect to t to obtain velocity components.
dx
= 10 - 5cos t and vy = 5sin t
dt
v2 = vx2 + vy2 = (10 - 5 cos t )2 + 25sin 2 t = 125 -100 cos t
vx =
d ( v )2
= 100sin t = 0 t = 0.  p.  2p...  N p
dt
(b)
When t = 2 N p .
cos t = 1.
and
v is minimum.
When t = (2 N + 1)p .
cos t = 1.
and
v is maximum.
2
2
(v2 )min = 125 -100 = 25(ft/s)2
vmin = 5 ft/s ◀
(v2 )max =125 +100 = 225(ft/s)2
vmax = 15 ft/s ◀
When v = vmax .
When N = 0,1, 2,...
x = 10(2 p N ) - 5 sin(2 p N )
x = 20 p N ft ◀
y = 10 - 5cos(2p N )
y = 5 ft ◀
vx = 10 - 5cos(2p N )
v x = 5 ft/s ◀
vy = 5sin(2pN )
tan q =
vy
vx
= 0,
vy = 0 ◀
q=0 ◀
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written consent of McGraw-Hill Education.
1988
PROBLEM 11.92 (CONTINUED)
t = (2 N + 1)p s ◀
When v = vmax .
x = 10[2 p ( N - 1)] - 5 sin[2 p ( N + 1)]
y = 10 - 5cos[2p( N + 1)]
vx = 10 - 5cos[2p( N + 1)]
vy = 5sin[2p(N +1)]
tan q =
vy
vx
= 0,
x = 20 p ( N + 1) ft ◀
y = 15 ft ◀
v x = 15 ft/s ◀
vy = 0 ◀
q=0 ◀
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1989
PROBLEM 11.93
Test engineers are examining how shock absorber designs affect the displacement of a mountain biker’s hip
after she lands from a jump. Immediately after the jump, the x and y locations of her hip can be described
by x (t ) = 20t + 0.2 sin( 43p t ) and y(t ) = 3.5 + 0.1e-0.3t sin( 43p t ) where x and y are expressed in feet and t is
expressed in seconds. Plot the path of the hip for the time interval 0 £ t £ 3 s and determine the hip’s
vertical position and the components of the hip velocity at t = 2 s.
SOLUTION
Graph y(t) vs x(t)
y(2) = 3.55 ft ◀
Evaluate y(t) at t = 2 seconds
Next, differentiate v and y with respect to t to get the velocities and evaluate at t=2 sec,
vx = 20 +
vy =
4p
cos(4pt / 3)
15
1 é
2pe(-0.3t ) cos(4pt / 3)ùú
û
15 ëê
v x (2) = 19.58 ft/s ◀
v y (2) = -0.129 ft/s ◀
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1990
Problem 11.94
y (m)
6
0
x (m)
2
A girl operates a radio-controlled model car in a vacant
parking lot. The girl’s position is at the origin of the xy
coordinate axes, and the surface of the parking lot lies in the
x-y plane.
 The motion of the car is defined by the position
vector r = (2 + 2t 2 )iˆ + (6 + t 3 ) ˆj where r and t are expressed
in meters and seconds, respectively. Determine (a) the distance
between the car and the girl when t = 2 s, (b) the distance the
car traveled in the interval from t = 0 to t = 2 s, (c) the speed
and direction of the car’s velocity at t = 2 s, (d) the magnitude
of the car’s acceleration at t = 2 s.
SOLUTION
Given:
r = (2 + 2t 2 )i + (6 + t 3 ) j
(a) At t = 2s
r (2) = 10 i + 14 j m
r (2) = 10 2 + 14 2 m
r(2) = 17.20 m ◀
(b) The car is traveling on a curved path. The distance that the car travels during any infinitesimal interval
is given by:
ds = dx 2 + dy2
where:
dx = 4tdt and dy = 3t 2 dt
Substituting
ds = 16 t 2 + 9t 4 dt
Integrating:
ò
s
ds =
0
s=
(c) Velocity can be found by
ò
2
t 16 + 9t 2 dt
0
2
1
(16 + 9t 2 )3/ 2
27
0
v=
s = 11.52 m ◀
dr
dt
v = 4 ti + 3t 2 j m/s
At t = 2 s
v = 8i + 12 j m/s
(d) Acceleration can be found by a =
v = 14.42 m/s
56.31 ◀
dv
dt
a = 4 i + 6tj m/s2
At t = 2 s
a = 4 i + 12 j m/s2
a = 12.65 m/s2 ◀
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1991
PROBLEM 11.95
The three-dimensional motion of a particle is defined by the position vector r = (Rt cos wnt)i + ctj +
(Rt sin wnt)k. Determine the magnitudes of the velocity and acceleration of the particle. (The space curve
described by the particle is a conic helix.)
SOLUTION
We have
r = ( Rt cos w n t ) i + ctj + ( Rt sin w n t ) k
Then
v=
and
a=
dr
= R (cos wnt - wnt sin wnt )i + cj + R (sin wnt + wnt cos wnt )k
dt
dv
dt
= R (-wn sin wnt - wn sin wnt - wn2t cos wnt )i
+ R (wn cos wnt + wn cos wnt - wn2t sin wnt )k
= R (-2wn sin wnt - wn2t cos wnt )i + R (2wn cos wnt - wn2t sin wnt )k
Now
v 2 = vx2 + vy2 + vz2
= [R (cos wnt - wnt sin wnt )]2 + (c )2 + [R (sin wnt + wnt cos wnt )]2
= R 2 [(cos2 wnt - 2wnt sin wnt cos wnt + wn2t 2 sin 2 wnt )
+ (sin2 wnt + 2wnt sin wnt cos wnt + wn2t 2 cos2 wnt )] + c 2
= R 2 (1 + wn2t 2 ) + c 2
v = R2 (1 + wn2 t 2 ) + c2 ◀
or
Also,
a 2 = a x2 + a y2 + a z2
= [ R(-2wn sin wn t - wn2 t cos wn t )]2 + (0)2
+ [ R(2wn cos wn t - wn2 t sin wn t )]2
= R2 [(4wn2 sin2 wn t + 4wn3t sin wn t cos wn t + wn4 t 2 cos2 wn t )
+ (4wn2 cos2 wn t - 4wn3t sin wn t cos wn t + wn4 t 2 sin 2 wn t )]
= R2 (4wn2 + wn4 t 2 )
a = Rwn 4 + wn2 t 2 ◀
or
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written consent of McGraw-Hill Education.
1992
PROBLEM 11.96
y
The three-dimensional motion of a particle is defined by the position
vector r = ( At cos t ) i + ( A t 2 + 1) j + ( Bt sin t ) k , where r and t are
expressed in feet and seconds, respectively. Show that the curve
2
2
described by the particle lies on the hyperboloid (y/A) - (x/A) 2
(z/B) = 1. For A = 3 and B = 1, determine (a) the magnitudes of the
velocity and acceleration when t = 0, (b) the smallest nonzero value
of t for which the position vector and the velocity are perpendicular to
each other.
2
2
y2
– x – z =1
2
2
A
B2
A
z
x
SOLUTION
We have
r = ( At cos t ) i + ( A t 2 + 1) j + ( Bt sin t ) k
or
x = At cos t
x
At
y = A t 2 + 1)
z
Bt
æ y ö2
t 2 = çç ÷÷ -1
è Aø
Then
cos t =
Now
æ x ö2 æ z ö2
cos2 t + sin 2 t = 1  çç ÷÷÷ + çç ÷÷÷ = 1
è At ø è Bt ø
æ x ö2 æ z ö2
t 2 = çç ÷÷ + çç ÷÷
è Aø è B ø
or
æ y ÷ö2
æ x ö2 æ z ö2
ç ÷ -1 = ç ÷÷ + ç ÷÷
èç A ø
èç A ø èç B ø
Then
æ y ÷ö2 æ x ÷ö2 æ z ÷ö2
ç ÷ -ç ÷ -ç ÷ = 1
çè A ø çè A ø çè B ø
or
(a)
sin t =
z = Bt sin t
Q.E.D.
With A = 3 and B = 1, we have
v=
dr
t
j + (sin t + t cos t )k
= 3(cos t - t sin t )i + 3
2
dt
t +1
æ t ö÷
çç
÷÷
t + 1 - t çç 2
è t + 1 ÷ø÷
dv
a=
= 3(- sin t - sin t - t cos t )i + 3
j
dt
(t 2 + 1)
+ (cos t + cos t - t sin t )k
2
and
= -3(2sin t + t cos t )i + 3
1
j + (2 cos t - t sin t )k
(t + 1)3/ 2
2
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written consent of McGraw-Hill Education.
1993
◀
PROBLEM 11.96 (CONTINUED)
v = 3(1 - 0)i + (0) j + (0)k
At t = 0:
v = v x2 + v y2 + vz2
v = 3 ft/s ◀
or
a = -3(0)i + 3(1) j + (2 - 0)k
and
a 2 = (0)2 + (3)2 + (2)2 = 13
Then
a = 3.61 ft/s2 ◀
or
(b)
If r and v are perpendicular, r ⋅ v = 0
æ
ö÷
t
[(3t cos t )i + (3 t 2 + 1) j + (t sin t )k ] ⋅ [3(cos t - t sin t )i + ççç3
÷÷ j + (sin t + t cos t )k ] = 0
çè t 2 + 1 ø÷÷
or
æ
÷÷ö
t
ç
(3t cos t )[3(cos t - t sin t )] + (3 t 2 + 1) çç3
÷÷ + (t sin t )(sin t + t cos t ) = 0
ççè t 2 + 1) ÷ø
Expanding
(9t cos2 t - 9t 2 sin t cos t ) + (9t ) + (t sin 2 t + t 2 sin t cos t ) = 0
10 + 8cos2 t - 8t sin t cos t = 0
or (with t ¹ 0)
7 + 2 cos 2 t - 2 t sin 2 t = 0
or
Using “trial and error” or numerical methods, the smallest root is
t = 3.82 s ◀
Note: The next root is t = 4.38 s.
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written consent of McGraw-Hill Education.
1994
PROBLEM 11.97
An airplane used to drop water on brushfires is flying
horizontally in a straight line at 130 knots (1 knot = 1.15 mph)
at an altitude of 175 ft. Determine the distance d at which the
pilot should release the water so that it will hit the fire at B.
SOLUTION
First note
æ1.15mph öæ
÷÷çç 1hr öæ
÷÷çç 5280ft ö÷÷ = 219.27ft/s
vo = 130knots çç
çè knot øè
÷÷ç 3600s øè
÷÷ç mi ø÷÷
Place origin of coordinates at Point A.
Vertical motion. (Uniformly accelerated motion)
1
y = 0 + (0)t - gt 2
2
1
-175 ft = - (32.2 ft/s2 )t 2
2
At B:
t B = 3.297s
or
Horizontal motion. (Uniform)
x = 0 + (vx )0 t
At B:
d = (219.27 ft/s)(3.297 s) = 722.9 ft
d = 723 ft ◀
or
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written consent of McGraw-Hill Education.
1995
PROBLEM 11.98
25 m/s
A ski jumper starts with a horizontal take-off velocity of
25 m/s and lands on a straight landing hill inclined at 30
Determine (a) the time between take-off and landing, (b) the
length d of the jump, (c) the maximum vertical distance
between the jumper and the landing hill.
d
30°
SOLUTION
y = -x tan 30
(a) At the landing point,
x = x0 + (vx )0 t = v0 t
Horizontal motion:
Vertical motion:
1
1
y = y0 + (vy )0 t - gt 2 = - gt 2
2
2
from which
t2 = -
t=
Rejecting the t = 0 solution gives
(b) Landing distance:
2 y 2 x tan 30 2v0 t tan 30
=
=
g
g
g
d=
2 v0 tan 30 (2)(25) tan 30
=
g
9.81
v0 t
x
(25)(2.94)
=
=
cos30 cos30
cos30
t = 2.94 s ◀
d = 84.9 m ◀
h = x tan 30 + y
(c) Vertical distance:
1
h = v0 t tan30 - gt 2
2
or
Differentiating and setting equal to zero,
dh
= v0 tan 30 - gt = 0
dt
Then,
hmax =
=
or
t=
v0 tan 30
g
2
(v0 )(v0 tan30)tan30 1 æç v0 tan30 ÷ö
÷÷
- gç
÷ø
g
2 çè
g
v02 tan 2 30 (25)2 (tan 30)2
=
2g
(2)(9.81)
hmax = 10.62 m ◀
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written consent of McGraw-Hill Education.
1996
PROBLEM 11.99
12.2 m
A
v0
α
1.5 m
B
h
A baseball pitching machine
“throws” baseballs with a horizontal
velocity v0. Knowing that height h
varies between 788 mm and
1068 mm, determine (a) the range of
values of v0, (b) the values of a
corresponding to h = 788 mm and
h = 1068 mm.
SOLUTION
(a)
y0 = 1.5 m,
Vertical motion:
1
y = y0 + (vy )0 t - gt 2
2
t=
or
( v y )0 = 0
2( y0 - y)
g
tB =
2( y0 - h)
g
When h = 788 mm = 0.788 m,
tB =
(2)(1.5 - 0.788)
= 0.3810s
9.81
When h = 1068 mm = 1.068 m,
tB =
(2)(1.5 - 1.068)
= 0.2968s
9.81
At Point B,
Horizontal motion:
y=h
or
x 0 = 0, ( v x ) 0 = v0 ,
x = v0 t
or
v0 =
x xB
=
t
tB
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written consent of McGraw-Hill Education.
1997
PROBLEM 11.99 (CONTINUED)
With x B = 12.2 m,
v0 =
and
(b)
v0 =
we get
12.2
= 32.02 m/s
0.3810
12.2
= 41.11 m/s
0.2968
32.02 m/s £ v0 £ 41.11 m/s
or
Vertical motion:
vy = (vy )0 - gt =-gt
Horizontal motion:
v x = v0
tan a = -
115.3 km/h £ v0 £ 148.0 km/h ◀
(vy ) B gt B
dy
==
dx
v0
(vx ) B
For h = 0.788 m,
tan a =
(9.81)(0.3810)
= 0.11673,
32.02
a = 6.66  ◀
For h = 1.068 m,
tan a =
(9.81)(0.2968)
= 0.07082,
41.11
a = 4.05  ◀
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written consent of McGraw-Hill Education.
1998
PROBLEM 11.100
While delivering newspapers, a
girl throws a newspaper with a
horizontal velocity v0. Determine
the range of values of v0 if the
newspaper is to land between
Points B and C.
14 in.
36 in.
v0
A
C
4 ft
8 in.
8 in.
8 in.
B
7 ft
SOLUTION
Vertical motion. (Uniformly accelerated motion)
1
y = 0 + (0)t - gt 2
2
Horizontal motion. (Uniform)
x = 0 + (vx )0 t = v0 t
At B:
y:
t B = 0.455016 s
or
Then
x:
y:
or
1
-2 ft = - (32.2 ft/s2 )t 2
2
t c = 0.352454 s
or
Then
7 ft = ( v0 ) B (0.455016 s)
( v0 ) B = 15.38 ft/s
or
At C:
1
1
-3 ft = - (32.2 ft/s2 )t 2
3
2
x:
1
12 ft = (v0 )c (0.352454 s)
3
( v0 )c = 35.0 ft/s
15.38 ft/s < v0 < 35.0 ft/s ◀
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written consent of McGraw-Hill Education.
1999
PROBLEM 11.101
A pump is located near the edge of the horizontal platform shown. The
nozzle at A discharges water with an initial velocity of 25 ft/s at an
angle of 55° with the vertical. Determine the range of values of the
height h for which the water enters the opening BC.
SOLUTION
v0 = 25 ft/s, a = 90 - 55 = 35, g = 32.2 ft/s2
Data:
Horizontal motion.
Vertical motion.
Eliminate t.
x = (v0 cos a) t
1
y = h + (v0 sin a) t - gt 2
2
t=
x
v0 cos a
y = h + x tan a -
Solve for h.
To hit point B.
h = y - x tan a +
gx 2
2v02 cos2 a
gx 2
2v02 cos2 a
x = 20 ft, y = 0
(32.2)(20)
h = 0 - 20 tan 35 +
2
(2)(25cos35)
2
To hit point C.
= 1.352 ft
x = 24 ft, y = 0
(32.2)(24)
2
(2)(25cos35)
2
h = 0 - 24 tan 35 +
= 5.31 ft
1.352 ft < h < 5.31 ft ◀
Range of values of h.
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2000
PROBLEM 11.102
v0
In slow pitch softball the underhand
pitch must reach a maximum height
of between 1.8 m and 3.7 m above
the ground. A pitch is made with an
initial velocity v0 of magnitude
13 m/s at an angle of 33 with the
horizontal. Determine (a) if the
pitch meets the maximum height
requirement, (b) the height of the
ball as it reaches the batter.
33°
0.6 m
15.2 m
SOLUTION
v0 = 13 m/s, a = 33 , x 0 = 0, y0 = 0.6 m
v y = v0 sin a - gt
Vertical motion:
1
y = y0 + (v0 sin a)t - gt 2
2
vy = 0
At maximum height,
t=
(a)
or
t=
v0 sin a
g
13sin 33
= 0.7217 s
9.81
1
ymax = 0.6 + (13sin 33)(0.7217) - (9.81)(0.7217)2
2
ymax = 3.16 m ◀
1.8 m < 3.16 m < 3.7 m
Horizontal motion:
x = x0 + (v0 cos a )t
At x = 15.2 m,
t=
(b) Corresponding value of y :
yes ◀
or
t=
x - x0
v0 cos a
15.2 - 0
= 1.3941s
13cos33
1
y = 0.6 + (13sin 33)(1.3941) - (9.81)(1.3941)2
2
y = 0.937 m ◀
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written consent of McGraw-Hill Education.
2001
PROBLEM 11.103
v0
C
20°
A
A volleyball player serves the ball
with an initial velocity v0 of
magnitude 13.40 m/s at an angle of
20° with the horizontal. Determine
(a) if the ball will clear the top of
the net, (b) how far from the net the
ball will land.
2.43 m
2.1 m
9m
SOLUTION
First note
(vx )0 = (13.40 m/s) cos 20 = 12.5919 m/s
(vy )0 = (13.40 m/s)sin 20 = 4.5831m/s
(a)
Horizontal motion. (Uniform)
x = 0 + ( v x )0 t
At C
9 m = (12.5919 m/s)t
or
t c = 0.71475 s
Vertical motion. (Uniformly accelerated motion)
1
y = y0 + (vy )0 t - gt 2
2
At C:
yc = 2.1m + (4.5831m/s)(0.71475s)
1
- (9.81m/s2 )(0.71475s)2
2
= 2.87 m
yc > 2.43 m (height of net)  ball clears net ◀
(b)
At B, y = 0
1
0 = 2.1m + (4.5831m/s)t - (9.81m/s2 )t 2
2
Solving
t B = 1.2471175 s (the other root is negative)
Then
d = (vx )0 t B = (12.5919m/s)(1.271175s)
= 16.01m
b = (16.01 - 9.00) m = 7.01m from the net ◀
The ball lands
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written consent of McGraw-Hill Education.
2002
PROBLEM 11.104
A golfer hits a golf ball with an initial velocity of 140 ft/s at an angle of 25° with the horizontal. Knowing
that the fairway slopes downward at an average angle of 5°, determine the distance d between the golfer and
Point B where the ball first lands.
SOLUTION
vx 0 = 140 cos25
First note
vy 0 = 140sin 25
x B = d cos 5 yB = -d sin 5
and at B
Now
Horizontal motion. (Uniform)
x = 0 + ( v x )0 t
d cos(5 ) = (140 cos(25 )t or t B =
At B
cos(5 )
140 cos(25 )
d
Vertical motion. (Uniformly accelerated motion)
1
y = 0 + (vy )0 t - gt 2 (g = 32.2 ft/s2 )
2
1
-d sin 5 = (140sin 25 )t B - gt B 2
2
At B:
Substituting for t B
or
-d sin 5 = (140 sin 25 )
d=
2
32.2 cos 5
d = 555.89 ft
1 æç cos(5 ) ö÷÷ 2
d
gç
÷d
2 ççè140 cos(25 ) ÷÷ø
140 cos(25 )
cos(5 )
(140 cos 25 )2 (tan 5 + tan 25 )
d = 185 yd ◀
or
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2003
PROBLEM 11.105
v0
B
A
A homeowner uses a snowblower to clear his
driveway. Knowing that the snow is discharged
at an average angle of 40° with the horizontal,
determine the initial velocity v0 of the snow.
40°
3.5 ft
2 ft
14 ft
SOLUTION
First note
(vx )0 = v0 cos 40
(vy )0 = v0 sin 40
Horizontal motion. (Uniform)
x = 0 + ( v x )0 t
14 = ( v0 cos 40)t
At B:
or
tB =
14
v0 cos 40
Vertical motion. (Uniformly accelerated motion)
1
y = 0 + (vy )0 t - gt 2
2
( g = 32.2 ft/s2 )
1
1.5 = (v0 sin 40)t B - gt B2
2
At B:
Substituting for t B
æ 14 ö÷ 1 æ 14 ö÷2
1.5 = (v0 sin 40)çç
÷ - gç
÷
çè v0 cos40 ÷ø 2 ççè v0 cos40 ÷ø
or
v02 =
1
2
(32.2)(196) / cos2 40
-1.5 + 14 tan 40
v0 = 22.9 ft/s ◀
or
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written consent of McGraw-Hill Education.
2004
PROBLEM 11.106
7m
C
At halftime of a football game souvenir balls are
thrown to the spectators with a velocity v0.
Determine the range of values of v0 if the balls are
to land between Points B and C.
B
v0
10
40°
A
m
35°
2 m 1.5 m
8m
SOLUTION
The motion is projectile motion. Place the origin of the xy-coordinate system at ground level just below Point A.
The coordinates of Point A are x 0 = 0, y 0 = 2m. The components of initial velocity are ( v x ) 0 = v0 cos 40  m/s
and ( v y )0 = v0 sin 40.
Horizontal motion:
x = x0 + (vx )0 t = (v0 cos 40)t
(1)
Vertical motion:
From (1),
1
y = y0 + (vy )0 t - gt 2
2
1
= 2 + (v0 sin 40) - gt 2
2
v0 t =
x
cos40
(2)
(3)
y = 2 + x tan 40 - 4.905t 2
Then
t2 =
Point B:
2 + x tan 40 - y
4.905
(4)
x = 8 + 10 cos35 = 16.1915 m
y = 1.5 + 10sin 35 = 7.2358 m
16.1915
= 21.1365 m
cos 40
2 + 16.1915tan 40 - 7.2358
t2 =
t = 1.3048 s
4.905
21.1365
v0 =
v0 = 16.199 m/s
1.3048
v0 t =
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written consent of McGraw-Hill Education.
2005
PROBLEM 11.106 (CONTINUED)
Point C:
x = 8 + (10 + 7)cos35 = 21.9256 m
y = 1.5 + (10 + 7)sin 35 = 11.2508 m
v0 t =
21.9256
= 28.622 m
cos 40
t2 =
2 + 21.9256 tan 40 - 11.2508
4.905
v0 =
28.622
1.3656
t = 1.3656 s
v0 = 20.96 m/s
16.20 m/s < v 0 < 21.0 m/s ◀
Range of values of v0.
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2006
PROBLEM 11.107
16 ft
d
B
v0
30°
A
A basketball player shoots when she is 16 ft from the
backboard. Knowing that the ball has an initial velocity v0 at an
angle of 30° with the horizontal, determine the value of v0 when
d is equal to (a) 9 in., (b) 17 in.
10 ft
6.8 ft
SOLUTION
First note
( v x ) 0 = v0 cos 30 
( v y ) 0 = v0 sin 30 
Horizontal motion. (Uniform) x = 0 + (vx )0 t
(16 - d ) = (v0 cos30)t or t B =
At B:
16 - d
v0 cos30
1
y = 0 + (vy )0 t - gt 2 ( g = 32.2 ft/s2 )
2
Vertical motion. (Uniformly accelerated motion)
At B:
1
3.2 = (v0 sin 30)t B - gt B2
2
Substituting for tB
æ 16 - d ö÷ 1 æ 16 - d ö÷2
3.2 = (v0 sin30)çç
÷ - gç
÷
çè v0 cos30 ÷ø 2 èçç v0 cos30 ø÷
or
v02 =
(a)
d = 9 in. :
æ
9 ö2
2(32.2) çç16 - ÷÷
è
12 ø
v02 =
é1 æ
ù
9 ÷ö
3 ê 3 çç16 - ÷ - 3.2 ú
è
ø
12
ëê
ûú
v0 = 29.8 ft/s ◀
d = 17 in. :
æ
17 ö2
2(32.2) çç16 - ÷÷
è
12 ø
v02 =
é1 æ
ù
17 ö÷
3 ê 3 çç16 - ÷ - 3.2 ú
êë è
úû
12 ø
v0 = 29.6 ft/s ◀
(b)
2 g(16 - d )2
3 éê 13 (16 - d ) - 3.2ùú
ë
û
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2007
PROBLEM 11.108
A tennis player serves the ball at a height h = 2.5 m with an initial velocity of v0 at an angle of 5° with the
horizontal. Determine the range for which of v0 for which the ball will land in the service area which extends
to 6.4 m beyond the net.
5°
v0
h
0.914 m
12.2 m
6.4 m
SOLUTION
The motion is projectile motion. Place the origin of the xy-coordinate system at ground level just below the
point where the racket impacts the ball. The coordinates of this impact point are x 0 = 0, y0 = h = 2.5m. The
components of initial velocity are (vx )0 = v0 cos 5 and (vy )0 = v0 sin5.
Horizontal motion:
x = x0 + ( v x )0 t = ( v0 cos 5)t
Vertical motion:
y = y0 + (vy )0 t =
(1)
1 2
gt
2
1
= 2.5 - (v0 sin 5)t = - (9.81)t 2
2
From (1),
Then
v0 t =
x
cos5
(2)
(3)
y = 2.5 - x tan 5 - 4.905t 2
t2 =
2.5 - x tan 5 - y
4.905
(4)
At the minimum speed the ball just clears the net.
x = 12.2 m,
y = 0.914 m
12.2
= 12.2466 m
cos5
2.5 -12.2 tan 5 - 0.914
t2 =
4.905
12.2466
v0 =
0.32517
v0 t =
t = 0.32517 s
v0 = 37.66 m/s
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2008
PROBLEM 11.108 (CONTINUED)
At the maximum speed the ball lands 6.4 m beyond the net.
x = 12.2 + 6.4 = 18.6 m
18.6
= 18.6710 m
cos 5
2.5 - 18.6 tan 5 - 0
t2 =
4.905
18.6710
v0 =
0.42181
y=0
v0 t =
t = 0.42181 s
v0 = 44.26 m/s
37.7 m/s < v0 < 44.3 m/s ◀
Range for v0.
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2009
PROBLEM 11.109
20 mm
10°
6°
A
B
v0
205 mm
C
The nozzle at A discharges cooling water with an initial
velocity v0 at an angle of 6° with the horizontal onto a grinding
wheel 350 mm in diameter. Determine the range of values of
the initial velocity for which the water will land on the
grinding wheel between Points B and C.
30°
200 mm
SOLUTION
First note
(vx )0 = v0 cos6
(vy )0 = -v0 sin6
Horizontal motion. (Uniform)
x = x 0 + ( v x )0 t
Vertical motion. (Uniformly accelerated motion)
1
y = t0 + (vy )0 t - gt 2 ( g = 9.81 m/s2 )
2
x = (0.175m)sin10
At Point B:
y = (0.175m) cos10
x : 0.175 sin10  = - 0.020 + ( v0 cos 6  ) t
tB =
or
0.050388
v0 cos6
1
y : 0.175cos10 = 0.205 + (-v0 sin 6)t B - gt B2
2
Substituting for t B
æ 0.050388 ö÷ 1
æ 0.050388 ö÷2
çç
-0.032659 = (-v0 sin6)çç
(9.81)
÷
÷
çè v0 cos6 ÷ø 2
çè v0 cos6 ÷ø
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2010
PROBLEM 11.109 (CONTINUED)
v02
or
=
2
1
2 (9.81)(0.050388)
cos2 6(0.032659 - 0.050388tan6
( v0 ) B = 0.678 m/s
or
x = (0.175m)cos30
At Point C:
y = (0.175m)sin 30
x:
0.175 cos 30  = - 0.020 + ( v0 cos 6 ) t
tc =
or
y:
0.171554
v0 cos 6
1
0.175sin 30 = 0.205 + (-v0 sin 6)tc - gtc2
2
Substituting for t c
æ 0.171554 ö÷ 1
æ 0.171554 ö÷2
-0.117500 = (-v0 sin6)çç
÷÷ - (9.81)çç
÷
çè v0 cos6 ø 2
çè v0 cos6 ø÷
or
or
v02 =
2
1
2 (9.81)(0.171554)
cos2 6(0.117500 - 0.171554 tan6)
( v 0 ) c = 1.211 m/s
0.678 m/s < v 0 < 1.211m/s ◀
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2011
PROBLEM 11.110
C
B
0.9 m
While holding one of its ends, a worker lobs a coil of rope over the
lowest limb of a tree. If he throws the rope with an initial velocity v0
at an angle of 65° with the horizontal, determine the range of values
of v0 for which the rope will go over only the lowest limb.
5.7 m
v0
65°
A
0.7 m
5m
SOLUTION
First note
(vx )0 = v0 cos65
(vy )0 = -v0 sin 65
Horizontal motion. (Uniform)
x = 0 + ( v x )0 t
At either B or C, x = 5 m
s = (v0 cos65)t B,C
t B,C =
or
5
(v0 cos65)
Vertical motion. (Uniformly accelerated motion)
1
y = 0 + (vy )0 t - gt 2
2
( g = 9.81 m/s2 )
At the tree limbs, t = t B ,C
æ
ö÷ 1 æ
ö2
5
5
÷
ç
yB,C = (v0 sin65)çç
g
÷
÷
çè v0 cos65 ÷ø 2 ççè v0 cos65 ÷ø
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2012
PROBLEM 11.110 (CONTINUED)
1
(9.81)(25)
2
v02 =
cos2 65(5tan65- yB,C )
or
=
686.566
5tan65 - yB,C
At Point B:
v02 =
686.566
5tan 65 - 5
At Point C:
v02 =
686.566
5tan 65 - 5.9
or
or
(v0 ) B = 10.95 m/s
(v0 )C = 11.93 m/s
10.95 m/s > v0 < 11.93 m/s ◀
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2013
PROBLEM 11.111
The pitcher in a softball game throws a ball with an initial velocity v0 of 108 km/h at an angle a with the
horizontal. If the height of the ball at Point B is 0.68 m, determine (a) the angle a, (b) the angle q that the
velocity of the ball at Point B forms with the horizontal.
SOLUTION
First note
v0 = 108 km/h = 30 m/s
(vx )0 = v0 cos a = (30 m/s)cos a
and
(vy )0 = v0 sin a = (30 m/s)sin a
(a) Horizontal motion. (Uniform)
x = 0 + (vx )0 t = (30 cos a )t
At Point B:
14 = (30 cos a )t
or
tB =
7
15cos a
Vertical motion. (Uniformly accelerated motion)
1
1
y = 0 + (vy )0 t - gt 2 = (30 sin a)t - gt 2
2
2
At Point B:
( g = 9.81 m/s2 )
1
0.08 = (30sin a )tB - gt B2
2
Substituting for t B
æ 7 ÷ö 1 æ 7 ÷ö2
0.08 = (30sin a)ççç
÷- g çç
÷
è15cos a ÷ø 2 çè15cos a ÷ø
or
1
49
0.08 = 14 tan a - g
2 225cos2 a
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2014
PROBLEM 11.111 (CONTINUED)
Solving
a = 4.718 and a = 85.61
a = 4.72 ◀
Rejecting the second root because it is not physically reasonable.
(b) We have
vx = (vx )0 = 30 cos a
vy = (vy )0 - gt = 30sin a - gt
and
(vy ) B = 30 sin a - gt B
At Point B:
= 30 sin a -
7g
15cos a
Noting that at Point B, vy < 0, we have
tan q =
=
=
(vy ) B
vx
7g
15cos a
- 30sin a
30 cos a
- 30sin 4.718
7
9.81
15 cos4.718
30 cos 4.718
q = 4.07 ◀
Or
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2015
PROBLEM 11.112
A model rocket is launched from Point A with an initial velocity v0 of
75 m/s. If the rocket’s descent parachute does not deploy and the
rocket lands a distance d = 100 m from A, determine (a) the angle a
that v0 forms with the vertical, (b) the maximum height above Point A
reached by the rocket, and (c) the duration of the flight.
v0
α
30°
A
100 m
B
SOLUTION
Set the origin at Point A.
x 0 = 0,
y0 = 0
Horizontal motion:
x = v0 t sin a
Vertical motion:
1
y = v0 t cos a - gt 2
2
cos a =
sin2 a + cos2 a =
sin a =
x
v0 t
ö
1 æ
1
çç y + gt 2 ÷÷
è
ø÷
v0 t
2
(2)
2ù
é
ê x2 + æç y + 1 gt 2 ö÷÷ ú = 1
çè
ê
2 ø÷ úû
ë
1
x 2 + y2 + gyt 2 + g2 t 4 = v02 t 2
4
1
(v0 t )2
1 2 4
g t - (v02 - gy)t 2 + ( x 2 + y2 ) = 0
4
At Point B,
(1)
x 2 + y2 = 100 m, x = 100 cos30 m
y = -100sin 30 = -50 m
1
(9.81)2 t 4 - [752 - (9.81)(-50)]t 2 + 1002 = 0
4
24.0590 t 4 - 6115.5 t 2 + 10000 = 0
t 2 = 252.54 s2
and
1.6458s2
t = 15.8916 s
and
1.2829s
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2016
(3)
PROBLEM 11.112 (CONTINUED)
Restrictions on a :
0 < a < 120 
tan a =
x
100 cos30
=
= 0.0729
y + 12 gt 2 -50 + (4.905)(15.8916)2
a = 4.1669
and
100 cos30
= 2.0655
-50 + (4.905)(1.2829)2
a = 115.8331
Use a = 4.1669  corresponding to the steeper possible trajectory.
(a)
Angle a .
(b)
Maximum height.
a = 4.17  ◀
vy = 0
at
y = ymax
vy = v0 cos a - gt = 0
t=
v0 cos a
g
v 2 cos2 a
1
ymax = v0 t cos a - gt = 0
2
2g
(75)2 cos2 4.1669
=
(2)(9.81)
(c)
Duration of the flight.
ymax = 285 m ◀
(time to reach B)
t = 15.89 s ◀
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2017
PROBLEM 11.113
The initial velocity v0 of a hockey puck is 90 mi/h. Determine (a) the largest value (less than 45°) of the angle
a for which the puck will enter the net, (b) the corresponding time required for the puck to reach the net.
SOLUTION
v0 = 90 mi/h = 132 ft/s
First note
(vx )0 = v0 cos a = (132 ft/s)cos a
and
(vy )0 = v0 sin a = (132 ft/s)sin a
(a) Horizontal motion. (Uniform)
x = 0 + (vx )0 t = (132 cos a)t
At the front of the net,
x = 16ft
Then
16 = (132 cos a)t
or
tenter =
8
66 cos a
Vertical motion. (Uniformly accelerated motion)
1
y = 0 + (vy )0 t - gt 2
2
1
= (132 sin a )t - gt 2
2
( g = 32.2 ft/s2 )
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2018
PROBLEM 11.113 (CONTINUED)
At the front of the net,
1 2
yfront = (132sin a)tenter - gtenter
2
æ 8 ö÷ 1 æ 8 ö÷2
32 g
= (132sin a) ççç
÷- g çç
÷ = 16 tan a è 66 cos a ÷ø 2 èç 66 cos a ø÷
4356 cos2 a
1
Now
cos a
2
= sec 2 a = 1 + tan 2 a
yfront = 16 tan a -
Then
tan 2 a -
or
tan a =
Then
32 g
(1 + tan 2 a)
4356
æ
ö
4356
4356
tan a + çç1 +
yfront ÷÷÷ = 0
çè
2g
32 g
ø÷
4356
2g
é
 ê - 4356
2g
ë
(
)
2
ù1/2
y
- 4 1 + 4356
front ú
2g
û
2
(
)
or
1/2
éæ
ö÷2 æ
ö÷ùú
4356
4356
4356
ê
 êçç÷ - çç1 +
tan a =
y ÷ú
4 ´ 32.2 êëçè 4 ´ 32.2 ÷ø çè
32 ´ 32.2 front ÷øúû
or
tan a = 33.82  [(33.82) - (1 + 4.2275 yfront )]1/2
2
Now 0 < yfront < 4 ft so that the positive root will yield values of a > 45 for all values of yfront.
When the negative root is selected, a increases as yfront is increased. Therefore, for a max , set
yfront = yc = 4 ft
tan a = 33.82 - [(33.82) - (1 + 16.910 yfront )]1/2
2
Then
a max = 14.887
or
(b) We had found
a max = 14.89 ◀
8
66 cos a
8
=
66 cos14.887
tenter =
tenter = 0.1254 s ◀
or
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2019
PROBLEM 11.114
A worker uses high-pressure water to clean the inside of a long drainpipe. If the water is discharged with an
initial velocity v0 of 11.5 m/s, determine (a) the distance d to the farthest Point B on the top of the pipe that the
worker can wash from his position at A, (b) the corresponding angle a.
d
B
v0
A
1.1 m
α
C
SOLUTION
First note
(vx )0 = v0 cos a = (11.5 m/s)cos a
(vy )0 = v0 sin a = (11.5 m/s)sin a
By observation, dmax occurs when
ymax = 1.1m.
Vertical motion. (Uniformly accelerated motion)
1
y = 0 + (vy )0 t - gt 2
2
1
= (11.5sin a ) - gt
= (11.5sin a )t - gt 2
2
vy = (vy )0 - gt
y = ymax
When
Then
at
(vy )B = 0
B,
( v y ) B = 0 = (11.5 sin a ) - gt
11.5sin a
( g = 9.81m/s2 )
g
or
tB =
and
1
yB = (11.5sin a )t B - gt B2
2
Substituting for t B and noting y B = 1.1 m
æ11.5sin a ÷ö 1 æ11.5sin a ÷ö2
1.1 = (11.5sin a ) çç
÷÷ - g ççç
÷÷
g
g
èç
ø 2 è
ø
=
or
sin 2 a =
1
(11.5)2 sin 2 a
2g
2.2 ´ 9.81
11.52
a = 23.8265
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2020
PROBLEM 11.114 (CONTINUED)
(a)
Horizontal motion. (Uniform)
x = 0 + (vx )0 t = (11.5cos a )t
At Point B:
where
Then
x = d max
tB =
and
11.5
sin 23.8265 = 0.47356s
9.81
dmax = (11.5)(cos 23.8265)(0.47356)
d max = 4.98 m ◀
or
(b)
t = tB
a = 23.8  ◀
From above
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2021
PROBLEM 11.115
v0
h
α
A
B
1.5 m
d
An oscillating garden sprinkler which
discharges water with an initial velocity
v0 of 8 m/s is used to water a vegetable
garden. Determine the distance d to the
farthest Point B that will be watered and
the corresponding angle a when (a) the
vegetables are just beginning to grow,
(b) the height h of the corn is 1.8 m.
SOLUTION
First note
(vx )0 = v0 cos a = (8 m/s)cos a
(vy )0 = v0 sin a = (8 m/s)sin a
Horizontal motion. (Uniform)
x = 0 + ( v x )0 t = (8cos a )t
At Point B:
x = d : d = (8 cos a )t
tB =
or
d
8cos a
Vertical motion. (Uniformly accelerated motion)
At Point B:
1
y = 0 + (vy )0 t - gt 2
2
1 2
= (8sin a )t - gt
(g = 9.81m/s2 )
2
1
0 = (8sin a )t B - gt B2
2
Simplifying and substituting for t B
or
(a)
1 æ d ÷ö
0 = 8sin a - g çç
÷
2 è 8cos a ÷ø
64
d=
sin 2a
g
When h = 0, the water can follow any physically possible trajectory. It then follows from
(1)
Eq. (1) that d is maximum when 2 a = 90 
a = 45  ◀
or
Then
d=
64
sin(2 ´ 45)
9.81
d max = 6.52 m ◀
or
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2022
PROBLEM 11.115 (CONTINUED)
(b)
Based on Eq. (1) and the results of Part a, it can be concluded that d increases in value as a increases
in value from 0 to 45° and then d decreases as a is further increased. Thus, dmax occurs for the value
of a closest to 45° and for which the water just passes over the first row of corn plants. At this row,
xcorn = 1.5 m
tcorn =
so that
1.5
8cos a
Also, with ycorn = h, we have
1 2
h = (8sin a )tcorn - gtcorn
2
Substituting for t co rn and noting h = 1.8 m,
æ 1.5 ö÷ 1 æ 1.5 ö÷2
1.8 = (8sin a ) çç
÷ - gç
÷
è 8cos a ø÷ 2 çè 8cos a ø÷
or
Now
Then
or
1.8 = 1.5tan a -
2.25g
128cos2 a
1
= sec2 a = 1 + tan 2 a
cos2 a
1.8 = 1.5tan a -
2.25(9.81)
(1 + tan 2 a )
128
0.172441tan 2 a - 1.5 tan a + 1.972441 = 0
a = 58.229  and a = 81.965 
Solving
From the above discussion, it follows that d = dmax when
a = 58.2  ◀
Finally, using Eq. (1)
d=
64
sin(2 ´ 58.229)
9.81
d max = 5.84 m ◀
or
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2023
PROBLEM 11.116*
A nozzle at A discharges water with
an initial velocity of 36 ft/s at an angle
a with the horizontal. Determine
(a) the distance d to the farthest point
B on the roof that the water can reach,
(b) the corresponding angle a Check
that the stream will clear the edge of
the roof.
B
18 ft
α
A
A
3.6 ft
d
13.5 ft
SOLUTION
x = ( v0 cos a )t
Vertical motion:
1
1
gx 2
y = y0 + (v0 sin a)t - gt 2 = y0 + x tan a 2
2 (v0 cos a)2
or
y = y0 + x tan a Let
u = x tan a
x2 =
gx 2
(1 + tan2 a)
2
2v0
y = y0 + u -
so that
Solving for x 2 :
t=
x
v0 cos a
Horizontal motion:
2v02
(u + y0 - y) - u2
g
The maximum value of x2 is required:
d(x 2 )
= 0.
du
d( x 2 ) 2v02
=
- 2u = 0
du
g
Data:
v0 = 36 ft/s,
(xmax )2 =
g
( x 2 + u2 )
2 v02
y0 = 3.6 ft,
or
y B = 18 ft
u=
u=
v02
g
(36)2
= 40.2484 ft
32.2
(2)(36)2
(40.2484 + 3.6 - 18) - (40.2484)2 = 460.78ft 2
32.2
(a) Maximum distance:
d = xmax -13.5
xmax = 21.466 ft
d = 7.97 ft ◀
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2024
PROBLEM 11.116* (CONTINUED)
(b) Angle a.
tan a =
xmax tan a
u
40.2484
=
=
= 1.875
x max
xmax
21.466
a = 61.93 
Check the edge.
y = y0 + x tan a +
a = 61.9  ◀
gx 2
2(v0 cos a)2
y = 3.6 + (13.5)(1.875) -
(32.2)(13.5)2
2[36 cos61.93]2
y = 18.69 ft ◀
Since y > 18 ft, the stream clears the edge.
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2025
PROBLEM 11.117
45 ft/s
B
The velocities of skiers A and B are
as shown. Determine the velocity
of A with respect to B.
30 ft/s
25°
A
10°
SOLUTION
We have
v A = v B + v A/B
The graphical representation of this equation is then as shown.
Then
v 2A / B = 30 2 + 452 - 2(30)(45) cos15
or
v A / B = 17.80450 ft/s
30
17.80450
=
sin a
sin15
and
a = 25.8554 
or
a + 25  = 50.8554 
v A / B = 17.8 ft/s
50.9° ◀
Alternative solution.
vA/ B = vA + vB
= 30 cos 10 i - 30 sin 10 j - (45 cos 25i - 45 sin 25 j)
= 11.2396 i + 13.8084 j
= 5.05 m/s = 17.8 ft/s
50.9°
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2026
PROBLEM 11.118
D
A
B
The three blocks shown move with constant velocities. Find the velocity of each
block, knowing that the relative velocity of A with respect to C is 300 mm/s upward
and that the relative velocity of B with respect to A is 200 mm/s downward.
C
SOLUTION
From the diagram
Cable 1:
y A + yD = constant
Then
v A + vD = 0
Cable 2:
(1)
( yB + yD ) + ( yC + yD ) = constant
vB + vC - 2 vD = 0
Then
(2)
Combining Eqs. (1) and (2) to eliminate v D ,
2 v A + vB + vC = 0
(3)
Now
v A /C = v A - vC = - 300 mm/s
(4)
and
v B /A = v B - v A = 200 mm/s
(5)
(3) + (4) - (5) 
Then
(2 v A + vB + vC ) + (v A - vC ) - (vB - v A ) = (-300) - (200)
v A = 125 mm/s ◀
or
vB - (-125) = 200
and using Eq. (5)
v B = 75 mm/s ◀
or
-125 - vC = -300
Eq. (4)
v C = 175 mm/s ◀
or
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2027
PROBLEM 11.119
N
Three seconds after automobile B passes through the intersection shown,
automobile A passes through the same intersection. Knowing that the speed
of each automobile is constant, determine (a) the relative velocity of B with
respect to A, (b) the change in position of B with respect to A during a 4-s
interval, (c) the distance between the two automobiles 2 s after A has
passed through the intersection.
70°
S
A
45 mi/h
B
30 mi/h
SOLUTION
vA = 45 mi/h = 66 ft/s
vB = 30 mi/h = 44 ft/s
Law of cosines
vB2 /A = 662 + 44 2 - 2(66)(44)cos110
vB/A = 90.99 ft/s
v B = v A + v B/A
Law of sines
sin b sin110
=
66
90.99
b = 42.97
a = 90 - b = 90 - 42.97 = 47.03
(a)
Relative velocity:
(b)
Change in position for Dt = 4 s.
D rB /A = rB /A D t = (91.0 ft/s)(4 s)
(c)
v B /A = 91.0 ft/s
47.0° ◀
rB /A = 364 ft
47.0° ◀
Distance between autos 2 seconds after auto A has passed intersection.
Auto A travels for 2 s.
v A = 66 ft/s
20°
rA = v A t = (66 ft/s)(2 s)=132 ft/s
rA = 132 ft
20°
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2028
PROBLEM 11.119 (CONTINUED)
Auto B
v B = 44 ft/s
rB = v B t = (44 ft/s)(5 s) = 200 ft
rB = rA + rB/A
Law of cosines
rB2/A = (132)2 + (220)2 - 2(132)(220) cos110
rB/A = 292.7 ft
Distance between autos = 293 ft ◀
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2029
PROBLEM 11.120
Shore-based radar indicates that a ferry leaves its slip with a
70°, while instruments aboard the
velocity v = 18 km/h
ferry indicate a speed of 18.4 km/h and a heading of 30° west
of south relative to the river. Determine the velocity of the
river.
SOLUTION
We have
v F = v R + v F /R
or
v F = v F /R + v R
The graphical representation of the second equation is then as shown.
We have
v R2 = 182 + 18.4 2 - 2(18)(18.4) cos10
or
v R = 3.1974 km/h
and
or
18
3.1974
=
sin a sin10
a = 77.84 
Noting that
v R = 3.20 km/h
17.8° ◀
Alternatively one could use vector algebra.
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2030
PROBLEM 11.121
C
Airplanes A and B are flying at the same altitude and are
tracking the eye of hurricane C. The relative velocity of C
with respect to A is v C A = 350 km/h
75°, and the
relative velocity of C with respect to B is v C B = 400 km/h
40°. Determine (a) the relative velocity of B with respect
to A, (b) the velocity of A if ground-based radar indicates that
the hurricane is moving at a speed of 30 km/h due north, (c)
the change in position of C with respect to B during a 15-min
interval.
N
A
B
SOLUTION
(a)
We have
v C = v A + v C/A
and
v C = v B + v C/B
Then
v A + v C /A = v B + v C /B
or
v B - v A = v C/A - v C/B
Now
v B - v A = v B/A
so that
v B / A = vC/A - vC/B
or
v C /A = v C /B + v B/ A
The graphical representation of the last equation is then as shown.
We have
v B2 /A = 350 2 + 400 2 - 2(350)(400) cos 65
or
v B /A = 405.175 km/h
and
400
405.175
=
sin a
sin 65
or
a = 63.474 
75  - a = 11.526 
v B / A = 405 km/h
11.53° ◀
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2031
PROBLEM 11.121 (CONTINUED)
(b)
We have
v C = v A + v C /A
or
v A = (30 km/h) j - (350 km/h)(- cos 75  i - sin 75  j)
v A = (90.587 km/h) i + (368.07 km/h) j
v A = 379 km/h
or
(c)
76.17° ◀
Noting that the velocities of B and C are constant, we have
rB = ( rB ) 0 + v B t
Now
rC = ( rC ) 0 + v C t
rC/B = rC - rB = [(rC )0 - (rB )0 ] + (vC - v B )t
= [(rC )0 - (rB )0 ] + vC / B t
Then
DrC/B = (rC/B )t2 - (rC/B )t1 = v C/B (t2 - t1 ) = vC/B Dt
For D t = 15 min :
æ1 ö
DrC/B = (400 km/h) çç h÷÷÷ = 100 km
è4 ø
D rC /B = 100 km
40° ◀
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2032
PROBLEM 11.122
Instruments in an airplane indicate that with respect to the air, the plane is moving north at a speed of 500
km/h. At the same time ground-based radar indicates that the plane is moving at a speed of 530 km/h in a
direction 5 east of north. Determine the magnitude and direction of the velocity of the air.
SOLUTION
Starting with
v Plane = v Air + v Plane / Air
or
v Air = v Plane - v Plane / Air
v Air = 530(sin(5)i + cos(5) j) - 500 j
v Air = 46.19i + 527.9 j - 500 j
v Air = 46.19i + 27.9 j
v Air = 5431.2 km/h
v Air = 54
31.2 km/h ◀
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2033
Problem 11.123
aB = 3 ft/s2
vB = 2 ft/s
B
Knowing that at the instant shown
block B has a velocity of 2 ft/s to the
2
right and an acceleration of 3 ft/s to
the left, determine (a) the velocity of
block A, (b) the acceleration of
block A.
A
SOLUTION
Given:
v B = 2 ft/s, a B = - 3 ft/s
Length of Cable
L = y A + 2 x B + constants
Differentiate Length Equation
0 = v A + 2 vB
(1)
Differentiate again
0 = a A + 2 aB
(2)
(e) Rearrange (1)
vA = -2vB
= -2(2) ft/s
v A = - 4 ft/s
(f) Rearrange (2)
v A = 4 ft/s  ◀
a A = -2 aB
= - 2(- 3) ft/s2
aA = 6 ft/s2
a A = 6 ft/s2  ◀
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2034
PROBLEM 11.124
Knowing that at the instant shown block A has a velocity of
2
8 in./s and an acceleration of 6 in./s both directed down the
incline, determine (a) the velocity of block B, (b) the
acceleration of block B.
B
15°
25°
A
SOLUTION
From the diagram
2 x A + x B / A = constant
Then
2 v A + vB / A = 0
v B / A = 16 in./s
or
2 a A + aB / A = 0
and
aB / A = 12 in./s2
or
Note that v B / A and a B / A must be parallel to the top surface of block A.
(a) We have
vB = vA + vB / A
The graphical representation of this equation is then as shown. Note that because A is moving
downward, B must be moving upward relative to A.
We have
v B2 = 82 + 16 2 - 2(8)(16) cos15
or
v B = 8.5278 in./s
and
or
8
8.5278
=
sin a sin15
a = 14.05 
v B = 8.53 in./s
(b)
54.1° ◀
The same technique that was used to determine v B can be used to determine a B . An alternative method
is as follows.
aB = a A + aB / A
We have
= (6i) + 12(- cos15i + sin15 j) *
= -(5.5911 in./s2 )i + (3.1058 in./s2 ) j
a B = 6.40 in./s2
or
54.1° ◀
* Note the orientation of the coordinate axes on the sketch of the system.
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2035
PROBLEM 11.125
A boat is moving to the right with a constant
2
deceleration of 0.3 m/s when a boy standing on the
deck D throws a ball with an initial velocity relative to
the deck which is vertical. The ball rises to a maximum
height of 8 m above the release point and the boy must
step forward a distance d to catch it at the same height
as the release point. Determine (a) the distance d,
(b) the relative velocity of the ball with respect to the
deck when the ball is caught.
8m
d
D
vD
aD = 0.3 m/s2
SOLUTION
Horizontal motion of the ball:
v x = ( v x ) 0 , x ball = ( v x ) 0 t
Vertical motion of the ball:
v y = ( v y )0 - gt
1
yB = (vy )0 t - gt 2 , (vy )2 - (vy )20 = -2 gy
2
(vy ) = 0
At maximum height,
and
y = ymax
( v y )2 = 2 gymax = (2)(9.81)(8) = 156.96 m 2 /s2
( v y )0 = 12.528 m/s
At time of catch,
1
y = 0 = 12.528 - (9.81)t 2
2
t catch = 2.554 s
or
Motion of the deck:
v x = ( v x )0 + a D t ,
and
v y = 12.528 m/s
1
xdeck = (vx )0 t + aD t 2
2
Motion of the ball relative to the deck:
( vB / D ) x = ( v x )0 - [( v x )0 + a D t ] = -a D t
é
ù
1
1
x B / D = ( v x )0 t - ê ( v x )0 t + a D t 2 ú = - a D t 2
êë
úû
2
2
( vB / D ) y = ( v y )0 - gt ,
yB / D = yB
(a)
(b)
At time of catch,
1
d = x D / B = - (-0.3)(2.554)2
2
( v B / D ) x = -(- 0.3)(2.554) = + 0.766 m/s
( v B / D ) y = 12.528 m/s
d = 0.979 m ◀
or 0.766 m/s
v B / D = 12.55 m/s
86.5 ◀
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2036
PROBLEM 11.126
C
The assembly of rod A and wedge B starts from rest and moves
2
to the right with a constant acceleration of 2 mm/s . Determine
(a) the acceleration of wedge C, (b) the velocity of wedge C
when t = 10 s.
20°
A
75°
B
SOLUTION
(a)
We have
aC = a B + a C / B
The graphical representation of this equation is then as shown.
First note
a = 180 - (20 + 105)
= 55
Then
(b)
aC
2
=
sin 20 sin 55
aC = 0.83506 mm/s2
aC = 0.835 mm/s2
75° ◀
v C = 8.35 mm/s
75° ◀
For uniformly accelerated motion
vC = 0 + aC t
At t = 10 s :
vC = (0.83506 mm/s2 )(10 s)
= 8.3506 mm/s
or
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2037
PROBLEM 11.127
Coal discharged from a dump truck with an initial velocity (vC )0 =
50 falls onto conveyor belt B. Determine the required
5 ft/s
velocity v B of the belt if the relative velocity with which the coal
hits the belt is to be (a) vertical, (b) as small as possible.
SOLUTION
First determine the velocity vC of the coal at the point where the coal impacts on the belt.
(vC )x = éëê(vC )x ùûú 0 = -5cos 50
Horizontal motion:
= -3.2139 m/s
(vC )y = éêë(vC )y ùúû
2
Vertical motion:
2
0
- 2 g ( y - y0 )
= (5sin 50) - (2)(32.2)(-4)
2
= 272.27 ft 2 /s2
(vC )y = -16.501 ft/s
tan b =
vC =
vcy
vcx
=
-16.5006
= 5.134,
-3.2139
(vC )x + (vC )y
2
2
b = 79.98
= 16.81 ft/s
vC = 16.81 ft/s
79.98
vC = (-3.214 ft/s) i + (-16.501 ft/s) j
or
Velocity of the belt:
v B = vB (- cos10i + sin10 j)
Relative velocity:
vC/B = vC - v B = vC + (-v B )
(v )
(a) vC/B is vertical.
C /B
(v )
C /B
y
=0
x
j = (-3.214 i - 16.501j) - (-.9848vB i + .1736 vB j)
i : 0 = -3.214 + 0.9848vB
vB = 3.264
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2038
PROBLEM 11.127 (CONTINUED)
vB = 3.26 ft/s
10 ◀
vB = 0.300 ft/s
10 ◀
(b) for minimum vC/B
vC/B = vC - v B
(v ) i + (v )
C /B
x
C /B
y
j = (-3.214 i - 16.501j) - (-0.9848vB i + 0.1736vB j)
i : vC / Bx = -3.214 + 0.9848vB
j : vC / By = -16.501 - 0.1736 vB
(vC / B )2 = (vC / B ) x 2 + (vC / B ) y 2 = 282.6 - 0.6 vB + vB 2
To find the minimum
d (vC / B )2
= -0.6 + 2vB = 0
dvB
vB = 0.300
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2039
PROBLEM 11.128
vA
B
A
(vB)0
Conveyor belt A, which forms a 20° angle
with the horizontal, moves at a constant speed
of 4 ft/s and is used to load an airplane.
Knowing that a worker tosses duffel bag B
with an initial velocity of 2.5 ft/s at an angle
of 30° with the horizontal, determine the
velocity of the bag relative to the belt as it
lands on the belt.
20°
30°
1.5 ft
SOLUTION
First determine the velocity of the bag as it lands on the belt. Now
[(vB ) x ]0 = (vB )0 cos30
= (2.5 ft/s)cos 30
[(vB ) y ]0 = (vB )0 sin 30
= (2.5 ft/s)sin 30
Horizontal motion. (Uniform)
x = 0 + [(vB ) x ]0 t
(vB ) x = [(vB ) x ]0
= (2.5cos30)t
= 2.5 cos 30 
Vertical motion. (Uniformly accelerated motion)
1
y = 0 + [(vB ) y ]0 t - gt 2
2
( v B ) y = [( v B ) y ]0 - gt
1
= 1.5 + (2.5sin 30)t - gt 2
2
= 2.5sin 30 - gt
The equation of the line collinear with the top surface of the belt is
y = x tan 20
Thus, when the bag reaches the belt
1
= 1.5 + (2.5sin 30)t - gt 2 = [(2.5cos30)t ]tan 20
2
1
(32.2)t 2 + 2.5(cos30 tan 20 - sin 30)t - 1.5 = 0
2
or
16.1t 2 - 0.46198t - 1.5 = 0
or
Solving
t = 0.31992 s
and
t = - 0.29122 s
(Reject)
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2040
PROBLEM 11.128 (CONTINUED)
The velocity v B of the bag as it lands on the belt is then
v B = (2.5cos30)i + [2.5sin 30 - 32.2(0.31992)]j
= (2.1651 ft/s)i - (9.0514 ft/s) j
Finally
or
vB = vA + vB / A
v B / A = (2.1651i - 9.0514 j) - 4(cos20i + sin 20 j)
= (1.59367 ft/s)i - (10.4195 ft/s) j
v B / A = 10.54 ft/s
or
81.3° ◀
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2041
PROBLEM 11.129
During a rainstorm the paths of the raindrops appear to form an angle of 30° with the vertical and to be directed
to the left when observed from a side window of a train moving at a speed of 15 km/h. A short time later, after
the speed of the train has increased to 24 km/h, the angle between the vertical and the paths of the drops
appears to be 45°. If the train were stopped, at what angle and with what velocity would the drops be observed
to fall?
SOLUTION
vrain = vtrain + vrain/train
Case ①:
vT = 15 km/h
;
vR / T
30°
Case ②:
vT = 24 km/h
;
vR / T
45°
Case ①:
( vR ) y tan 30 = 15 - (vR ) x
(1)
Case ②:
( vR ) y tan 45 = 24 - ( vR ) x
(2)
Substract (1) from (2)
Eq. (2):
(vR ) y (tan 45 - tan 30) = 9
(vR ) y = 21.294 km/h
21.294 tan 45 = 25 - (vR ) x
( v R ) x = 3.706 km/h
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written consent of McGraw-Hill Education.
2042
PROBLEM 11.129 (CONTINUED)
3.706
21.294
b = 7.24
tan b =
vR =
21.294
= 21.47 km/h = 5.96 m/s
cos 7.24
v R = 5.96 m/s
82.8° ◀
Alternate solution
Alternate, vector equation
v R = vT + v R / T
For first case,
v R = 15i + vR / T -1 (- sin 30i - cos30 j)
For second case,
v R = 24 i + vR / T -2 (- sin 45i - cos 45 j)
Set equal
15i + vR / T -1 (- sin 30i - cos30 j) = 24 i + vR / T -2 (- sin 45i - cos 45 j)
Separate into components:
i:
15 - vR / T -1 sin 30 = 24 - vR / T -2 sin 45
-vR / T -1 sin 30 + vR / T -2 sin 45 = 9
j:
(3)
-vR / T -1 cos30 = -vR / T -2 cos 45
-vR / T -1 cos30 + vR / T -2 cos 45 = 0
(4)
Solving Eqs. (3) and (4) simultaneously,
v R / T -1 = 24.5885 km/h
v R / T - 2 = 30.1146 km/h
Substitute v R / T - 2 back into equation for v R .
v R = 24 i + 30.1146(- sin 45i - cos 45 j)
v R = 2.71i - 21.29 j
æ -21.29 ÷ö
= -82.7585
q = tan-1 çç
è 2.71 ÷÷ø
v R = 21.4654 km/hr = 5.96 m/s
v R = 5.96 m/s
82.8° ◀
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2043
PROBLEM 11.130
N
A
30°
Instruments in airplane A indicate that with respect to the air the plane is headed 30
north of east with an airspeed of 300 mi/h. At the same time radar on ship B indicates
that the relative velocity of the plane with respect to the ship is 280 mi/h in the
direction 33 north of east. Knowing that the ship is steaming due south at 12 mi/h,
determine (a) the velocity of the airplane, (b) the wind speed and direction.
12 mi/h
B
SOLUTION
Let the x-axis be directed east, and the y-axis be directed north.
Airspeed:
v A /W = 300 mi/h
30  = 300(cos 30  i + sin 30  j) mi/h
v A /B = 280(cos 33  i + sin 33  j) mi/h
Plane relative to ship:
v B = 12 mi/h
Ship:
= -12 j
(a) Velocity of airplane.
v A = v B + v A/B = -12 j + 280(cos33i + sin 33 j)
= 234.83i + 140.50 j
v A = 274 mi/h
30.9 ◀
v W = 26.7 mi/h
20.8 ◀
(b) Wind velocity.
v W = v A + v W /A = v A - v A /W
= 234.83i + 140.50 j - 300(cos30i + sin 30 j)
= -24.98i - 9.50 j
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2044
PROBLEM 11.131
When a small boat travels north at 15 km/h, a flag mounted on
its stern forms an angle q = 50  with the centerline of the boat
as shown. A short time later, when the boat travels east at 25
km/h, angle q is again 50°. Determine the speed and the
direction of the wind.
SOLUTION
vw = vB + vB/ W
For both cases we have
Case 1:
vwx i + vwy j = 15 j + (vw / B )1 sin(50)i - (vw / B )1 cos(50) j
Separate i and j components to get
i : vwx = 0 + (vw / B )1 sin(50)
j : vwy = 15 - (vw / B )1 cos(50)
Case 2:
vwx i + vwy j = 25i - (vw / B )2 cos(50)i - (vw / B )2 sin(50) j
Separate i and j components to get
i : vwx = 25 - (vw / B )2 cos(50)
j : vwy = 0 - (vw / B )2 sin(50)
Solving the 4 scalar equations to get,
(vw / B )1 = 28.79 km/h
(vw / B )2 = 4.579 km/h
vwx = 22.057 km/h
vwy = -3.5078 km/h
vw = 22.33 km/h
9.04  ◀
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2045
PROBLEM 11.132
30°
C
As part of a department store
display, a model train D runs on a
slight incline between the store’s
up and down escalators. When the
train and shoppers pass Point A,
the train appears to a shopper on
the up escalator B to move
downward at an angle of 22° with
the horizontal, and to a shopper on
the down escalator C to move
upward at an angle of 23° with the
horizontal and to travel to the left.
Knowing that the speed of the
escalators is 3 ft/s, determine the
speed and the direction of the train.
vC
D
A
vB
30°
B
SOLUTION
We have
vD = vB + vD / B
v D = vC + v D / C
The graphical representations of these equations are then as shown.
Then
vD
3
=
sin8 sin(22 +a)
Equating the expressions for
vD
3
=
sin 7 sin(23-a)
vD
3
sin8
sin 7
=
sin(22 +a) sin(23-a)
sin8(sin 23 cos a - cos23 sin a)
or
or
or
Then
= sin 7(sin 22 cos a + cos 22 sin a )
tan a =
sin8 sin 23 - sin 7 sin 22
sin8 cos 23 + sin 7 cos22
a = 2.0728 
vD =
3sin8
= 1.024 ft/s
sin(22 + 2.0728)
v D = 1.024 ft/s
2.07° ◀
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2046
PROBLEM 11.132 (CONTINUED)
Alternate solution using components.
v B = (3 ft/s)
30  = (2.5981 ft/s) i + (1.5 ft/s) j
v C = (3 ft/s)
30  = (2.5981 ft/s) i - (1.5 ft/s) j
v D / B = u1
22 = -(u1 cos 22)i - (u1 sin 22) j
v D / C = u2
23 = -(u2 cos 23)i + (u2 sin 23) j
v D = vD
a = -(vD cos a )i + (vD sin a ) j
v D = v B + v D /B = v C + v D / C
2.5981i + 1.5 j - (u1 cos 22)i - (u1 sin 22) j = 2.5981i - 1.5 j - (u2 cos 23)i + (u2 sin 23) j
Separate into components, transpose, and change signs.
u1 cos22 - u2 cos23 = 0
u1 sin 22 + u1 sin 23 = 3
Solving for u1 and u 2 ,
u1 = 3.9054 ft/s
u2 = 3.9337 ft/s
v D = 2.5981i + 1.5 j - (3.9054 cos22)i - (3.9054 sin 22)j
= -(1.0029 ft/s)i + (0.0370 ft/s) j
or
v D = 2.5981i + 1.5 j - (3.9337cos23)i - (3.9337sin 23)j
= -(1.0029 ft/s)i + (0.0370 ft/s) j
v D = 1.024 ft/s
2.07° ◀
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2047
PROBLEM 11.133
Determine the normal component of acceleration of a car
traveling at 72 km/h on an exit ramp of radius r = 150 m.
SOLUTION
an =
an =
v2
r
2
æ
æ 1 hr ö÷ 1000 m ö÷
çç
çç
÷
÷
÷
÷
çç75 km/h ç
ççè 3600 s ÷÷ø km ÷÷÷
çè
ø
150 m
= 2.89 m/s2
an = 2.89 m/s2 ◀
◀
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2048
PROBLEM 11.134
ρ
A
B
Determine the maximum speed that the cars of the roller-coaster
can reach along the circular portion AB of the track if r is 25 m
and the normal component of their acceleration cannot exceed 3 g.
SOLUTION
We have
an =
v2
r
Then
(vmax )2AB = (3´ 9.81 m/s2 )(25 m)
or
( v m ax ) AB = 27.124 m/s
( vmax ) AB = 97.6 km/h ◀
or
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2049
PROBLEM 11.135
Human centrifuges are often used to simulate different
acceleration levels for pilots and astronauts. Space shuttle
pilots typically face inwards towards the center of the gondola
in order to experience a simulated 5g forward acceleration.
Knowing that the astronaut sits 5 m from the axis of rotation
and experiences 5 g’s inward, determine her velocity.
SOLUTION
Given:
an = 5g, r = 5 m
an =
Rearrange
v2
r
v = ran
v = r 5g
Substitute in known values
v = 5 * 5 * 9.81
vA = 15.66 m/s ◀
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2050
PROBLEM 11.136
r
The diameter of the eye of a stationary hurricane is 20 mi and the
maximum wind speed is 100 mi/h at the eye wall, r = 10 mi.
Assuming that the wind speed is constant for constant r and
decreases uniformly with increasing r to 40 mi/h at r = 110 mi,
determine the magnitude of the acceleration of the air at
(a) r = 10 mi, (b) r = 60 mi, (c) r = 110 mi.
20 mi
220 mi
SOLUTION
(a)
r = 10 mi = 52800 ft,
a=
(b)
r = 60 mi = 316800 ft,
a=
(c)
v 2 (146.67)2
=
52800
r
v = 100 -
a = 0.407 ft/s2 ◀
40 - 100
(60 -10) = 70 mi/h = 102.67 ft/s
110 - 10
v 2 (102.67)2
=
316800
r
r = 110 mi = 580800 ft,
a=
v = 100 mi/h = 146.67 ft/s
a = 0.0333 ft/s2 ◀
v = 40 mi/h = 58.67 ft/s
v2
(58.67)2
=
580800
r
a = 0.00593 ft/s2 ◀
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2051
PROBLEM 11.137
The peripheral speed of the tooth of a 10-in.-diameter circular saw blade is 150 ft/s when the power to the saw
is turned off. The speed of the tooth decreases at a constant rate, and the blade comes to rest in 9 s. Determine
the time at which the total acceleration of the tooth is 130 ft/s2 .
SOLUTION
v = v0 + a t t
For uniformly decelerated motion:
At t = 9 s,
0 = 150 - at (9),
or
at = -16.667 ft/s2
a 2 = at2 + an2
Total acceleration:
1/ 2
2
2 1/ 2
a n = éë a 2 - at2 ùû = éë(130 ) - (- 16.667 ) ùû = 128.93 ft/s 2
Normal acceleration:
an =
v2
,
r
where
1
5
r = diameter = ft
2
12
æ5ö
v 2 = r an = çç ÷÷÷ (128.93) = 53.72 ft 2 /s2 ,
è12 ø
Time:
t=
v - v0 7.329 - 150
=
at
-16.667
v = 7.329 ft/s
t = 8.56 s ◀
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2052
PROBLEM 11.138
P
A robot arm moves so that P travels in a circle about Point B, which
is not moving. Knowing that P starts from rest, and its speed
2
increases at a constant rate of 10 mm/s , determine (a) the magnitude
of the acceleration when t = 4 s, (b) the time for the magnitude of
2
the acceleration to be 80 mm/s .
0.8 m
B
O
SOLUTION
Tangential acceleration:
Speed:
at = 10 mm/s2
v = at t
v2 at2 t 2
=
r
r
Normal acceleration:
an =
where
r = 0.8 m = 800 mm
(a)
When t = 4 s
v = (10)(4) = 40 mm/s
an =
Acceleration:
(40)2
= 2 mm/s2
800
a = at2 + an2 = (10)2 + (2)2
a = 10.20 mm/s2 ◀
(b)
Time when a = 80 mm/s2
a 2 = an2 + at2
é (10)2 t 2 ù 2
ú + 10 2
(80) = ê
êë 800 úû
2
t 4 = 403200 s 4
t = 25.2 s ◀
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2053
PROBLEM 11.139
A monorail train starts from rest on a curve of radius 400 m and accelerates at the constant rate a t . If the
maximum total acceleration of the train must not exceed 1.5 m/s2 , determine (a) the shortest distance in which
the train can reach a speed of 72 km/h, (b) the corresponding constant rate of acceleration a t .
SOLUTION
When v = 72 km/h = 20 m/s and r = 400 m,
an =
v2 (20)2
=
= 1.000 m/s2
400
r
a = an2 + at2
But
at = a 2 - an2 = (1.5)2 - (1.000)2 = 1.11803 m/s2
Since the train is accelerating, reject the negative value.
(a)
Distance to reach the speed.
v0 = 0
Let
x0 = 0
v12 = v02 + 2 at ( x1 - x0 ) = 2 at x1
x1 =
(b)
v12
(20)2
=
2at (2)(1.11803)
x1 = 178.9 m ◀
Corresponding tangential acceleration.
at = 1.118 m/s2 ◀
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2054
PROBLEM 11.140
B
C
A motorist starts from rest at Point A on a circular entrance ramp
when t = 0, increases the speed of her automobile at a constant
rate and enters the highway at Point B. Knowing that her speed
continues to increase at the same rate until it reaches 100 km/h at
Point C, determine (a) the speed at Point B, (b) the magnitude of
the total acceleration when t = 20 s.
100 m
150 m
A
SOLUTION
v0 = 0
Speeds:
v1 = 100 km/h = 27.78 m/s
p
s = (150) +100 = 335.6 m
2
Distance:
v12 = v02 + 2 at s
Tangential component of acceleration:
at =
At Point B,
v12 - v02 (27.78)2 - 0
=
= 1.1495 m/s2
2s
(2)(335.6)
vB2 = v02 + 2 at s B
where
sB =
p
(150) = 235.6 m
2
vB2 = 0 + (2)(1.1495)(235.6) = 541.69 m2 /s2
v B = 23.27 m/s
(a)
At t = 20 s,
v B = 83.8 km/h ◀
v = v0 + a t t = 0 + (1.1495)(20) = 22.99 m/s
Since v < v B , the car is still on the curve.
(b)
r = 150 m
v2 (22.99)2
=
= 3.524 m/s2
150
r
Normal component of acceleration:
an =
Magnitude of total acceleration:
a = at2 - an2 = (1.1495)2 + (3.524)2
a = 3.71 m/s2 ◀
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2055
PROBLEM 11.141
A
240 km/h
Racecar A is traveling on a straight portion of the track while
racecar B is traveling on a circular portion of the track. At the
instant shown, the speed of A is increasing at the rate of
2
2
10 m/s , and the speed of B is decreasing at the rate of 6 m/s .
For the position shown, determine (a) the velocity of B relative
to A, (b) the acceleration of B relative to A.
B
50°
300 m
200 km/h
SOLUTION
vA = 240 km/h = 66.67 m/s
Speeds:
vB = 200 km/h = 55.67 m/s
v A = 66.67 m/s
Velocities:
(a)
and v B = 55.56 m/s
50°
vB/ A = vB - vA
Relative velocity:
v B / A = (55.56 cos 50) ¬ +55.56sin 50  +66.67
= 30.96  + 42.56
= 52.63 m/s
53.96°
v B / A = 189.5 km/h
Tangential accelerations:
(a A )t = 10 m/s2
(a B )t = 6 m/s2
an =
Normal accelerations:
(r = ¥)
Car B:
( r = 300 m )
Acceleration of B relative to A:
50°
v2
r
Car A:
(a B ) n =
(b)
54.0° ◀
(a A ) n = 0
(55.56)2
= 10.288
300
(a B )n = 10.288 m/s2
40°
a B /A = a B - a A
a B/A = (a B )t + (a B )n - (a A )t - (a A )n
=6
50  + 10.228
40  + 10  + 0
= (6 cos 50 + 10.288cos 40 + 10)
+(6 sin 50 - 10.288sin 40)
= 21.738  + 2.017
a B/A = 21.8 m/s2
5.3° ◀
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2056
PROBLEM 11.142
400 m
A
450 km/h
B
300 m
30°
540 km/h
At a given instant in an airplane race, airplane A is flying
horizontally in a straight line, and its speed is being increased
at the rate of 8 m/s2. Airplane B is flying at the same altitude
as airplane A and, as it rounds a pylon, is following a circular
path of 300-m radius. Knowing that at the given instant the
speed of B is being decreased at the rate of 3 m/s2, determine,
for the positions shown, (a) the velocity of B relative to A,
(b) the acceleration of B relative to A.
SOLUTION
First note
v A = 450 km/h
(a)
v B = v A + v B /A
We have
v B = 540 km/h = 150 m/s
The graphical representation of this equation is then as shown.
We have
v B2 /A = 450 2 + 540 2 - 2(450)(540) cos 60
vB/A = 501.10 km/h
and
540
501.10
=
sin a sin 60
a = 68.9 
v B / A = 501 km/h
(b)
First note
Now
a A = 8 m/s2
(aB )n =
(a B )t = 3 m/s2
68.9 ◀
60
vB2 (150 m/s)2
=
300 m
rB
(aB )n = 75 m/s2
30
a B = ( a B )t + ( a B ) n
Then
= 3(- cos60i + sin 60 j) + 75(- cos30i - sin 30 j)
= -(66.452 m/s2 )i - (34.902 m/s2 ) j
Finally
aB = a A + aB / A
a B / A = -(66.452 i - 34.902 j) - (8i )
= -(74.452 m/s2 )i - (34.902 m/s2 ) j
a B / A = 82.2 m/s2
25.1 ◀
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2057
PROBLEM 11.143
120 km/h
y
θ
P
x
A race car enters the circular portion of a track that has a radius
of 70 m. When the car enters the curve at point P, it is travelling
2
with a speed of 120 km/h that is increasing at 5 m/s . Three
seconds later, determine (a) the total acceleration of the car in xy
components, (b) the linear velocity of the car in xy components.
r = 70 m
SOLUTION
Speeds:
vt0 = 120 km/h = 33.33 m/s
vt = vt0 + at t
vt = 33.33 + 5(3) = 48.33 m/s
Distance:
Location:
1 2
at t
2
1
s = 33.33(3) + (5)(3)2 = 122.5 m
2
s = Rq
s = s0 + vt0 t +
q=
s 122.5
=
= 1.75 rad
R
70
q = 100.3 
Tangential Acceleration:
at = 5 m/s2
Normal Acceleration:
an =
v2
r
48.332
= 33.37 m/s2
70
a x = -5cos(10.3) + 33.37sin(10.3)
an =
(a)
Accelerations:
ax = 1.047 m/s2 ◀
a y = - 5sin(10.3) + 33.37 cos(10.3)
ay = -33.726 m/s2 ◀
(b)
Velocities:
v x = -48.33cos(10.3)
v x = - 47.55 m/s ◀
v y = -48.33sin(10.3)
v y = - 8.64 m/s ◀
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2058
PROBLEM 11.144
Pin A, which is attached to link AB, is constrained to move in the
circular slot CD. Knowing that at t = 0 the pin starts from rest and
moves so that its speed increases at a constant rate of 0.8 in./s2 ,
determine the magnitude of its total acceleration when (a) t = 0, (b)
t = 2 s.
SOLUTION
Given:
(vA )0 = 0,
(aA )t =
dvA
= 0.8 s
dt
Then,
vA = (vA )0 + (a A )t t = 0.8 t
(a) t = 0,
v A = 0,
(aA )n =
v 2A
=0
r
a A = (aA )t
(b) t = 2 s,
a A = 0.800 in./s2 ◀
vA = 0 + (0.8)(2) = 1.6 in./s
v 2 (1.6)
= A=
= 0.731 in./s2
r
3.5
2
(aA )n
1/2
1/2
2
2ù
2
2ù
é
é
a A = ê(a A )t + (a A )n ú = ê(0.8) + (0.731) ú
ë
û
ë
û
a A = 1.084 in./s2 ◀
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2059
vA
A
25°
PROBLEM 11.145
A golfer hits a golf ball from Point A with an initial velocity of 50 m/s at an angle of
25° with the horizontal. Determine the radius of curvature of the trajectory described
by the ball (a) at Point A, (b) at the highest point of the trajectory.
SOLUTION
(a)
We have
v2
(aA )n = rA
A
or
rA =
(50 m/s)2
(9.81 m/s2 )cos25
r A = 281 m ◀
or
(b)
We have
v2
(aB )n = rB
B
where Point B is the highest point of the trajectory, so that
vB = (v A ) x = v A cos 25
Then
rB =
[(50 m/s)cos25]2
9.81 m/s2
r B = 209 m ◀
or
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2060
PROBLEM 11.146
A nozzle discharges a stream of water in the direction shown with an
initial velocity of 24 m/s. Determine the radius of curvature of the
stream (a) as it leaves the nozzle, (b) at the maximum height of the
stream.
SOLUTION
(a) As water leaves nozzle.
v = 24 ft/s
an = g sin 55 = 32.2 sin 55 = 26.377 ft/s2
an =
v2
r
(24)
v2
r=
=
an
26.377
2
r = 21.8 ft ◀
(b) At maximum height of stream.
v = (vx )0 = 24 sin 55 =19.660 ft/s
an = g = 32.2 ft/s2
an =
v2
r
(19.660)
v2
r=
=
an
32.2
2
r = 12.0 ft ◀
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2061
PROBLEM 11.147
50°
vA
A
Coal is discharged from the tailgate A of a dump truck with an
initial velocity v A = 2 m/s 50 . Determine the radius of
curvature of the trajectory described by the coal (a) at Point A,
(b) at the point of the trajectory 1 m below Point A.
SOLUTION
a A = g = 9.81 m/s
2
(a) At Point A.
Sketch tangential and normal components of acceleration at A.
( a A ) n = g cos 50 
rA =
v2A
(2)2
=
(aA )n 9.81cos50
r A = 0.634 m ◀
(b) At Point B, 1 meter below Point A.
Horizontal motion: (vB )x = (vA )x = 2cos50 =1.286m/s
Vertical motion:
(vB )2y = (vA )2y + 2ay ( yB - yA )
= (2 cos 40)2 + (2)(- 9.81)(- 1)
= 21.97 m 2 /s 2
( v B ) y = 4.687 m/s
tan q =
( vB ) y
( vB ) x
=
4.687
,
1.286
or
q = 74.6
a B = g cos 74.6 
rB =
=
(vB )2x + (vB )2y
vB2
=
(aB )n
g cos74.6
(1.286)2 + 21.97
9.81cos 74.6
r B = 9.07 m ◀
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2062
PROBLEM 11.148
B
From measurements of a photograph, it has been found that as the
stream of water shown left the nozzle at A, it had a radius of
curvature of 25 m. Determine (a) the initial velocity vA of the
stream, (b) the radius of curvature of the stream as it reaches its
maximum height at B.
vA
3
A
4
SOLUTION
(a)
We have
(aA )n =
v2A
rA
or
é4
ù
v 2A = ê (9.81 m/s2 )ú (25m)
ëê 5
ûú
or
v A = 14.0071 m/s
v A = 14.01 m/s
(b)
We have
Where
Then
(aB )n =
vB2
rB
vB = ( v A ) x =
rB =
36.9° ◀
4
vA
5
( 45 ´14.0071
m/s)
2
9.81 m/s2
r B = 12.80 m ◀
or
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2063
3°
PROBLEM 11.149
v0
A
B
A child throws a ball from point A with an initial
velocity v0 at an angle of 3 with the horizontal.
Knowing that the ball hits a wall at point B, determine
(a) the magnitude of the initial velocity, (b) the
minimum radius of curvature of the trajectory.
1.5 m
0.97 m
6m
SOLUTION
vx = v0 cos a
Horizontal motion.
x = v0 t cos a
v y = v0 sin a - gt
Vertical motion.
1
y = y0 + v0 t sin a - gt 2
2
y = y0 + x tan a -
Eliminate t.
gx 2
2 v02 cos 2 a
(1)
Solving (1) for v0 and applying result at point B
v0 =
gx 2
(9.81)(6)2
=
2
2( y0 + x tan a - y)cos a
(2)(1.5 + 6 tan 3 - 0.97)(cos2 3)
v0 = 14.48 m/s ◀
(a)
Magnitude of initial velocity.
(b)
Minimum radius of curvature of trajectory.
an = g =
v2
r
r=
v2
v2
=
an
g cos q
(2)
where q is the slope angle of the trajectory.
The minimum value of r occurs at the highest point of the trajectory where cos q = 1
and v = v x = v 0 cos a
Then
rmin =
v02 cos 2 a (14.48)2 cos 2 3
=
g
9.81
r min = 21.3 m ◀
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2064
PROBLEM 11.150
x
B
v0
θ
C
ρmin
A projectile is fired from Point A
with an initial velocity v 0 . (a) Show
that the radius of curvature of the
trajectory of the projectile reaches its
minimum value at the highest Point
B of the trajectory.(b) Denoting by
q the angle formed by the trajectory
and the horizontal at a given Point C,
show that the radius of curvature of
the trajectory at C is
r = rmin / cos3 q.
α
A
ρ
SOLUTION
For the arbitrary Point C, we have
( aC )n =
rC =
or
vC2
rC
vC2
g cos q
Noting that the horizontal motion is uniform, we have
( v A ) x = ( vC ) x
( v A ) x = v0 cos a
where
( vC ) x = vC cos q
v 0 cos a = vC cos q
Then
or
vC =
cos a
v0
cos q
so that
rC =
1
cos a
v0
g cos q cos q
(a)
(
2
2
0
2
3
In the expression for rC , v0 , a, and g are constants, so that rC is minimum where cos q is maximum.
By observation, this occurs at Point B where q = 0.
rmin = r B =
(b)
cos a
) = vg cos
q
v02 cos 2 a
g
rC =
1 çæ v02 cos2 a ÷ö
÷÷
ç
÷ø
g
cos3 q çè
rC =
rmin
cos3 q
Q.E.D.
Q.E.D.
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2065
◀
◀
PROBLEM 11.151*
Determine the radius of curvature of the path described by the particle of Problem 11.95 when t = 0.
PROBLEM 11.95 The three-dimensional motion of a particle is defined by the position vector
r = (Rt cos wnt)i + ctj + (Rt sin wnt)k. Determine the magnitudes of the velocity and acceleration of the
particle. (The space curve described by the particle is a conic helix.)
SOLUTION
We have
v=
a=
and
dr
= R(cos wn t - wn t sin wn t )i + cj + R(sin wn t + wn t cos wn t )k
dt
dv
= R(-wn sin wn t - wn sin wn t - wn2 t cos wn t )i
dt
+ R(wn cos wn t + wn cos wn t - wn2 t sin wn t )k
a = w n R[-(2 sin w n t + w n t cos w n t ) i + (2 cos w n t - w n t sin w n t ) k ]
or
v 2 = R 2 (cos w n t - w n t sin w n t )2 + c 2 + R 2 (sin w n t + w n t cos w n t )2
Now
= R 2 (1 + w n2 t 2 ) + c 2
1/ 2
v = éêë R 2 (1 + w n2 t 2 ) + c 2 ùúû
Then
and
R 2 w n2 t
dv
=
dt
é R 2 (1 + w 2 t 2 ) + c 2 ù1 / 2
n
ëê
ûú
Now
a =
At t = 0 :
dv
=0
dt
2
at2
+ an2
2
æ dv ö÷2 æç v 2 ö÷
= çç ÷÷ + çç ÷÷
è dt ø è r ø÷
a = w n R (2 k )
or
a = 2w n R
v2 = R 2 + c2
Then, with
we have
or
dv
= 0,
dt
a=
2wn R =
v2
r
R2 + c2
r
r=
R2 + c2
◀
2wn R
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2066
PROBLEM 11.152*
Determine the radius of curvature of the path described by the particle
of Problem 11.96 when t = 0, A = 3, and B = 1.
y
PROBLEM 11.96
2
2
y2
– x – z =1
A2
B2
A2
z
The three-dimensional motion of a particle is defined by the position
2
vector r = ( At cos t )i + ( A t +1) j + ( Bt sin t )k, where r and t are
expressed in feet and seconds, respectively. Show that the curve
2
2
described by the particle lies on the hyperboloid (y/A) - (x/A) 2
(z/B) = 1. For A = 3 and B = 1, determine (a) the magnitudes of the
velocity and acceleration when t = 0, (b) the smallest nonzero value
of t for which the position vector and the velocity are perpendicular to
each other.
x
SOLUTION
With
A = 3,
B =1
we have
r = (3t cos t )i + (3 t 2 + 1) j + (t sin t )k
æ 3t ÷ö
dr
= 3(cos t - t sin t )i + çç
÷ j + (sin t + t cos t )k
dt
çè t 2 +1 ÷÷ø
Now
v=
and
æ t2 +1 - t
çç
dv
a=
= 3(- sin t - sin t - t cos t )i + 3çç
çè
dt
t 2 +1
+ (cos t + cos t = t sin t )k
(
= -3(2sin t + t cos t )i + 3
1
(t 2 + 1)
1/ 2
t
t 2 +1
ö
)÷÷÷÷ j
÷ø÷
j
+ (2 cos t - t sin t )k
v 2 = 9(cos t - t sin t )2 + 9
Then
t2
+ (sin t + t cos t )2
t2 +1
Expanding and simplifying yields
v2 = t 4 + 19t 2 + 1 + 8(cos2 t + t 4 sin2 t ) - 8(t 3 + t )sin 2t
Then
and
v = [t 4 + 19t 2 + 1 + 8(cos2 t + t 4 sin 2 t ) - 8(t 3 + t )sin 2t ]1/ 2
dv 4t 3 + 38t + 8(-2cos t sin t + 4t 3 sin2 t + 2t 4 sin t cos t - 8[(3t 2 +1)sin 2t + 2(t 3 + t )cos2t ]
=
dt
2[t 4 +19t 2 +1 + 8(cos2 t + t 4 sin2 t ) - 8(t 3 + t )sin 2t ]1/ 2
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2067
PROBLEM 11.152* (CONTINUED)
a
Now
2
= at2
+ an2
2
æ dv ÷ö2 æç v2 ö÷
ç
÷
= ç ÷÷ + ç ÷
è dt ø çè r ÷ø
At t = 0 :
a = 3 j + 2k
or
a = 13 ft/s2
dv
=0
dt
v 2 = 9(ft/s)2
Then, with
dv
= 0,
dt
we have
a=
or
r=
v2
r
9 ft 2 /s2
r = 2.50 ft ◀
13 ft/s2
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2068
PROBLEM 11.153
A satellite will travel indefinitely in a circular orbit around a planet if the normal component of the
acceleration of the satellite is equal to g( R / r )2, where g is the acceleration of gravity at the surface of the
planet, R is the radius of the planet, and r is the distance from the center of the planet to the satellite. Knowing
that the diameter of the sun is 1.39 Gm and that the acceleration of gravity at its surface is 274 m/s2 ,
determine the radius of the orbit of the indicated planet around the sun assuming that the orbit is circular.
Earth: (umean )orbit = 107 Mm/h.
SOLUTION
g = 274 m/s2 ,
For the sun,
R=
and
Given that an =
æ1ö
1
D = çç ÷÷÷ (1.39 ´10 9 ) = 0.695 ´10 9 m
è2ø
2
gR 2
v2
and that for a circular orbit an =
2
r
r
r=
Eliminating an and solving for r,
v = 107 ´106 m/h = 29.72 ´103 m/s
For the planet Earth,
Then
gR 2
v2
r=
(274)(0.695´109 )2
= 149.8´109 m
3 2
(29.72 ´10 )
r = 149.8 Gm ◀
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2069
PROBLEM 11.154
A satellite will travel indefinitely in a circular orbit around a planet if the normal component of the
acceleration of the satellite is equal to g( R / r )2, where g is the acceleration of gravity at the surface of the
planet, R is the radius of the planet, and r is the distance from the center of the planet to the satellite. Knowing
that the diameter of the sun is 1.39 Gm and that the acceleration of gravity at its surface is 274 m/s2,
determine the radius of the orbit of the indicated planet around the sun assuming that the orbit is circular.
Saturn: (umean )orbit = 34.7 Mm/h.
SOLUTION
g = 274 m/s2
For the sun,
R=
and
Given that an =
æ1ö
1
D = çç ÷÷÷ (1.39 ´10 9 ) = 0.695 ´10 9 m
è2ø
2
gR 2
v2
and
that
for
a
circular
orbit:
=
a
n
r
r2
gR 2
v2
Eliminating an and solving for r,
r=
For the planet Saturn,
v = 34.7 ´106 m/h = 9.639 ´103 m/s
Then,
r=
(274)(0.695´109 )2
= 1.425´1012 m
(9.639´103 )2
r = 1425 Gm ◀
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2070
PROBLEM 11.155
Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a
circular orbit 100 mi above the surface of the planet. (See information given in Problems 11.153–11.154).
Venus: g = 29.20 ft/s2 , R = 3761 mi.
SOLUTION
From Problems 11.153 and 11.154,
an =
gR 2
r2
For a circular orbit,
an =
v2
r
v=R
Eliminating an and solving for v,
For Venus,
g
r
g = 29.20 ft/s2
R = 3761mi = 19.858´106 ft.
r = 3761 + 100 = 3861mi = 20.386 ´106 ft
Then,
v = 19.858 ´106
29.20
= 23.766 ´103 ft/s
6
20.386 ´10
v = 16200 mi/h ◀
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2071
PROBLEM 11.156
Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a
circular orbit 100 mi above the surface of the planet. (See information given in Problems 11.153–11.154).
Mars: g = 12.17 ft/s2 , R = 2102 mi.
SOLUTION
From Problems 11.153 and 11.154,
an =
gR 2
r2
For a circular orbit,
an =
v2
r
v=R
Eliminating an and solving for v,
For Mars,
g
r
g = 12.17 ft/s2
R = 2102 mi = 11.0986 ´106 ft
r = 2102 + 100 = 2202 mi = 11.6266 ´106 ft
Then,
v = 11.0986 ´106
12.17
= 11.35 ´103 ft/s
11.6266 ´106
v = 7740 mi/h ◀
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2072
PROBLEM 11.157
Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a
circular orbit 100 mi above the surface of the planet. (See information given in Problems 11.153–11.154).
Jupiter: g = 75.35 ft/s2 , R = 44, 432 mi.
SOLUTION
From Problems 11.153 and 11.154,
an =
gR 2
r2
For a circular orbit,
an =
v2
r
v=R
Eliminating an and solving for v,
For Jupiter,
g
r
g = 75.35 ft/s2
R = 44432 mi = 234.60 ´106 ft.
r = 44432 + 100 = 44532 mi = 235.13´106 ft
Then,
v = (234.60 ´106 )
75.35
= 132.8 ´103 ft/s
235.13´106
v = 90600 mi/h ◀
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2073
PROBLEM 11.158
A satellite will travel indefinitely in a circular orbit around the earth if the normal component of its
acceleration is equal to g(R/r )2, where g = 9.81 m/s2, R = radius of the earth = 6370 km, and r = distance
from the center of the earth to the satellite. Assuming that the orbit of the moon is a circle of radius
384 ´103 km, determine the speed of the moon relative to the earth.
SOLUTION
Normal acceleration:
an =
gR 2
r2
and
gR 2
r
Solve for v :
v2 = ran =
Data:
g = 9.81 m/s2,
2
v2 v 2
an = r =
r
R = 6370 km = 6.370 ´106 m
r = 384 ´103 km = 384 ´106 m
(9.81)(6.370 ´106 )
2
v2 =
384 ´106
v = 1.018 m/s
= 1.0366 ´106 m 2 /s2
v = 3670 km/h ◀
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2074
PROBLEM 11.159
Knowing that the radius of the earth is 6370 km, determine the time of one orbit of the Hubble Space
Telescope, knowing that the telescope travels in a circular orbit 590 km above the surface of the earth.
(See information given in Problems 11.153–11.155.)
SOLUTION
an = g
We have
Then
g
an =
and
v2
r
R2
v2
=
r2
r
v=R
or
R2
r2
g
r
where
r = R+h
The circumference s of the circular orbit is equal to
s = 2pr
Assuming that the speed of the telescope is constant, we have
s = vt orbit
Substituting for s and v
2 pr = R
torbit =
or
=
g
torbit
r
2p r 3 / 2
R g
1h
2p
[(6370 + 590)km]3 / 2
´
6370 km [9.81´10-3 km/s2 ]1/ 2 3600 s
t orbit = 1.606 h ◀
or
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2075
B
A
rB
rA
PROBLEM 11.160
Satellites A and B are traveling in the same plane in circular orbits around the
earth at altitudes of 120 and 200 mi, respectively. If at t = 0 the satellites are
aligned as shown and knowing that the radius of the earth is R = 3960 mi,
determine when the satellites will next be radially aligned. (See information
given in Problems 11.153–11.155.)
SOLUTION
an = g
We have
Then
g
R2
r2
R2
v2
=
r
r2
and
or
an =
v2
r
v=R
g
r
r=R+h
where
The circumference s of a circular orbit is equal to
s = 2pr
Assuming that the speeds of the satellites are constant, we have
s = vT
Substituting for s and v
2 pr = R
or
Now
T=
g
T
r
2p r 3/ 2 2p ( R + h)3/ 2
=
R g
R
g
hB > hA  (T ) B > (T ) A
Next let time TC be the time at which the satellites are next radially aligned. Then, if in time TC satellite B
completes N orbits, satellite A must complete ( N + 1) orbits.
Thus,
TC = N (T ) B = ( N + 1)(T ) A
or
3/ 2 ù
é 2p ( R + hB )3/ 2 ù
é
ú = ( N +1) ê 2p ( R + hA ) ú
Nê
êR
ú
êR
ú
g
g
ë
û
ë
û
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2076
PROBLEM 11.160 (CONTINUED)
N=
( R + hA )3/ 2
1
=
3/ 2
3/ 2
3/ 2
R
h
+
( R + hB ) - ( R + hA )
( B ) -1
R + hA
or
=
1
+200
( 3960
3960+120 )
3/ 2
TC = N (T ) B = N
Then
= 33.835
-1
= 33.835 orbits
2p ( R + hB )3/ 2
R
g
é(3960 + 200)miù 3/ 2
1h
2p
ë
û
´
1/
2
1mi
3960 mi
3600s
32.2 ft/s2 ´ 5280 ft
(
)
TC = 51.2 h ◀
or
Alternative solution
From above, we have (T ) B > (T ) A . Thus, when the satellites are next radially aligned, the angles qA and qB
swept out by radial lines drawn to the satellites must differ by 2 p . That is,
qA = qB + 2p
For a circular orbit
s = rq
From above
s = vt
Then
R g
R g
s vt 1 æ g ö÷
q = = = çç R
t
÷ t = 3/ 2 t =
( R + h)3/ 2
r
r
r çè r ø÷
r
At time TC :
R g
3/ 2
( R + h)
TC =
R g
( R + hB )
TC =
g
r
v=R
and
3/ 2
TC + 2p
2p
R g
1
3/2
( R + hA )
-
1
( R+hB )
3/ 2
2p
=
(
1mi
(3960 mi) 32.2 ft/s2 ´ 5280ft
or
´
´
1
[(3960+120)mi]3 / 2
1
-
)
1/ 2
1
[(3960+200)mi]3 / 2
1h
3600s
TC = 51.2 h ◀
or
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2077
PROBLEM 11.161
Robots with telescoping arms are sometimes used to perform tasks (e.g., welding
or placing screws) where access may be difficult for other robotic types. During a
test run, a robot arm is programmed to extend according to the relationship
r = 3 + 0.5cos (4q )
and the arm rotates according to the relationship
p
q = - t 2 + pt , where r is in feet, q is in radians, and t is in seconds. Determine
4
(a) the velocity and acceleration of the robot tip A at t = 3 s and (b) use a
computer program to plot the path of tip A in x and y coordinates for 0 £ t £ 4 s.
SOLUTION
Differentiate both functions with respect to time
r = 3 + 0.5cos (4q )
r = - sin(4q )2q
r = -8q 2 (4q ) - 2q sin(4q )
p
q = - t 2 + pt
4
q = - p t + p
2
p
q = 2
Next, evaluate the results at t=2 to find
p
q = p, q = 0, q = 2
r = 3.5, r = 0, r = 0
To find the velocity and acceleration,
v = rer + rqeθ
a = (r - rq 2 )er + (rq + 2rq)eθ
Using the values above
v=0
p
a = 3.5(- )e θ = -5.5 ft/s e θ
2
v=0◀
a = -5.5 ft/s e θ ft/s2 ◀
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2078
PROBLEM 11.161 (CONTINUED)
Similarly, at t=3
3 
p
p
p, q = - , q = 4
2
2
r = 2.5, r = 0, r = 19.739
q=
And
p
v = 2.5(- )e θ = -3.93e θ
2
a = -3.927e θ + 13.571e r
v = -3.93e θ ft/s ◀
a = -3.927eθ + 13.571e r ft/s2 ◀
A graph of the path of the robot grip is found by graphing a parametric plot of x vs y for
values of t between 0 and 4 seconds, using
x = cos(q )
y = sin(q )
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written consent of McGraw-Hill Education.
2079
PROBLEM 11.162
The angular displacement of the robotic arm is programmed according to the relationship
q = (1 / p)(sin pt) , where q and t are expressed in radians and seconds, respectively. Simultaneously, the
arm is programmed to extend so that the distance to A follows the relationship r = 4 1 + e-2 t , where r
and t are expressed in feet and seconds, respectively. When t = 1.5 s, determine (a) the velocity of point A,
(b) the acceleration of point A.
(
)
SOLUTION
Evaluate the angle at t=1.5 s to get,
q = (1 / p)(sin p1.5) = -1 / p
Differentiate to get angular velocity and acceleration and evaluate at t = 1.5 seconds,
q = cos(pt ) = cos(p1.5) = 0
q = -p sin(pt ) = -p sin(p1.5) = p
Next, differentiate r(t) with respect to time and evaluate at t = 1.5 seconds
(
) (
)
r = 8(e ) = 8(e
) = -0.398
r = 16 (e ) = 16 (e
) = 0.800
r = 4 1 + e-2 t = 4 1 + e-2*1.5 = 4.200
-2 t
-2*1.5
-2 t
-2*1.5
Next compute and evaluate the velocity in radial and traverse directions,
v = reˆ r + rqeˆ q
v = -.398eˆ r + 4.200(0)eˆ q
v = -.398eˆ r ft/s ◀
v = -.398eˆ r
Next compute and evaluate the acceleration in radial and traverse directions
a = (r - rq 2 )eˆ r + (rq + 2rq)eˆ q
a = (0.797 - 0)eˆ r + (4.200(p ) + 2(-.398)p )eˆ q
a = (0.797)eˆ r + (13.192)eˆ q
a = (0.797)eˆ r + (13.19)eˆ q ft/s2 ◀
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written consent of McGraw-Hill Education.
2080
PROBLEM 11.163
During a parasailing ride, the boat is traveling at a
constant 30 km/hr with a 200 m long tow line. At
the instant shown, the angle between the line and
the water is 30and is increasing at a constant rate
of 2/s. Determine the velocity and acceleration of
the parasailer at this instant.
r
θ
SOLUTION
Given:
v B = 30i km/hr = 8.333i m/s
aB = 0
r = 200 m, r = 0, r = 0
q = 30
q = 2 /s = 0.0349 rad/s
Relative Motion relations:
ê
eˆr
P
r
v (m/s)
60
26
–5
–20
10
q = 0
vP = vB + vP / B
41
46
t
B
aP = aB + aP / B
Using Radial and Transverse components:
v P / B = rer + rqeq
a P / B = (r + rq 2 )er + (rq + 2rq)eq
Substitute in known values:
v P / B = 6.981eq m/s
a P / B = -0.2437er m/s2
Change to rectangular coordinates:
v P / B = 6.981* sin 30i + 6.981* cos30 j m/s
= 3.491i = 6.046 j m/s
a P / B = -0.2437 * (- cos30) i - 0.2437 * sin 30 j m/s2
= 0.2111i - 0.1219 j m/s2
Substitute into Relative Motion relations:
v P = 8.33i + 3.491i + 6.046 j m/s
= 11.824 i = 6.046 j m/s
a P = 0 + 0.2111i - 0.1219 j m/s2
= 0.2111i - 0.1219 j m/s2
Velocity:
v P = 13.280 m/s
27.08 ◀
Acceleration:
aP = 0.2437 m/s2
30.00 ◀
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written consent of McGraw-Hill Education.
2081
PROBLEM 11.164
Some parasailing systems use a winch to pull the rider back to the boat. During the interval when q is between
20 and 40, (where t = 0 at q = 20) the angle increases at the constant rate of 2/s. During this time, the length
1
of the rope is defined by the relationship r = 600 - t 5 / 2, where r and t are expressed in ft and s, respectively.
8
Knowing that the boat is travelling at a constant rate of 15 knots (where 1 knot = 1.15 mi/h), (a) plot the
magnitude of the velocity of the parasailer as a function of time (b) determine the magnitude of the acceleration
of the parasailer when t = 5 s.
SOLUTION
Given:
v B = 15i knots*
aB = 0
1.15mph 5280 ft / mi
*
= 25.30 i ft/s
knot
3600 s / hr
1 5
r = 600 - t 2 m
8
20 £ q £ 40
q = 2 /s = 0.0349 rad/s
ê
eˆr
P
q = 0
Take Derivatives of r(t):
Relative Motion relations:
r
v (m/s)
60
26
–5
–20
10
41
46
t
B
5 3
r = - t 2 m/s
16
15 1
r = - t 2 m/s2
32
vP = vB + vP / B
aP = aB + aP / B
Using Radial and Transverse components:
v P /B = rer + rqeq
a P /B = r + rq2 er + (rq + 2rq) eq
(
)
Change velocity to rectangular coordinates:
v P /B = r (- cos q i + sin q j) + r q (sin q i + cos q j)
Substitute into Relative Motion relations:
v P = (vB - r cos q + rq sin q ) i + (r sin q + rq cos q ) j
a = r - rq 2 e + (rq + 2rq ) e
P
Note that:
(
)
q
r
v P = vPx i + vPy j
a P = ar e r + aq e q
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written consent of McGraw-Hill Education.
2082
PROBLEM 11.164 (CONTINUED)
Where:
vPx = v B - r cos q + r q sin q
v = r sin q + r q cos q
Py
ar = r - r q 2
a = r q + 2rq
q
2
2
v P = vPx
+ vPy
Magnitude of velocity:
a P = ar2 + aq2
(a) Plot of Velocity as a function of theta:
Magnitude of Velocity as Theta Changes
47
46
45
Velocity, (ft/s)
44
43
42
41
40
39
38
37
20
22
24
26
28
30
32
theta, (deg)
34
36
38
40
(b) Magnitude of Acceleration at t = 5s:
1 5
r = 600 - (5) 2 = 593.012 m
8
5 32
r = (5) = -3.494 m/s
16
15 1 2
r = (5) = -1.048 m/s2
32
a = (r - rq 2 ) = -1.771 m/s2
r
a q = (rq + 2rq ) = -0.2439 m/s2
aP = -1.7712 +-0.24392
aP = 1.787 m/s2 ◀
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2083
PROBLEM 11.165
D
As rod OA rotates, pin P moves along the parabola BCD. Knowing that the
equation of this parabola is r = 2 b(1 + cos q ) and that q = kt , determine the
velocity and acceleration of P when (a) q = 0, (b) q = 90  .
A
r
θ
O
P
C
b
B
SOLUTION
r=
r =
r =
(a)
2b
1 + cos kt
2bk sin kt
(1 + cos kt )2
2bk
(1 + cos kt )4
q = kt
q=k
q = 0
[(1 + cos kt )2 k cos kt + (sin kt )2(1 + cos kt )(k sin kt )]
When q = kt = 0:
r = b r = 0 r =
q = 0 q = k
2bk
(2)

q =0
4
[(2)2 k (1) + 0] =
1 2
bk
2
vr = r = 0 vq = rq = bk
v = bk e q ◀
üï
1 2
1
bk - bk 2 = - bk 2 ïï
2
2
ý
ïï
aq = r q + 2rq = b(0) + 2(0) = 0
ïïþ
ar = r - r q 2 =
(b)
1
a = - bk 2 er ◀
2
When q = kt = 90°:
r = 2b
r = 2bk
q = 90 q = k
2bk
[0 + 2 k ] = 4 bk 2
19
q = 0
r =
vr = r = 2bk vq = rq = 2bk
v = 2bke r + 2bke q ◀
ar = r - rq 2 = 4bk 2 - 2bk 2 = 2bk 2
a = rq + 2rq = 2b(0) + 2(2bk )k = 4bk 2
a = 2 bk 2 e r + 4 bk 2 e q ◀
q
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2084
PROBLEM 11.166
D
r
B
C
The pin at B is free to slide along the circular slot DE and
along the rotating rod OC. Assuming that the rod OC
rotates at a constant rate q, (a) show that the acceleration
of pin B is of constant magnitude, (b) determine the
direction of the acceleration of pin B.
θ
b
O
A
b
E
SOLUTION
From the sketch:
r = 2b cos q
r = -2b sin q q
Since q = constant, q = 0
r = -2b cos q q 2
ar = r - rq 2 = -2b cos q q 2 - (2b cos q )q 2
a = -4 b cos q q 2
r
aq = rq + 2rq = (2b cos q )(0) + 2(-2b sin q )q 2
a = -4 b sin q q 2
q
Since both b and q are constant, we find that
a = ar2 + aq2 = 4 bq 2 (- cos q )2 + (- sin q )2
a = 4 bq 2
a = constant ◀
g = tan-1
æ
ö
aq
ç -4b sin q q2 ÷÷
÷
= tan-1 çç
çç -4b cos q q2 ÷÷÷
ar
è
ø
g = tan-1 (tan q)
g =q
Thus, a is directed toward A ◀
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2085
PROBLEM 11.167
To study the performance of a racecar, a high-speed camera is
positioned at Point A. The camera is mounted on a mechanism
which permits it to record the motion of the car as the car
travels on straightway BC. Determine (a) the speed of the car in
terms of b, q, and q, (b) the magnitude of the acceleration in

terms of b, q , q, and q.
SOLUTION
Velocity, sketch the directions of the vectors v and eθ .
vθ = v ⋅ eθ = - v cos q
vθ = rq
But
rq = - v cos q
Hence,
r=
But from geometry,
bq
= - v cos q
cos q
b
cos q
v=-
or
bq
cos2 q
Speed is the absolute value of v.
v=
Acceleration, from geometry,
Differentiating with respect to time,
bq
cos2 q
r=
r =
◀
b
cos q
b sin qq
cos2 q
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2086
PROBLEM 11.167 (CONTINUED)
Transverse component of acceleration
bq
2bq 2 sin q
aq = rq + 2rq =
+
cos q
cos2 q
(1)
Sketch the directions of the vectors a and e θ referring to the figure above,
aq = a ⋅ eq = - a cos q
(2)
Matching from (1) and (2) and solving for a,
a=-
bq
-
2bq 2 sin q
cos2 q
cos3 q
b
=q + 2q 2 tan q
2
cos q
(
)
|a| =
Since magnitude of a is sought,
b
cos2 q
q + 2q 2 tan q ◀
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2087
v
After taking off, a helicopter climbs in
a straight line at a constant angle b.
Its flight is tracked by radar from
Point A. Determine the speed of the
helicopter in terms of d, b, q , and q.
B
A
PROBLEM 11.168
β
θ
d
SOLUTION
From the diagram
r
d
=
sin(180 - b ) sin(b - q )
d sin b = r (sin b cos q - cos b sin q )
or
tan b
tan b cos q - sin q
or
r=d
Then
r = d tan b
-(- tan b sin q - cos q ) 
q
(tan b cos q - sin q )2
tan b sin q + cos q
= dq tan b
(tan b cos q - sin q )2
From the diagram
vr = v cos( b - q )
where
vr = r
Then
d q tan b
tan b sin q + cos q
(tan b cos q - sin q )2
= v(cos b cos q + sin b sin q )
= v cos b (tan b sin q + cos q )
v=
or
dq tan b sec b
(tan b cos q - sin q)2
Alternative solution.
We have
v 2 = vr2 + vq2 = (r )2 + (r q )2
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2088
◀
PROBLEM 11.168 (Continued)
Using the expressions for r and r from above
2
é
tan b sin q + cos q ù
ú
v = ê dq tan b
ê
(tan b cos q- sin q )2 úû
ë
or
é (tan b sin q + cos q )2
ù1/2
dq tan b
ê
ú
+
1
ú
(tan b cos q - sin q ) êëê (tan b cos q - sin q )2
ûú
1/2
2
é
ù
dq tan b
(tan b + 1
ê
ú
=
ê
2ú
(tan b cos q - sin q ) ëê (tan b cos q - sin q ) ûú
v=
Note that as q increases, the helicopter moves in the indicated direction. Thus, the positive root is chosen.
v=
dq tan b sec b
(tan b cos q- sin q)2
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2089
◀
PROBLEM 11.169
At the bottom of a loop in the vertical plane, an
airplane has a horizontal velocity of 315 mi/h
2
and is speeding up at a rate of 10 ft/s . The
radius of curvature of the loop is 1 mi. The
plane is being tracked by radar at O. What are
the recorded values of r, r, q and q for this
instant?
1 mi
315 mi/hr
1800 ft
r
θ
O
2400 ft
SOLUTION
Geometry. The polar coordinates are
r = (2400)2 + (1800)2 = 3000 ft
Velocity Analysis.
æ 1800 ö÷
q = tan-1 çç
= 36.87
è 2400 ÷÷ø
v = 315 mi/h = 462 ft/s
vr = 462 cos q = 369.6 ft/s
vq = -462sin q = -277.2 ft/s
vr = r
r = 370 ft/s ◀
v
277.2
q = q = r
3000
vq = rq
q = -0.0924 rad/s ◀
at = 10 ft/s2
Acceleration analysis.
an =
v 2 (462)2
=
= 40.425 ft/s2
5280
r
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written consent of McGraw-Hill Education.
2090
PROBLEM 11.169 (CONTINUED)
ar = at cos q + an sin q = 10 cos36.87 + 40.425sin 36.87 = 32.225 ft/s2
aq = at sin q + an cos q = 10sin 36.87 + 40.425cos36.87 = 26.34 ft/s2
a = r - rq 2 r = a + rq 2
r
r
r = 57.9 ft/s2 ◀
r = 32.255 + (3000)(-0.0924)2
aq = r q + 2rq
a
2rq
q = q r
r
26.34 (2)(369.6)(-0.0924)
=
3000
3000
q = 0.0315 rad/s2 ◀
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written consent of McGraw-Hill Education.
2091
PROBLEM 11.170
An airplane passes over a radar tracking station at A and continues to
fly due east. When the plane is at P, the distance and angle of elevation
of the plane are, respectively, r = 12,600 ft and q = 31.2. Two
seconds later the radar station sights the plane at r = 13,600 ft and
q = 28.3. Determine approximately the speed and the angle of dive α
of the plane during the 2-s interval.
SOLUTION
Dr = 13600 - 12600 = 1000 ft
Changes in values over the interval
Dq = 28.3 - 31.2 = - 2.9 = - 5.0615 ´ 10-2 rad
Dt = 2s
Average rates of change.
r =
Dr 1000
=
= 500 ft/s
Dt
2
- 5.0615 ´ 10-2
q = Dq =
= - 2.5307 ´ 10-2 rad/s
Dt
2
Mean values.
r=
12600 + 13600
q=
2
= 13100 ft
31.2 + 28.3
2
= 29.75
vr = r = 500 ft/s
Velocity components.
(
)
vq = rq = (13100) -2.5307 ´ 10-2 = - 331.53 ft/s
v = vr2 + vq2 =
(500)
2
+ (-331.53) = 600 ft/s
2
v = 409 mi/h ◀
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written consent of McGraw-Hill Education.
2092
PROBLEM 11.170 (CONTINUED)
vy
vθ
vr
vx
vx = vr cos q - vq sin q
= 500 cos29.75 - (-331.53) sin 29.75 = 598.61 ft/s
vy = vr sin q + vq cos q
= 500sin 29.75 + ((-331.53) cos 29.75) = - 39.73 ft/s
tan a =
-vy
vx
=
39.73
= 0.06636
598.61
a = 3.80 ◀
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written consent of McGraw-Hill Education.
2093
PROBLEM 11.171
For the racecar of Problem 11.167, it was found that it took 0.4
s for the car to travel from the position q = 60 to the position
q = 35. Knowing that b = 25 m, determine the average speed
of the car during the 0.4-s interval.
SOLUTION
From the diagram:
Dr12 = 25 tan 60 - 25 tan 35
= 25.796 m
Dr12
Dt12
25.796 m
=
0.4 s
= 46.490 m/s
vavg =
Now
vavg
vavg
vavg = 232.2 km/h ◀
or
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2094
PROBLEM 11.172
For the helicopter of Problem 11.168, it was found that when the helicopter was at B, the distance and the
angle of elevation of the helicopter were r = 3000 ft and q = 20, respectively. Four seconds later, the radar
station sighted the helicopter at r = 3320 ft and q = 23.1  . Determine the average speed and the angle of
climb b of the helicopter during the 4-s interval.
PROBLEM 11.168 After taking off, a helicopter climbs in a straight line at a constant angle b. Its flight is
tracked by radar from Point A. Determine the speed of the helicopter in terms of d, b, q , and q.
v
B
β
A
θ
d
SOLUTION
We have
r0 = 3000 ft
q0 = 20
r4 = 3320 ft
q4 = 23.1
From the diagram:
Dr 2 = 30002 + 33202
-2(3000)(3320)cos(23.1 - 20)
or
D r = 362.70 ft
Now
vave =
=
Dr
Dt
362.70 ft
4s
= 90.675 ft/s
vave = 61.8 mi/h ◀
or
Also,
or
Dr cos b = r4 cos q4 - r0 cos q0
cos b =
3320 cos 23.1 - 3000 cos 20
362.70
b = 49.7 ◀
or
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written consent of McGraw-Hill Education.
2095
PROBLEM 11.173
1 2
θ
r = be2
A particle moves along the spiral shown. Determine the magnitude
of the velocity of the particle in terms of b, q, and q.
O
SOLUTION
1
q2
Given:
r = be 2
Take time derivative
q
r = be 2 qq
1
2
1
Radial and Transverse Components
q
vr = r = be 2 qq
2
1
q
vq = r q = be 2 q
Magnitude of Velocity
2
v 2 = vr2 + vq2
æ 1 q 2 ö÷2
çç 2 ÷
= ççbe ÷÷÷ (q 2 + 1) q 2
çç
÷÷ø
è
1
q2
v = be 2 (q 2 + 1)1/2 q ◀
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written consent of McGraw-Hill Education.
2096
PROBLEM 11.174
2
rθ =b
A particle moves along the spiral shown. Determine the magnitude of the
velocity of the particle in terms of b, q, and q.
O
SOLUTION
Given:
r=
b
q2
Take time derivative
r =
2b 
q
q3
Radial and Transverse Components
vr = r = -
2b 
q
q3
b
vq = r q = 2 q
q
Magnitude of Velocity
v 2 = vr2 + vq2
4b2 2 b2 2
q + 4q
q6
q
b2
= 6 (4 + q 2 )q 2
q
=
v=
b
q
3
(4 + q )
2
1/2
q ◀
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written consent of McGraw-Hill Education.
2097
PROBLEM 11.175
1 2
θ
r = be2
A particle moves along the spiral shown. Knowing that q is
constant and denoting this constant by w, determine the magnitude
of the acceleration of the particle in terms of b, q, and q .
O
SOLUTION
1
q2
r = be 2
q = w
and
Given:
1
Take time derivatives
q
r = be 2 qq
1
r = be 2
Radial Component of Acceleration
2
q2
é(  )2  2
ù
êë qq + q + qqúû
ar = r - rq 2
1
= be 2
1
= be 2
Transverse Component of Acceleration
q = 0
q2
é(  )2  2
  2 ù
ëê qq + q + qq - q ûú
q2
é 2
ù
ëê(qq ) + qq ûú
aq = rq + 2rq
1
1
q
q
= be 2 q + be 2 qq 2
2
1 2
q
= be 2
Substitute Given Values
q = w
éq + 2qq 2 ù
ë
û
and
1
q = 0
1
q2
ar = be 2 (qw )2
Magnitude of Acceleration
2
and
q2
aq = be 2 (2qw 2 )
a 2 = ar2 + aq2
æ 1 q 2 ö÷2
ç 2 ÷
= çççbe ÷÷÷ (q 4 + 4q 2 ) w 4
çç
÷
è
ø÷
1
q2
1
a = be 2 q (q 2 + 4) 2 w 2 ◀
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written consent of McGraw-Hill Education.
2098
PROBLEM 11.176
2
rθ =b
A particle moves along the spiral shown. Knowing that q is constant and
denoting this constant by w, determine the magnitude of the acceleration of
the particle in terms of b, q, and q .
O
SOLUTION
b
Given:
r=
Take time derivatives
r = -
Radial Component of Acceleration
Transverse Component of Acceleration
q2
, q = w and q = 0
2b 
q
q3
2b
6b
r = 3 q + 4 q 2
q
q
ar = r - r q 2
b
2b
6b
= - 3 q + 4 q 2 - 2 q 2
q
q
q
b(
2


= 4 -2qq + 6q - q 2 q 2 )
q
aq = rq + 2rq
æ 2b ö÷  2
b 
çç- ÷ q
+
q
2
(
)
çè q 3 ø÷÷
q2
b
= 3 (qq - 4q 2 )
q
=
b
(6 - q 2 )w 2
q4
Substitute Given Values
ar =
Magnitude of Acceleration
a 2 = ar2 + aq2
and
aq = -
4b 2
w
q3
b2
16b 2 4
2
4) 4
(
+
+
q
q
w
w
36
12
q8
q6
b2
= 8 (36 + 4q 2 + q 4 ) w 4
q
=
a=
1/ 2
b
(36 + 4q 2 + q 4 ) w 2 ◀
4
q
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written consent of McGraw-Hill Education.
2099
PROBLEM 11.177
The motion of a particle on the surface of a right circular cylinder is
defined by the relations R = A, q = 2pt, and z = At 2 / 4, where A is
a constant. Determine the magnitudes of the velocity and acceleration
of the particle at any time t.
SOLUTION
In cylindrical coordinates.
q = 2p t ,
z=
R = 0,
q = 2p,
z =
R = 0,
q = 0,
R = A,
At 2
4
Differentiating with respect to time,
Velocity vector:
vr = R = 0,
At
2
At
z =
2
vq = Rq = 2p A,
vz = z =
v 2 = vR2 + vq2 + vz2 = 0 + 4p 2 A2 +
Acceleration vector:
At
2
1 2 2
At
4
v=
1
A 16p 2 + t 2 ◀
2
a=
1
A 64p 4 + 1 ◀
2
ar = R - Rq 2 = 0 - 4p 2 A
aq = Rq + 2 R q = 0,
az = z = A/2
a2 = aR2 + aq2 + az2 = 16p 4 A2 + 0 +
1 2
A
4
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2100
PROBLEM 11.178
Show that r = hf sin q knowing that at the instant shown, step
AB of the step exerciser is rotating counterclockwise at a
constant rate f .
O
h
θ r
ϕ
B
P
d
A
SOLUTION
From the diagram
r 2 = d 2 + h 2 - 2 dh cos f
2rr = 2 dhf sin f
Then
r
d
=
sin f sin q
Now
r=
or
d sin f
sin q
Substituting for r in the expression for r
æ d sin f ö÷

çç

çè sin q ÷÷ø r = dhf sin f
r = hf sin q
or
Q.E.D.
Alternative solution.
First note
a = 180 -(f + q)
Now
v = vr + vq = rer + rqeq
With B as the origin
vP = df
(d = constant  d = 0)
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2101
◀
PROBLEM 11.178 (CONTINUED)
With O as the origin
(vP )r = r
where
(vP )r = vP sin a
r = df sin a
Then
h
d
=
sin a sin q
Now
d sin a = h sin q
or
substituting
r = hf sin q
Q.E.D.
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2102
PROBLEM 11.179
The three-dimensional motion of a particle is defined by the cylindrical coordinates
R = A / (t +1), q = Bt, and z = Ct / (t +1). Determine the magnitudes of the velocity and
acceleration when (a) t = 0, (b) t = ¥.
SOLUTION
Differentiating with respect to time,
R = R =
A
(t + 1)
2
2A
(t + 1)
3
q = B,
,
z =
q = 0,
,
z = -
(
)
C t + 1 - Ct
(t + 1)
2
=
C
(1 + t )
2
2C
(1 + t )
3
(a) Let t = 0 and use the above equations to get,
R = A,
q = 0,
z=0
R = - A,
R = 2 A,
q = B,
q = 0,
z = C
vq = Rq = AB,
vR = R = - A,
For velocity we get,
z = - 2C
vz = z = C
v2 = vR2 + vq2 + vz2 = A2 + A2 B 2 + C 2
So,
v=
A 2 + A2 B 2 + C 2 ◀
Also for acceleration we get,
aR = R - Rq 2 = 2 A - AB 2
aR2 = 4 A2 - 4 A2 B 2 + A2 B 4
aq = Rq + 2 R q = 0 - 2 AB and
az = z = - 2c
aq2 = 4 A2 B
and az2 = 4C 2
a 2 = aR2 + aq2 + az2 = 4 A2 + A2 B 4 + 4C 2
So
a = 4 A 2 + A 2 B 4 + 4C 2 ◀
(b) Let t = ¥ and use the above equations to get,
R = 0,
q = ¥,
z = C,
R = 0,
R = 0,
q = 0,
q = B,
z = 0,
z = 0
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2103
PROBLEM 11.179 (CONTINUED)
vr = R = 0,
ar = R - Rq 2 = 0,
vq = Rq = 0,
vz = z = 0,
aq = Rq - Rq 2 = 0,
a z = z = 0,
v=0 ◀
a=0 ◀
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2104
PROBLEM 11.180*
For the conic helix of Problem 11.95, determine the angle that the osculating plane forms with the y axis.
PROBLEM 11.95 The three-dimensional motion of a particle is defined by the position vector
r = (Rt cos wnt)i + ctj + (Rt sin wnt)k. Determine the magnitudes of the velocity and acceleration of the
particle. (The space curve described by the particle is a conic helix.)
SOLUTION
First note that the vectors v and a lie in the osculating plane.
Now
r = ( Rt cos w n t )i + ctj + ( Rt sin w n t )k
Then
v=
and
a=
dr
= R(cos wn t - wn t sin wn t )i + cj + R(sin wn t + wn t cos wn t )k
dt
dv
dt
= R(-wn sin w n t - wn sin wn t - w n2 t cos wn t )i
+ R(wn cos w n t + wn cos w n t - w n2 t sin w n t )k
= w n R[-(2 sin wn t + w n t cos wn t )i + (2 cos w n t - wn t sin w n t )k ]
It then follows that the vector ( v ´ a ) is perpendicular to the osculating plane.
i
(v ´ a) = wn R R(cos wn t - wn t sin wn t )
j
k
c R(sin wn t + wn t cos wn t )
-(2sin wn t + wn t cos wn t ) 0 (2cos wn t - wn t sin wn t )
= wn R{c(2 cos wn t - wn t sin wn t )i + R[- sin wn t + wn t cos wn t )(2 sin wn t + wn t cos wn t )
- (cos wn t - w n t sin wn t )(2 cos wn t - wn t sin wn t )]j + c(2 sin wn t + wn t cos wn t )k
= wn R éëê c(2 cos wn t - wn t sin wn t )i - R (2 + wn2 t 2 ) j + c(2 sin w n t + wn t cos wn t )k ùûú
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2105
PROBLEM 11.180* (CONTINUED)
The angle a formed by the vector ( v ´ a ) and the y axis is found from
cos a =
Where
(v ´ a) ⋅ j
(v ´ a) j
j =1
(v ´ a ) ⋅ j = -w n R 2 (2 + wn2 t 2 )
2
é
(v ´ a) = wn R êëc 2 (2 cos wn t - wn t sin wn t )2 + R 2 (2 + wn2 t 2 )
+c 2 (2 sin wn t + wn t cos wn t )2 ùû
1/ 2
2 ù1/ 2
é
= wn R êëc 2 (4 + wn2 t 2 ) + R 2 (2 + wn2 t 2 ) úû
Then
cos a =
=
-w n R 2 (2 + w n2 t 2 )
2 ù1/ 2
é
w n R êë c 2 (4 + w n2 t 2 ) + R 2 (2 + w n2 t 2 ) úû
-R (2 + w n2 t 2 )
1/ 2
é 2
2 2
2
2 2 2ù
êë c (4 + wn t ) + R (2 + w n t ) úû
The angle b that the osculating plane forms with y axis (see the above diagram) is equal to
b = a - 90
Then
cos a = cos(b + 90) = - sin b
- sin b =
Then
tan b =
-R (2 + w n2 t 2 )
1/ 2
é 2
2 2
2
2 2 2ù
êë c (4 + w n t ) + R (2 + w n t ) úû
R (2 + w n2 t 2 )
c 4 + w n2 t 2
é R (2 + w 2 t 2 ) ù
n
ú ◀
b = tan-1 êê
2 2 ú
êë c 4 + wn t úû
or
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2106
PROBLEM 11.181*
Determine the direction of the binormal of the path described by the particle of Problem 11.96 when
(a) t = 0, (b) t = p / 2 s.
SOLUTION
r = ( At cos t )i + ( A t 2 + 1) j + ( Bt sin t )k
Given:
r - ft, t - s; A - 3, B - 1
First note that e b is given by
eb =
v´a
v´a
Now
r = (3t cos t )i + (3 t 2 + 1) j + (t sin t )k
Then
v=
dr
dt
= 3(cos t - t sin t )i +
3t
t +1
2
t 2 + 1 - t ( t 2 t+1 )
dv
= 3(- sin t - sin t - t cos t )i + 3
a=
j
dt
t2 +1
+ (cos t + cos t - t sin t )k
and
= -3(2 sin t + t cos t )i +
(a)
j + (sin t + t cos t )k
At t = 0 :
3
(t + 1)
2
3/ 2
j + (2 cos t - t sin t )k
v = (3 ft/s) i
a = (3 ft/s2 ) j + (2 ft/s2 )k
Then
v ´a = 3i ´(3j + 2k)
= 3(-2 j + 3k)
and
Then
or
v ´a = 3 (-2)2 + (3)2 = 3 13
3(-2 j + 3k )
1
=
(-2 j + 3k )
3 13
13
2
3
cos qx = 0 cos qy = cos qz =
13
13
eb =
q x = 90 
q y = 123.7 
q z = 33.7 
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2107
◀
PROBLEM 11.181* (CONTINUED)
(b)
At t =
p
s:
2
æ 3p
ö æ 3p
ö
v = -çç ft/s÷÷÷ i + çç
ft/s÷÷ j + (1 ft/s)k
÷÷
2
è2
ø çè p + 4
ø
é
ù
æp
ö
24
ft/s2 ú j - çç ft/s2 ÷÷÷ k
a = -(6 ft/s2 )i + ê 2
3/2
è2
ø
êë (p + 4)
úû
i
Then
v´a = -
3p
2
-6
j
k
3p
(p + 4)1/ 2
24
2
(p + 4)3/ 2
2
1
-
p
2
2ö
é
ù æ
3p 2
24
ú i - çç6 + 3p ÷÷÷ j
=ê
+
ê 2(p 2 + 4)1/ 2 (p 2 + 4)3/ 2 ú èç
4 ø
ë
û
é
ù
36p
18p
+ ê- 2
+
úk
êë (p + 4)3/ 2 (p 2 + 4)1/ 2 úû
= -4.43984 i - 13.40220 j + 12.99459k
and
v ´ a = [(-4.43984)2 + (-13.40220)2 + (12.99459)2 ]1/ 2
= 19.18829
Then
or
1
(-4.43984 i -13.40220 j + 12.99459k )
19.1829
4.43984
13.40220
12.99459
cos qx = cos qy = cos qz =
19.18829
19.18829
19.18829
eb =
q x = 103.4 
q y = 134.3 
q z = 47.4 
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2108
◀
PROBLEM 11.182
Some students are testing their new drone to see if it can safely deliver packages to different departments
on
campus.
Position
data
can
be
approximated
using
the
expressions
x(t ) =-0.0000225t 4 + 0.003t 3 + 0.01t 2 and y(t ) = 300(1 - cos( 40p t )) , where x and y are expressed in
meters and t is expressed in seconds. Knowing that the take-off and landing altitudes are the same, plot the
path of the drone and determine (a) the duration of the flight, (b) its maximum speed in the x-direction, (c)
its maximum altitude and the horizontal distance travelled during the flight.
SOLUTION
(a) Plot x vs t, y vs t and x vs y
The duration is determined by the time to return to zero elevation
p
y(t ) = 300(1 - cos( 40
t )) = 0
t = 80 sec
t = 80 sec ◀
The maximum speed in the x direction is determined by differentiating to get velocity and
acceleration,
x (t ) = -0.0000225t 4 + 0.003t 3 + 0.01t 2
vx (t ) = -(4)0.0000225t 3 + (3)0.003t 2 + (2)0.01t
ax (t ) = -(3)(4)0.0000225t 2 + (2)(3)0.003t + (2)0.01
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2109
PROBLEM 11.182 (CONTINUED)
(b) Set the acceleration to zero to find the time of the maximum velocity
a x (t ) = -(3)(4)0.0000225t 2 + (2)(3)0.003t + (2)0.01 = 0
t = 67.76 sec
Evaluate the velocity at this time to get the max velocity
v x (67.76) = -(4)0.0000225(67.76)3 + (3)0.003(67.76)2 + (2)0.01(67.76) = 14.67
vx max = 14.67 m/s ◀
(c) The max elevation is found by
ymax = 300(1 - (-1)) = 600
ymax = 600 m ◀
The max distance is found by evaluating x(t) at 80 seconds,
xmax = x (80) = 678
xmax = 678 m ◀
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2110
PROBLEM 11.183
A drag car starts from rest and starts down the racetrack with and acceleration defined by a = 50 – 10 t, where
2
a and t are in m/s and s, respectively. After reaching a speed of 125 m/s, a parachute is deployed to help slow
2
down the dragster. Knowing that this deceleration is defined by the relationship a = - 0.02v , where v is the
velocity in m/s, determine (a) the total time from the beginning of the race until the car slows back down to
10 m/s, (b) the total distance the car travels during this time.
SOLUTION
Given:
a1-2 = 50 - 10 t m/s2 for 0 £ v £ 125 m/s
a2-3 = - 0.02 v 2 after v = 125 m/s
Find v(t) as car speeds up
a=
dv
 dv = adt
dt
v(t )
t
ò dv = ò (50 -10t )dt
0
0
v(t ) = 50t - 5t 2
Find time to reach v = 125 m/s
125 = 50t - 5t 2
Rearrange and solve for t:
t22 - 10t2 + 25 = 0
(t2 - 5)2 = 0
t2 = 5 s
Find distance to reach-125 m/s:
dx = vdt
x2
ò
t2
dx =
0
ò (50t - 5t
2
)dt
0
5
5
x2 = 25t 2 - t 3
3 0
x2 = 416.7 m
(a) Find total time to slow down 10 m/s:
a=
dv
dv
 dt =
dt
a
t3
ò
10
dt = -50
5
dv
òv
2
125
10
t3 - 5 =
50
v 125
t 3 = 9.6 s ◀
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2111
PROBLEM 11.183 (CONTINUED)
(b)
Find total distance to slow down to 10 m/s
adx = vdv  dx
x3
v3
x2
v2
ò dx = -50 ò
vdv
a
dv
v
x3 - 416.7 = -50 ln v
10
125
x 3 = 543.0 m ◀
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2112
PROBLEM 11.184
110
A driver is travelling at a speed of 72 km/h in car A
when he looks down to text a friend that he is
running late. Just before he looks down, he is 80 m
from an intersection, and a bicyclist, B, travelling at a
constant speed of 8 m/s is 32 m from that same
intersection. The light turns red during the 3 s text,
and when the driver looks up, he hits the brakes,
2
causing a constant deceleration of 5 m/s . Determine
(a) the distance between the car and the bike when
the bike reaches the center of the intersection, (b) the
velocity that the car appears to have to the cyclist at
this time, (c) where the car stops.
SOLUTION
Initial car velocity
v=72 km/h = 20 m/s
d = vt
Distance of car while texting is
d = 20(3) = 60 m
Time for bike to get to intersection
t = d / v = (32 m) / (8 m/s)=4 sec
So car has 1 second of deceleration until bike gets to intersection and the distance of car
during deceleration
1
s = v0 t - at 2
2
So
And
1
s(1) = 20(1) - (5)12 = 17.5 m
2
Stotal = 60 + 17.5 = 77.5 m
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2113
PROBLEM 11.184 (CONTINUED)
Distance between car and bike when the bike reaches the center of the intersection
100 - 77.5 = 2.5 m
2.5 m ◀
To find the velocity that the car appears to have to the cyclist at this time
vcar = 20 - 5(1) = 15 m/s
vA = vB + vA /B
vA / B = vA - vB
v A / B = 15i - 8(cos q i - sin q j)
= -7.518i + (15 + 2.736)j
v A / B = -7.518i + (15 + 2.736) j m/s ◀
Deceleration time for the car to come to a complete stop
v = v0 - at
0 = 20 - 5t
t = 4.0 s
Braking distance
1
s = v0 t - at 2
2
1
= 20(4) - (5)4 2
2
= 80 - 40 = 40 m
Total distance
60 + 40 = 100 m
100 m◀
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2114
PROBLEM 11.185
66 km/h
A
48 km/h
B
The velocities of commuter trains A
and B are as shown. Knowing that
the speed of each train is constant
and that B reaches the crossing 10
min after A passed through the same
crossing, determine (a) the relative
velocity of B with respect to A,
(b) the distance between the fronts
of the engines 3 min after A passed
through the crossing.
25°
SOLUTION
(a)
v B = v A + v B /A
We have
The graphical representation of this equation is then as shown.
Then
vB2 /A = 66 2 + 482 - 2(66)(48) cos155
or
v B /A = 111.366 km/h
and
48
111.366
=
sin a sin155
or
a = 10.50 
v B / A = 111.4 km /h
(b)
10.50° ◀
First note that
at t = 3 min, A is (66 km/h) ( 603 ) = 3.3 km west of the crossing.
at t = 3 min, B is (48 km/h) ( 607 ) = 5.6 km southwest of the crossing.
Now
rB = rA + rB/A
Then at t = 3 min, we have
v B2 /A = 3.32 + 5.6 2 - 2(3.3)(5.6) cos 25
rB /A = 2.96 km ◀
or
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2115
PROBLEM 11.186
Knowing that slider block A starts from rest and moves to the left with a
2
constant acceleration of 1 ft/s , determine (a) the relative acceleration of
block A with respect to block B, (b) the velocity of block B after 2 s.
A
B
SOLUTION
Given:
a A = 1 ft/s2
From the diagram can write an expression for the length of the cable
L = x A + 3yB + constants
Differentiate:
0 = x A + 3y B
0 = vA + 3vB
Differentiate again:
Find the acceleration of B:
Write accelerations as vectors:
0 = a A + 3aB
1
aB = - aA
3
1
= - ft/s2
3
a A = -1i ft/s2
aB =
(c) Find relative accelerations:
1
j ft/s2
3
aA/ B = a A - aB
1
= -1i - j ft/s2
3
a A / B = 1.054 ft/s2
(d) Find speed at t = 2 s:
18.43 ◀
v B = ( v B )0 + a B t
1
= 0 + (2)
3
2
= m/s
3
vb = 0.667 m/s ◀
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2116
PROBLEM 11.187
A roller-coaster car is
travelling at a speed of 20
m/s when it passes through
point B. At that point, it
enters a concave down
circular section of the track
that has a radius of
curvature of 60 m. After it
reaches point B, the car’s
speed increases at a rate of
2
4 m/s . Determine (a) the
time it takes the car to
reach point D, (b) the x
and y components of the overall acceleration of the car when it reaches D.
SOLUTION
The distance from B to D is found as,
x = rq = (60m)(50)(p / 180) = 52.36 m
For constant acceleration,
1
x = at 2 + v0 t + x0
2
Set initial position to zero, and x=52.36 m to find time,
1 2
4 t + 20t + 0
2
t = 2.154 sec
52.36 =
t = 2.154 sec ◀
Find the velocity at D,
v = at + v0 = 4(2.154) + 20 = 28.62 m/s
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2117
PROBLEM 11.187 (CONTINUED)
Use n-t coordinates to find the acceleration at D,
v2
eˆ
r n
28.622
eˆ n
+
60
+ 13.65eˆ n
a = veˆ t +
a = 4 eˆ t
a = 4 eˆ t
Resolve acceleration from the n-t to x-y coordinates
ay
ax
40°
50°
an
at
a x = at cos(50) - an cos(40)
ax
ax
ay
ay
ay
= 4 cos(50) - 13.65cos(40)
= -7.884
= at sin(50) - an sin(40)
= 4 sin(50) - 13.65sin(40)
= -11.837
ax = -7.88 m/s2 ◀
ay = -11.84 m/s2 ◀
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2118
PROBLEM 11.188
v0
α
14 m
12 m
30 m
A golfer hits a ball with an initial velocity of
magnitude v 0 at an angle a with the
horizontal. Knowing that the ball must clear the
tops of two trees and land as close as possible
to the flag, determine v 0 and the distance d
when the golfer uses (a) a six-iron with
a = 31, (b) a five-iron with a = 2 7  .
d
70 m
10 m
SOLUTION
The horizontal and vertical motions are
x
t cos a
1 2
1
y = (v0 sin a ) t - gt = x tan a - gt 2
2
2
x = ( v0 cos a)t
or
(1)
2( x tan a - y )
g
or
t2 =
At the landing Point C:
yC = 0,
And
xC = ( v0 cos a )t =
(a)
v0 =
(2)
t=
2 v0 sin a
g
2 v02 sin a cos a
g
(3)
a = 31
To clear tree A:
From (2),
From (1),
To clear tree B:
From (2),
From (1),
x A = 30m, yA = 12m
2 (30 tan 31 - 12)
= 1.22851 s2 ,
9.81
30
= 31.58 m/s
( v0 ) A =
1.1084 cos 31
t A2 =
xB = 100 m,
t A = 1.1084 s
yB = 14 m
2 (100 tan 31 - 14 )
= 9.3957 s2 ,
9.81
100
= 38.06 m/s
( v0 )B =
3.0652 cos31
(t B )2 =
The larger value governs,
v0 = 38.06 m/s
From (3),
xC =
t B = 3.0652 s
v0 = 38.1 m/s ◀
(2)(38.06)2 sin 31 cos31
9.81
d = xC - 110
= 130.38 m
d = 2 0 .4 m
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2119
◀
PROBLEM 11.188 (CONTINUED)
(b)
a = 27
By a similar calculation,
t A = 0.81846 s,
(v0 ) A = 41.138 m/s,
t B = 2.7447 s,
(v0 ) B = 40.890 m/s,
v0 = 41.138 m/s
v0 = 41.1 m/s ◀
xC = 139.56 m,
d = 2 9 .6 m
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2120
◀
PROBLEM 11.189
B
A
As the truck shown begins to back up with a constant
acceleration of 4 ft/s2 , the outer section B of its boom starts to
retract with a constant acceleration of 1.6 ft/s2 relative to the
truck. Determine (a) the acceleration of section B, (b) the
velocity of section B when t = 2 s.
50°
SOLUTION
For the truck,
a A = 4 ft/s2
For the boom,
a B/A = 1.6 ft/s2
50
(a) a B = a A + a B/A
Sketch the vector addition.
By law of cosines:
aB2 = a2A + aB2 / A - 2aAaB/A cos50
= 4 2 + 1.62 - 2 (4)(1.6) cos50
aB = 3.214 ft/s2
Law of sines:
sin a =
aB/A sin 50 1.6 sin 50
=
= 0.38131
aB
3.214
a = 22.4,
aB = 3.21 ft/s2
2 2 .4 
◀
2 2 .4 
◀
v B = (vB )0 + aB t = 0 + (3.214)(2)
(b)
vB = 6.43 ft/s2
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2121
PROBLEM 11.190
B
A velodrome is a specially designed track used in
bicycle racing. If a rider starts from rest and accelerates
between points A and B according to the relationship
2
2
at = (11.46 – 0.01878 v ) m/s , what is her acceleration
at point B?
28 m
.5
18
m
A
SOLUTION
at = (11.46 - 0.01878v2 ) m/s2
Given:
Distance traveled between points A and B:
sB = 18.5 + rqm
= 18.5 + 28 *
p
m
2
= 62.48 m
Find velocity at point B:
ads = vdv  ds =
vB
62.48
ò
ds =
0
vdv
a
vds
ò 11.46 - 0.01878v
2
0
v
62.48 =
B
1
1n (11.46 - 0.01878v 2 )
-(2)0.01878
0
-2.347 = ln (11.46 - 0.01878vB2 ) - ln(11.46)
e0.09211 = 11.46 - 0.01878vB2
vB = 23.49 m/s
Acceleration:
a B = at et +
vB2
en
r
= (11.46 - 0.01878vB2 ) et +
(23.49)2
28
en
aB =1.097et +19.71en m/s2 ◀
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2122
PROBLEM 11.191
v0
A
α
Sand is discharged at A from a conveyor belt and falls onto the
top of a stockpile at B. Knowing that the conveyor belt forms
an angle a = 2 5  with the horizontal, determine (a) the speed
v0 of the belt, (b) the radius of curvature of the trajectory
described by the sand at Point B.
18 ft
B
30 ft
SOLUTION
The motion is projectile motion. Place the origin at Point A. Then x0 = 0 and y0 = 0.
The coordinates of Point B are x B = 30 ft and yB = -18 ft.
Horizontal motion:
Vertical motion:
vx = v0 cos 25
(1)
x = v0 t cos 25
(2)
vy = v0 sin25- gt
(3)
1
y = v0 sin 25 - gt 2
2
(4)
At Point B, Eq. (2) gives
v0 t B =
xB
30
=
= 33.101 ft
cos 25 cos 25
Substituting into Eq. (4),
1
-18 = (33.101)(sin 25) - (32.2) t B2
2
t B = 1.40958 s
(a)
Speed of the belt.
v0 =
v0 t B
33.101
=
= 23.483
1.40958
tB
v0 = 23.4 ft/s ◀
Eqs. (1) and (3) give
vx = 23.483cos 25 = 21.283 ft/s
vy = (23.483)sin 25 - (32.2)(1.40958) = -35.464 ft/s
tan q
-vy
vx
= 1.66632
q = 59.03
v = 41.36 ft/s
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2123
PROBLEM 11.191 (CONTINUED)
Components of acceleration.
a = 32.2 ft/s2
at = 32.2 sin q
an = 32.2cos q = 32.2cos59.03 =16.57 ft/s2
(b)
Radius of curvature at B.
v2
an = r
2
( 41.36 )
v2
r= a =
n
16.57
r = 103.2 ft ◀
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2124
PROBLEM 11.192
B
A
The end Point B of a boom is originally 5 m from fixed Point A
when the driver starts to retract the boom with a constant radial
acceleration of r = -1.0 m/s2 and lower it with a constant
angular acceleration q = -0.5 rad/s2 . At t = 2 s, determine
(a) the velocity of Point B, (b) the acceleration of Point B,
(c) the radius of curvature of the path.
60°
SOLUTION
r0 = 5 m, r0 = 0, r = -1.0 m/s2
Radial motion.
1 2
 = 5 + 0 - 0.5t 2
r = r0 + r0 t + rt
2
 - 0 - 1.0t
r = r0 + rt
r = 5 - (0.5)(2)2 = 3 m
r = (-1.0)(2) = -2 m/s
At t = 2 s,
q0 = 60 =
Angular motion.
p
rad, q0 = 0, q = -0.5 rad/s2
3
p
1
q = q0 + q0 + qt 2 = + 0 - 0.25t 2
2
3
q = q0 + qt = 0 - 0.5t
p
+ 0 - (0.25)(2)2 = 0.047198 rad = 2.70
3
q = -(0.5)(2) = -1.0 rad/s
q=
At t = 2 s,
Unit vectors er and eq .
(a)
Velocity of Point B at t = 2 s.
v B = rer + rqeq
= -(2 m/s)er + (3 m)( - 1.0 rad/s)e q
v B = (-2.00 m/s) er + (-3.00 m/s) eq ◀
tan a =
vq -3.0
=
= 1.5
vr -2.0
a = 56.31
v = vr2 + vq2 = (-2)2 + (-3)2 = 3.6055 m/s
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2125
PROBLEM 11.192 (CONTINUED)
Direction of velocity.
et =
v -2er - 3eq
=
= -0.55470er - 0.83205eq
3.6055
v
q + a = 2 .7 0 + 5 6 .3 1  = 5 9 .0 1 
v B = 3.61 m/s
(b)
59.0° ◀
Acceleration of Point B at t = 2 s.
a B = (r - rq 2 )er + (rq + 2rq)eq
= [-1.0 - (3)(-1)2 ]er + [(3)(-0.5) + (2)(-1.0)(-0.5)]eq
a B = (-4.00 m/s2 ) er + (2.50 m/s2 ) eq ◀
tan b =
aq
2.50
=
= -0.625
-4.00
ar
b = -32.00
2
2
a = ar2 + aq2 = (-4) + (2.5) = 4.7170 m/s2
q + b = 2.70 - 32.00 = -29.30
aB = 4.72 m/s2
Tangential component:
29.3° ◀
a t = (a ⋅ e t ) e t
a t = (-4er + 2.5eq ) ⋅ (-0.55470er - 0.83205eq ) et
= [(-4)(-0.55470) + (2.5)(-0.83205)]e t
= (0.138675 m/s2 ) et = 0.1389 m/s2
Normal component:
59.0°
an = a - at
a n = -4er + 2.5eq - (0.138675)(-0.55470er - 0.83205eq )
= (-3.9231 m/s2 ) er + (2.6154 m/s2 ) eq
2
an = (3.9231)2 + (2.6154) = 4.7149 m/s2
(c)
Radius of curvature of the path.
an =
v2
r
r=
v2 (3.6055 m/s2 )
=
an
4.7149 m/s
r = 2.76 m ◀
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2126
PROBLEM 11.193
A telemetry system is used to quantify
kinematic values of a ski jumper immediately
before she leaves the ramp. According to the
r = -105 ft/s,
system
r = 500 ft,
2


r = -10 ft/s , q = 25, q = 0.07 rad/s,
q = 0.06 rad/s2 , Determine (a) the velocity
of the skier immediately before she leaves
the jump, (b) the acceleration of the skier at
this instant, (c) the distance of the jump d
neglecting lift and air resistance.
30°
10 ft
r
θ
d
SOLUTION
(a)
(r = 500 ft, q = 25)
Velocity of the skier.
v = vr e r + vq e q = re r + r qe q
= (-105 ft/s)er + (500 ft)(0.07 rad/s)e q
v = (-105 ft/s) er + (35 ft/s) eq ◀
Direction of velocity:
v = (-105cos 25 - 35cos65) i + (35sin 65 -105sin 25) j
= (-109.95 ft/s) i + (-12.654 ft/s) j
tan a =
vy
vx
=
-12.654
-109.95
a = 6.565
2
2
v = (105) + (35) = 110.68 ft/s
v = 1 1 0 .7 f t/s
6.57° ◀
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2127
PROBLEM 11.193 (CONTINUED)
(b)
Acceleration of the skier.
a = ar er + aq eq = (r - rq 2 )er + (rq + 2rq)eq
ar = -10 - (500)(0.07)2 = -12.45 ft/s2
aq = (500)(0.06) + (2)(-105)(0.07) = 15.30 ft/s2
a = (-12.45 ft/s2 ) er + (15.30 ft/s2 ) eq ◀
a = (-12.45)(i cos25 + j sin 25) + (15.30)(-i cos65 + j sin 65)
= (-17.750 ft/s2 ) i + (8.6049 ft/s2 ) j
tan b =
ay
ax
=
8.6049
-17.750
b = -25.9
2
2
a = (12.45) + (15.30) = 19.725 ft/s2
a = 19.73 ft/s2
(c)
25.9° ◀
Distance of the jump d.
Projectile motion. Place the origin of the xy-coordinate system at the end of the ramp with the
x-coordinate horizontal and positive to the left and the y-coordinate vertical and positive downward.
Horizontal motion: (Uniform motion)
x0 = 0
x 0 = 109.95 ft/s
(from part a )
x = x0 + x 0 t = 109.95t
Vertical motion:
(Uniformly accelerated motion)
y0 = 0
y 0 = 12.654 ft/s
y = 32.2 ft/s
(from part a )
2
y = y0 + y 0 t +
1 2
 = 12.654t -16.1t 2
yt
2
At the landing point,
x = d cos 30
(1)
y = 10 + d sin30 or y -10 = d sin30
(2)
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2128
PROBLEM 11.193 (CONTINUED)
Multiply Eq. (1) by sin 30° and Eq. (2) by cos 30° and subtract
x sin 30 - ( y -10) cos30 = 0
(109.95t ) sin30 - (12.654t + 16.1t 2 -10) cos30 = 0
-13.943t 2 + 44.016t + 8.6603 = 0
t = -0.1858 s
and
3.3427 s
Reject the negative root.
x = (109.95 ft/s)(3.3427 s) = 367.53 ft
d=
x
cos30
d = 424 ft ◀
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2129
PROBLEM 11.CQ1
B
A bus travels the 100 miles between A and B at 50 mi/h and then another 100
miles between B and C at 70 mi/h. The average speed of the bus for the
entire 200-mile trip is:
(a) more than 60 mi/h
A
C
(b) equal to 60 mi/h
(c) less than 60 mi/h
SOLUTION
The time required for the bus to travel from A to B is 2 h and from B to C is 100/70 = 1.43 h, so the total time
is 3.43 h and the average speed is 200/3.43 = 58 mph.
Answer: (c) ◀
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2130
PROBLEM 11CQ2
Position
Two cars A and B race each other down a straight
road. The position of each car as a function of time
is shown. Which of the following statements are
true (more than one answer can be correct)?
B
A
(a) At time t2 both cars have traveled the same
distance
(b) At time t1 both cars have the same speed
t1
t2
time
(c) Both cars have the same speed at some time t < t1
(d) Both cars have the same acceleration at some
time t < t1
(e) Both cars have the same acceleration at some
time t1 < t < t2
SOLUTION
The speed is the slope of the curve, so answer c) is true.
The acceleration is the second derivative of the position. Since A’s position increases linearly the second
derivative will always be zero. The second derivative of curve B is zero at the pont of inflection which occurs
between t1 and t2.
Answers: (c) and (e) ◀
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2131
PROBLEM 11.CQ3
A
B
Two model rockets are fired simultaneously from a ledge and follow
the trajectories shown. Neglecting air resistance, which of the rockets
will hit the ground first?
(a) A
h
(b) B
(c) They hit at the same time.
(d) The answer depends on h.
SOLUTION
The motion in the vertical direction depends on the initial velocity in the y-direction. Since A has a larger
initial velocity in this direction it will take longer to hit the ground.
Answer: (b) ◀
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2132
PROBLEM 11.CQ4
Ball A is thrown straight up. Which of the following statements about the ball are true at the
highest point in its path?
h
v0
A
(a) The velocity and acceleration are both zero.
y
(b) The velocity is zero, but the acceleration is not zero.
(c) The velocity is not zero, but the acceleration is zero.
(d) Neither the velocity nor the acceleration are zero.
SOLUTION
At the highest point the velocity is zero. The acceleration is never zero.
Answer: (b) ◀
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2133
PROBLEM 11.CQ5
Ball A is thrown straight up with an initial speed v0 and reaches a maximum elevation h before falling back
down. When A reaches its maximum elevation, a second ball is thrown straight upward with the same initial
speed v0. At what height, y, will the balls cross paths?
(a)
y=h
(b)
y > h/2
(c)
y = h/2
(d)
y < h/2
(e)
y=0
SOLUTION
When the ball is thrown up in the air it will be constantly slowing down until it reaches its apex, at which
point it will have a speed of zero. So, the time it will take to travel the last half of the distance to the apex will
be longer than the time it takes for the first half. This same argument can be made for the ball falling from the
maximum elevation. It will be speeding up, so the first half of the distance will take longer than the second
half. Therefore, the balls should cross above the half-way point.
Answer: (b) ◀
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2134
PROBLEM 11.CQ6
Two cars are approaching an intersection at constant speeds as shown. What velocity will
car B appear to have to an observer in car A?
(a)
(b)
(c)
(d )
(e)
SOLUTION
Since vB = vA+ vB/A we can draw the vector triangle and see
v B = v A + v B/A
Answer: (e) ◀
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2135
PROBLEM 11.CQ7
Blocks A and B are released from rest in the positions shown. Neglecting friction between all surfaces, which
figure below best indicates the direction a of the acceleration of block B?
SOLUTION



Since aB = a A + aB / A we get
Answer: (d ) ◀
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2136
PROBLEM 11.CQ8
The Ferris wheel is rotating with a constant angular velocity w. What is
the direction of the acceleration of Point A?
(a)
(b)
(c)
A
(d )
(e) The acceleration is zero.
SOLUTION
The tangential acceleration is zero since the speed is constant, so there will only be normal acceleration
pointed upwards.
Answer: (b) ◀
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2137
C
v
B
A
PROBLEM 11.CQ9
A racecar travels around the track shown at a constant speed. At
which point will the racecar have the largest acceleration?
(a) A
(b) B
D
(c) C
(d ) The acceleration will be zero at all the points.
SOLUTION
The tangential acceleration is zero since the speed is constant, so there will only be normal acceleration. The
normal acceleration will be maximum where the radius of curvature is a minimum that is at Point A.
Answer: (a) ◀
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2138
PROBLEM 11.CQ10
u
A
A child walks across merry-go-round A with a constant speed u relative to
A. The merry-go-round undergoes fixed axis rotation about its center with a
constant angular velocity w counterclockwise.When the child is at the
center of A, as shown, what is the direction of his acceleration when viewed
from above.
(a)
ω
(b)
(c)
(d )
(e) The acceleration is zero.
SOLUTION
Polar coordinates are most natural for this problem, that is,

a = (r - rq 2 )er + (rq + 2rq )eq
(1)
From the information given, we know r = 0, q = 0, r = 0, q = w, r = -u. When we substitute
these values into (1), we will only have a term in the -q direction.
Answer: (d ) ◀
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2139