PROBLEM 11.1
A snowboarder starts from rest at the top of a double black diamond hill. As he rides down the slope, GPS
coordinates are used to determine his displacement as a function of time: x= 0.5t3 + t2 + 2t where x and t are
expressed in meters 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
dv
 3t  2
dt
Acceleration:
a
At t  5 s,
x  0.5 5  52  2  5
x  97.5 m 
v  1.5 5  2  5  2
v  49.5 m/s 
a  35  2 
a  17 m/s2 
3
2
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PROBLEM 11.2
The motion of a particle is defined by the relation x = 2t 3 − 9t 2 + 12t + 10, where x and t are expressed in
meters and seconds, respectively. Determine the time, the position, and the acceleration of the particle when
v = 0.
SOLUTION
x = 2t 3 − 9t 2 + 12t + 10
Differentiating,
v=
a=
dx
= 6t 2 − 18t + 12 = 6(t 2 − 3t + 2)
dt
= 6(t − 2)(t − 1)
dv
= 12t − 18
dt
So v = 0 at t = 1 s and t = 2 s.
At t = 1 s,
x1 = 2 − 9 + 12 + 10 = 15
a1 = 12 − 18 = −6
t = 1.000 s W
x1 = 15.00 m W
a1 = −6.00 m/s 2 W
At t = 2 s,
x2 = 2(2)3 − 9(2) 2 + 12(2) + 10 = 14
t = 2.00 s W
x2 = 14.00 m W
a2 = (12)(2) − 18 = 6
a2 = 6.00 m/s 2 W
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4
PROBLEM 11.3
The vertical motion of mass A is defined by the relation x  10 sin 2t  15cos 2t  100,
where x and t are expressed in mm and seconds, respectively. Determine (a) the position,
velocity and acceleration of A when t 1 s, (b) the maximum velocity and acceleration of A.
SOLUTION
x  10sin 2t  15cos 2t  100
v
dx
 20cos 2t 30sin 2t
dt
a
dv
 40sin 2t 60cos 2t
dt
For trigonometric functions set calculator to radians:
(a) At t  1 s.
x1  10sin 2  15cos2  100  102.9
x1  102.9 mm 
v1  20cos2 30sin 2  35.6
v1  35.6 mm/s 
a1  40sin 2 60cos2  11.40
a1  11.40 mm/s 2 
(b) Maximum velocity occurs when a  0.
60 cos 2t  0
40sin 2t
tan 2t 
60
 1.5
40
2t  tan 1 ( 1.5)  0.9828 and 0.9828 
Reject the negative value. 2t  2.1588
t  1.0794 s
t  1.0794 s for vmax
so
vmax  20cos(2.1588) 30sin(2.1588)
vmax  36.1 mm/s 
 36.056
Note that we could have also used
vmax  202  302  36.056
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  402  602  72.1 mm/s 2
amax  72.1 mm/s 2 
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PROBLEM 11.4
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
4.8t
sin16t where x and t
defined by the relation x  60e
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.
SOLUTION
x  60e 4.8t sin16t
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,
(b) At t  0.3 s,
x0  0
x0  0 mm 
v0  960 mm/s
v0  960 mm/s

a0  9216 mm/s 2
a0  9220 mm/s 2

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  cos 4.8  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/s 2 
or 3.11 m/s
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2

PROBLEM 11.5
The motion of a particle is defined by the relation x  6t 4 2t 3 12t 2  3t  3, where x and t are expressed in
meters and seconds, respectively. Determine the time, the position, and the velocity when a  0.
SOLUTION
We have
x  6t 4 2t 3 12t 2  3t  3
Then
v
dx
 24t 3 6t 2
dt
and
a
dv
 72t 2 12t 24
dt
When a  0:
72t 2 12t
24  12(6t 2
or
t
At t 
2
s:
3
2)  0
t
2)(2t  1)  0
(3t
or
24t  3
2
1
s and t 
s (Reject)
3
2
x2/3  6
2
3
v2/3  24
2
3
4
2
2
3
6
2
3
3
3
12
2
24
2
3
2
3
2
3
3
t  0.667 s 
2
3
3
or
x2/3  0.259 m 
or v2/3  8.56 m/s 
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PROBLEM 11.6
The motion of a particle is defined by the relation x = t 3 - 9t 2 + 24t - 8, where x and t are expressed in meters
and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total distance
traveled when the acceleration is zero.
SOLUTION
We have
x = t 3 - 9t 2 + 24t - 8
Then
v=
dx
= 3t 2 - 18t + 24
dt
and
a=
dv
= 6 t - 18
dt
(a)
3 t 2 - 18t + 24 = 3(t 2 - 6t + 8) = 0
When v = 0:
(t - 2)(t - 4) = 0
or
or
(b)
When a = 0:
t = 2.00 s and t = 4.00 s !
6t - 18 = 0 or t = 3 s
x3 = (3)3 - 9(3)2 + 24(3) - 8
At t = 3 s:
or
x3 = 10.00 m !
v>0
First observe that 0 ≤ t < 2 s:
v<0
2 s < t ≤ 3 s:
Now
At t = 0:
x0 = -8 m
At t = 2 s:
x2 = (2)3 - 9(2)2 + 24(2) - 8 = 12 m
Then
x2 - x0 = 12 - ( -8) = 20 m
| x3 - x2 | = |10 - 12 | = 2 m
Total distance traveled = (20 + 2) m = 22.0 m
!
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8
PROBLEM 11.7
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 defined by the relation x(t) = 0.5t3 - 3t2 + 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.
SOLUTION
Position:
x  t   0.5t 3 3t 2  3t  2
Velocity:
v t  
dx
dt
v  t   1.5t 2
(a) Time when v=0
0  1.5t 2 6t  3
t
Acceleration:
6t  3
6
a t  
62
4 1.5  3
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
t2s
(b) Position at t=2 s
x  2   0.5  2 
3
3  2  3 2  2
2
x  2  0 m 
To find total distance note that car changes direction at t=0.586 s
Position at t=0 s
x  0   0.5  0 
3
3 0  3 0  2
2
x  0  2
Position at t=0.586 s
x  0.586   0.5  0.586 
3
3  0.586   3 0.586  2
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  0.828 m  2.828 m
Total distance  3.656 m 
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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..
3
2
SOLUTION
Position:
xt   t2
Velocity:
v t  
t
2
3
dx
dt
v  t   2t 3  t 2 

v  t   2t 3 t 2
2
4t  4

 3t 2  14t 12
Time when v(t)=0
0  3t 2  14t 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
x 1.131  1.131
1.1.31
2
2
3
x 1.131  1.935 m.
Position at t=3.535 s
x  3.535    3.535 
 3.535
2
2
3
x  3.531  8.879 m.
To find total distance traveled note that the particle changes direction at t=1.131 s and again at t=3.535 s.
Position at t=0 s
x  0   0
2
0
2
x  4   4
4
2
3
x  0  8 m
Position at t=4 s
2
3
x  4  8 m
(b)
Distances traveled:
From t=0 to t=1.131 s:
x 1.131 x  0   6.065 m.
From t=1.131 to t=3.531 s:
x  3.535  x 1.131  6.944 m.
From t=3.531 to t=4 s:
x  4  x  3.535   0.879 m.
Total distance traveled  6.065 m  6.944 m  0.879 m
Total distance  13.888 m.
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10
PROBLEM 11.9
The brakes of a car are applied, causing it to slow down at a
rate of 3 m/s2. Knowing that the car stops in 100 m, determine
(a) how fast the car was traveling immediately before the
brakes were applied, (b) the time required for the car to stop.
SOLUTION
a = −3 m/s 2
(a)
Velocity at x = 0.
v
dv
= a = −3
dx
0
vdv = −
v0
0−
xf
(−3)dx
0
v02
= −3x f = −(3)(100)
2
v0 = 24.5 m/s 2 W
v02 = 600
(b)
Time to stop.
dv
= a = −3
dx
0
dv = −
v0
tf
−3dt
0
0 − v0 = −3t f
tf =
v0
24.5
=
3
3
t f = 8.17 s W
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11
PROBLEM 11.10
The acceleration of a particle is defined by the relation a  3e 0.2t , where a andt are expressed in m/s2 and
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
Acceleration:
a  3e 0.2t ft/s2
Given:
v0  0 m/s, x0  0 m
Velocity:
a
dv
dt
v
dv  adt
t
dv  adt
v0
0
t
3e 0.2t dt
v vo 
0

t
v 0= 15 e 0.2t
0

v  15 1 e 0.2t m/s
Position:
v
dx
dt
dx  vdt
x
t
dx  vdt
x0
0
t
x

0

v  15 1 e
x  15 t  5 e 0.2t
Velocity at t=0.5 s
Position at t=0.5 s

xo  15 1 e 0.2t dt

0.2 0.5
 75 m
 m/s
x  15 0.5  5 e 0.2 0.5
 75 m
x

0=15 t  5 e 0.2t

t
0
v  0.5  1.427 m/s 
x  0.5  0.363 m. 
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12
PROBLEM 11.11
The acceleration of a particle is directly proportional to the square of the time t. When t  0, the particle is
at x  24 m. Knowing that at t  6 s, x  96 m and v  18 m/s, express x and v in terms of t.
SOLUTION
We have
a  kt 2
Now
dv
 a  kt 2
dt
At t  6 s, v  18 m/s:
v
18
dv 
t
6
k  constant
kt 2 dt
1
v 18  k (t 3
3
or
216)
or
1
v  18  k (t 3
3
Also
dx
1
 v  18  k (t 3
dt
3
At t  0, x  24 m:
x
24
dx 
216)(m/s)
216)
1
18  k (t 3 216) dt
0
3
t
1 1
x 24  18t  k t 4
3 4
or
216t
Now
At t  6 s, x  96 m:
1 1
96 24  18(6)  k (6)4
3 4
1
m/s 4
9
or
k
Then
x 24  18t 
or
x(t ) 
and
or
1 1
3 9
1 4
t
4
216t
1 4
t  10t  24
108
v  18 
v(t ) 
216(6)

1 1 3
(t 216)
3 9
1 3
t  10
27
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
PROBLEM 11.12
2
The acceleration of a particle is defined by the relation a  kt . (a) Knowing that v  8 m/s when t0
and that v  8 m/s when t  2 s, determine the constant k. (b) Write the equations of motion, knowing also
that x  0 when t  2 s.
SOLUTION
a  kt 2
(1)
dv
 a  kt 2
dt
t  0, v  8 m/s and t  2 s, v  8 ft/s
8
(a)
8
dv 
2
0
kt 2 dt
1
8 ( 8)  k (2)3
3
(b)
k  6.00 m/s4 
Substituting k  6 m/s 4 into (1)
dv
 a  6t 2
dt
t  0, v  8 m/s:
v
8
dv 
t
0
a  6t 2 
6t 2 dt
1
v ( 8)  6(t )3
3
v  2t 3 8 
dx
 v  2t 3 8
dt
t  2 s, x  0:
x
0
dx 
t
2
t
(2t
3
1
8) dt; x  t 4 8t
2
2
1 4
1 4
t 8t
(2)
2
2
1
x  t 4 8t 8  16
2
x
8(2)
1
x  t 4 8t  8 
2
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14
PROBLEM 11.13
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/s2and 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.
SOLUTION
Acceleration:
a  1.8sin kt m/s 2
Given:
v0  0.6 m/s, x0  0,
Velocity:
a
k  3 rad/s
v
dv
dt
dv  adt
v
v0 
t
a dt 
0
v
0.6 
1.8
(cos kt
3
t
dv  adt
v0
0
t
1.8 0 sin kt dt 
t
1.8
cos kt
0
k
1)  0.6cos kt
0.6
v  0.6 cos kt m/s
Position:
x
t
dx
dt
dx  vdt
x
x0 
t
0.6
t
t
v dt  0.6 0 cos kt dt 
sin kt
0
0
x
0
v
dx  vdt
x0
0
k
0.6
(sin kt
3
0)  0.2sin kt
x  0.2sin kt m
When t =0.5 s,
kt  (3)(0.5)  1.5 rad
v  0.6 cos1.5  0.0424 m/s
v  42.4 mm/s 
x  0.2 sin1.5  0.1995 m
x  199.5 mm 
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15
PROBLEM 11.14
For the scotch yoke mechanism shown, the acceleration of Point A is defined by the
relation a  1.08 sin kt 1.44 cos kt , where a and t are expressed in m/s2 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.
SOLUTION
1.44 cos kt m/s 2
Acceleration:
a  1.08sin kt
Given:
v0  0.36 m/s, x0  0.16,
Velocity:
a
v
dv  adt
1.44
t
cos kt dt
0
0.36 
t
1.08
cos kt
0
k

1.08
(cos 3t
3
1)
1.44
(sin 3t
3
 0.36 cos 3t
0.36
0.48sin 3t
Evaluate at t  0.5 s
 0.36 cos1.5
Position:
v
x
0
t
sin kt dt
0
v  0.36 cos 3t
x
t
dv  adt
v0
v0  1.08
v
Integrate:
v
dv
dt
k  3 rad/s
0.36
t
1.44
sin kt
0
k
0.48sin 3t m/s
dx  vdt
0.16 
t
dx  vdt
x0
x0 
v  453 mm/s 
0.48sin1.5
x
dx
dt
0)
t
v dt  0.36
0
0
t
cos kt dt
0
0.48
t
sin kt dt
0
t
t
0.36
0.48
sin kt 
cos kt
0
0
k
k
0.36
0.48
(sin 3t 0) 
(cos 3t
3
3
 0.12sin 3t  0.16 cos 3t 0.16

1)
x  0.12sin 3t  0.16 cos 3t m
Evaluate at t  0.5 s
x  0.12sin1.5  0.16 cos1.5  0.1310 m
x  131.0 mm 
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16
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 20 mm, determine
the maximum acceleration of the equipment.
SOLUTION
a
vdv
 kx
dx
Separate and integrate.
vf
v0
1 2
vf
2
vdv 
1 2
v0 
2
xf
0
k x dx
1 2
kx
2
xf

0
1 2
kxf
2
Use v0  4 m/s, x f  0.02 m, and v f  0. Solve for k.
0
1 2
1
(4) 
k (0.02)2
2
2
k  40,000 s 2
Maximum acceleration.
amax  kxmax : ( 40, 000)(0.02)  800 m/s 2
a  800 m/s 2 
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17
PROBLEM 11.16
A projectile enters a resisting medium at x  0 with an initial velocity
v0  900 ft/s and travels 4 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 3.9 in. into the
resisting medium.
SOLUTION
First note
When x 
4
ft, v  0:
12
0  (900 ft/s) k
k  2700
or
(a)
4
ft
12
1
s
We have
v  v0 kx
Then
a
or
a  k (v0 kx)
At t  0:
1
a  2700 (900 ft/s 0)
s
dv d
 (v0
dt dt
kx)  kv
a0  2.43 106 ft/s 2 
or
(b)
dx
 v  v0
dt
We have
t
dx
 dt
0 v
0
0 kx
x
At t  0, x  0:
or
kx
1
[ln(v0
k
or
kx)]0x  t
t
v0
1
1
1
ln
 ln
k
v0 kx
k
1 vk x
0
When x  3.9 in.:
t
1
ln
1
2700 1s
1
2700 1/s
900 ft/s
 3.912 ft 
t  1.366 10 3 s 
or
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18
PROBLEM 11.17
The acceleration of a particle is defined by the relation a = - k /x. It has been experimentally determined that
v = 5 m /s when x = 0.2 m and that v = 3 m /s when x = 0.4 m. Determine (a) the velocity of the particle when
x = 0.5 m, (b) the position of the particle at which its velocity is zero.
SOLUTION
a=
vdv - k
=
dx
x
Separate and integrate using x = 0.2 m, v = 5 m/s.
v
x
dx
Ú5 vdv = -k Ú0.2 x
x
v
1 2
= - k ln x
v
2 0.2
0.2
1 2 1 2
Ê x ˆ
v - (5) = - k ln Á
Ë 0.2 ˜¯
2
2
(1)
When v = 3 m/s, x = 0.4 m
1 2 1 2
Ê 0.4 ˆ
(3) - (5) = - k ln Á
Ë 0.2 ˜¯
2
2
Solve for k.
k = 11.5416 m2 /s2
(a)
Substitute
k = 11.5416 m2 /s2
and
x = 0.5 m into (1).
1 2 1 2
Ê 0.5 ˆ
v - (5) = -11.5416 ln Á
Ë 0.2 ˜¯
2
2
(b)
For v = 0,
1
Ê x ˆ
0 - (5)2 = -11.5416 ln Á
Ë 0.2 ˜¯
2
Ê x ˆ
ln Á
= 1.083
Ë 0.2 ˜¯
v = 1.962 m/s !
x = 0.591 m !
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19
PROBLEM 11.18
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
of block B is a  9.81  k /x 2 , where a and x are expressed in m/s2 and m,
4 3 2
respectively, and k  4 10 m /s . Determine the maximum velocity and
acceleration of B.
SOLUTION
0  9.81 
The maximum velocity occurs when a  0.
xm2 
k
4 10 4

 40.775 10 6 m2
9.81
9.81
k
xm2
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 
v
0
vdv  9.81
x
x0
k dx
x2
dx  k
x dx
x0 x 2
1 2
1
v  9.81( x x0 ) k
2
x
1 2
vm  9.81( xm
2
x0 ) k
1
x0
1
xm
1
x0
 9.81(0.0063855 0.004) (4 10 4 )
1
0.0063855
1
0.004
 0.023402  0.037358  0.013956 m 2 /s 2
Maximum velocity:
vm  0.1671m/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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20
PROBLEM 11.19
Based on experimental observations, the acceleration of a particle is defined by the relation a  (0.1  sin x/b),
where a andx are expressed in m/s2 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:
dv
a
dx
v
x
1
vdv 
0.1  sin
0.1  sin
0
1 2
(v 1) 
2
or
x
0.8
x
dx
0.8
x
0.1x 0.8 cos
x
0.8 0
1 2
x
v  0.1x  0.8 cos
0.3
2
0.8
or
(a)
v
1 2
1
v  0.1( 1)  0.8 cos
0.3
2
0.8
When x  1 m:
v  0.323 m/s 
or
(b)
When v  vmax, a  0:
or
(c)
0.1  sin
x
0
0.8
x  0.080 134 m
x  0.0801 m 
When x  0.080 134 m:
1 2
0.080134
vmax  0.1( 0.080134)  0.8 cos
2
0.8
2 2
 0.504 m /s
0.3
vmax  1.004 m/s 
or
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21
PROBLEM 11.20
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
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 of
the collar as it passes through Point C.
SOLUTION
av
Since a is function of x,
dv
 100 x
dx
lx
2
l  x2
Separate variables and integrate:
vf
v0
vdv  100
0
x0
1 2
vf
2
1 2
x2
v0  100
2
2
1 2
vf
2
0  100
x
x02
2
lx
dx
2
l  x2
0
2
l l x
2
x0
l 2  l l 2  x02
1 2 100
vf 
( l 2  x02 l 2 2l l 2  x02 )
2
2
100

( l 2  x02 l ) 2
2
v f  10( l 2  x02
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22
l) 
PROBLEM 11.21


The acceleration of a particle is defined by the relation a  k 1 e x , 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.
SOLUTION

e x

Acceleration:
a k 1
Given:
at x  3 m, v  9 m/s
at x  0 m, 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


v
k 1 e x dx  vdv
3
9

k xe x
Velocity:

k xe x

x
v
1
 v2
2 9
3
 3  e   12 v
3
2
1 2
(9)
2
(1)
(a) Now substitute v=0 m/s and x=0 m into (1) and solve for k

k 0e 0
 3  e   12 0
3
2
1 2
(9)
2
k  2.52 m2 /s2 
(b) Find velocity when x= -2 m using the equation (1) and the value of k

2.518 2  e2
 3  e   12 v
3
2
1 2
(9)
2
v  4.70 m/s
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23
PROBLEM 11.22
Starting from x = 0 with no initial velocity, a particle is given an acceleration a = 0.1 v 2 + 49 , where a and
v are expressed in m /s2 and m /s, respectively. Determine (a) the position of the particle when v = 24 m/s, (b)
the speed of the particle when x = 40 m.
SOLUTION
We have
v
v
dv
= a = 0.1 v 2 + 49
dx
vdv
x
When x = 0, v = 0:
Ú0 v 2 + 49 = Ú0 0.1 dx
or
È v 2 + 49 ˘ = 0.1 x
ÍÎ
˙˚0
or
v 2 + 49 - 7 = 0.1
. x
(a)
v
When v = 24 m/s:
242 + 49 - 7 = 0.1
. x
x = 180 m !
or
(b)
When x = 40 m:
v 2 + 49 - 7 = 0.1(40)
v = 8.49 m/ s !
or
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24
PROBLEM 11.23
A bowling ball is dropped from a boat so that it strikes the surface of a
lake with a speed of 8 m / s. Assuming the ball experiences a downward
acceleration of a = 3 - 0.1 v 2 when in the water, determine the velocity
of the ball when it strikes the bottom of the lake.
SOLUTION
v0 = 8 m/s, x - x0 = 10 m
a = 3 - 0.1v 2 = k (c 2 - v 2 )
k = 0.1 m -1 and c 2 =
where
3
m2
= 30 2
0.1
s
c = 5.4772 m/s
Since v0 > c, write
a=v
dv
= - k ( v 2 - c2 )
dx
vdv
v 2 - c2
Integrating,
= - k dx
v
1
ln( v 2 - c 2 ) = - k ( x - x0 )
2
v0
ln
v 2 - c2
= -2k ( x - x0 )
v02 - c 2
v 2 - c2
= e -2 k ( x - x0 )
2
2
v0 - c
(
)
v 2 = c 2 + v02 - c 2 e -2 k ( x - x0 )
= 30 + [(8)2 - 30] e -2( 0.1)(10 )
= 34.6014 m2 /s2
v = 5.88 m/s !
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25
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
dv
a k v
dx
v
v dv  k dx
so that
When x  0, v  81 m/s:
or
v
v dv 
81
x
k dx
0
2 3/2 v
[v ]81  kx
3
or
2 3/2
[v
3
729]  kx
When x  18 m, v  36 m/s:
2 3/2
(36
3
729)  k (18)
k  19 m/s 2
or
Finally
When x  20 m:
2 3/2
(v
3
729)  19(20)
v3/2  159
or
(b)
We have
At t  0, v  81 m/s:
or
or
When v  0:
v  29.3 m/s 
dv
 a  19 v
dt
v dv
81
v

t
0
19dt
v
2[ v ]81
 19t
2( v
9)  19t
2( 9)  19t
t  0.947 s 
or
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26
PROBLEM 11.25
The acceleration of a particle is defined by the relation a  kv2.5, 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, x  0 mm, v  16 mm/s
at x  6 mm, v  4 mm/s
adx  vdv
2.5
kv dx  vdv
Separate variables
kdx  v 3 2 dv
Integrate using x=-0 m and v=16 mm/s as the lower limits of the integrals
x
v
-kdx  v 3 2 dv
0
16
x
 2v 1 2
kx
0
Velocity and position:
kx  2v 1 2
v
16
1
2
(1)
Now substitute v=4 mm/s and x=6 mm into (1) and solve for k
1
2
k  0.0833 mm 3 2 s1 2
k 6  4 1 2
(a) Find velocity when x= 5 mm using the equation (1) and the value of k
k  5  2v 1 2
1
2
v  4.76 mm/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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27
PROBLEM 11.25 (Continued)
Integrate using t=0 and v=16 mm/s as the lower limits of the integrals
t
v
-kdt  v 5 2 dv
0
16
t
v

kt
0
Velocity and time:
kt 
2 32
v
3
16
2 32
v
3
1
96
(2)
Find time when v= 9 mm/s using the equation (2) and the value of k
kt 
2
 9 3 2
3
1
96
t  0.171 s
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28
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
flying start) using a = A – Cv2, where a is acceleration in m/s2 and v
is the velocity in m/s. From wind tunnel testing, they found that
C= 0.0012 m-1. Knowing that the cyclist is going 100 km/h at the
260 meter mark, what is the value of A?
SOLUTION
Cv2 m/s2
Acceleration:
aA
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
 A Cv  dx  vdv
2
Separate variables
dx 
vdv
A Cv 2
Integrate starting from rest and traveling a distance xf with a final velocity vf.
xf
vf
dx 
0
xf
x0 
Next solve for A
vdv
A Cv 2
0


v
f
1
ln A Cv 2
2C
0


xf 
1
1
ln A v 2f 
ln  A 
2C
2C
xf 
1
A
ln
2C
A Cv 2f
A
Cv 2f e
1 e
2Cx f
2Cx f
A  1.995 m/s2 
Now substitute values of C, xf and vf and solve for A
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29
PROBLEM 11.27
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 v0 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.
SOLUTION
(a)
dv
dx
0.18v0 d 0.18v0

x dx
x
av
We have

When x  2 m:
a
0.0324v02
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:
3
1
xdx 
t3
t1
0.18v0 dt
3
or
1 2
 0.18v0 (t3
x
2
1
or
(t3
t1 ) 
1
(9
2
t1 )
1)
0.18(3.6)
t3 t1  6.17 s 
or
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PROBLEM 11.28
Based on observations, the speed of a jogger can be approximated by the relation
v = 12(1 - 0.06 x )0.3 , where v and x are expressed in km / h and km, respectively. Knowing
that x = 0 at t = 0, determine (a) the distance the jogger has run when t = 1 h, (b) the
jogger’s acceleration in m /s2 at t = 0, (c) the time required for the jogger to run 9 km.
SOLUTION
(a)
dx
= v = 12(1 - 0.06 x )0.3
dt
We have
or
dx
x
t
Ú0 (1 - 0.06 x)0.3 = Ú 012dt
At t = 0, x = 0:
1 Ê
1 ˆ
0.7 x
˜¯ [(1 - 0.06 x ) ]0 = 12t
ÁË 0.7 0.06
1 - (1 - 0.06 x )0.7 = 0.504t
or
or
x=
1
[1 - (1 - 0.504t )1/ 0.7 ]
0.06
At t = 1 h:
x=
1
{1 - [1 - 0.504(1)]1/ 0.7 }
0.06
x = 10.55 km !
or
(b)
We have
(1)
a=v
dv
dx
= 12(1 - 0.06 x )0.3
d
[12(1 - 0.06 x )0.3 ]
dx
= 122 (1 - 0.06 x )0.3 [(0.3)( -0.06)(1 - 0.06 x ) -0.7 ]
= -2.592(1 - 0.06 x )-0.4
At t = 0, x = 0:
a0 =
-2.592 km
h2
¥
1000 m Ê 1 h ˆ
¥Á
Ë 3600 ˜¯
1 km
a0 = -2 ¥ 10 -4 m/s2 !
or
(c)
From Eq. (1)
t=
1
[1 - (1 - 0.06 x )0.7 ]
0.504
When x = 9 km:
t=
1
{1 - [1 - 0.06(9)]0.7 }
0.504
or
2
= 0.832 hrs
t = 49.9 min !
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or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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31
PROBLEM 11.29
The acceleration due to gravity at an altitude y above the surface of the earth can be
expressed as
-9.81
a=
[1 + ( y /6.37 ¥ 106 )]2
where a and y are expressed in m /s2 and m, 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) 540 m /s, (b) 900 m /s, (c) 11,180 m /s.
SOLUTION
We have
v
- 9.81
dv
=a=2
dy
y
Ê
ˆ
+
1
ÁË
˜
6.37 ¥ 106 ¯
y = 0,
When
v = v0
y = ymax, v = 0
0
- 9.81
ymax
Úv v dv = Ú0
Then
0
y
Ê
ˆ
ÁË1 +
˜
6.37 ¥ 106 ¯
2
dy
y
or
È
˘ max
Í
˙
1
1
- v02 = -9.81 Í- 6.37 ¥ 106
˙
y
2
Í
˙
1+
ÍÎ
6.37 ¥ 106 ˙˚0
or
ˆ
Ê
˜
Á
1
v02 = 124.9794 ¥ 106 Á1 ymax ˜
˜
Á 1+
Ë
6.37 ¥ 106 ¯
È
˘
v02
ymax = 6.37 ¥ 106 Í
6
2˙
ÍÎ124.9794 ¥ 10 - v0 ˙˚
or
(a)
v0 = 540 m/s:
È
˘
5402
ymax = 6.37 ¥ 106 ¥ Í
6
2˙
Î124.9794 ¥ 10 - 540 ˚
ymax = 14.90 ¥ 103 m !
or
(b)
v0 = 900 m/s:
È
˘
9002
ymax = 6.37 ¥ 106 Í
6
2˙
Î124.9794 ¥ 10 - 900 ˚
or
ymax = 41.6 ¥ 103 m !
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32
PROBLEM 11.29 (Continued)
(c)
v0 = 11,180 m/s :
È
˘
(11,180)2
ymax = 6.37 ¥ 106 Í
6
2 ˙
Î (124.9794 ¥ 10 - 11,180 ) ˚
ymax = -6.12 ¥ 1010 m !
or
The velocity 11,180 m /s is approximately the escape velocity vR from the earth. For vR
ymax Æ • !
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37
33
PROBLEM 11.30
The acceleration due to gravity of a particle falling toward the earth is a = - gR2 /r 2, 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 = 6370 km 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 = •.)
SOLUTION
We have
v
dv
gR2
=a=- 2
dr
r
r = R, v = ve
When
r = •, v = 0
Then
0
Úv
e
vdv = Ú R
gR2
r2
or
1
È1 ˘
- ve2 = gR2 Í ˙
2
Î r ˚R
or
ve = 2 gR
dr
1000 m ˆ
Ê
= Á 2 ¥ 9.81 m/s2 ¥ 6370 km ¥
˜
Ë
1 km ¯
1/ 2
ve = 11,180 m/s !
or
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34
PROBLEM 11.31
The velocity of a particle is v  v0[1 sin ( t/T )]. Knowing that the particle starts from the origin with an
initial velocity v0 , determine (a) its position and its acceleration at t  3T , (b) its average velocity during the
interval t  0 to t  T .
SOLUTION
(a)
dx
t
 v  v0 1 sin
dt
T
We have
At t  0, x  0:
x
0
dx 
t
0
x  v0 t 
At t  3T :
At t  3T :
(b)
T
x3T  v0 3T 
a
t
T
v0 1 sin
dt
t
t
cos
T
T
 v0 t 
T
T
dv d
t
v0 1 sin

dt dt
T
a3T  v0
T
cos
0
3T
cos
T
t
T
 v0 3T
 v0
T
cos
T
2T
(1)
x3T  2.36 v0T 
t
T
3T
T
cos
a3T 
v0

T
Using Eq. (1)
At t  0:
x0  v0 0 
At t  T :
xT  v0 T 
Now
vave 
xT
x0
t
T
T

cos(0)
cos
T
T
T
0
T
0.363v0T 0
T 0
 v0 T
2T
 0.363v0T
vave  0.363v0 
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35
Problem 11.32
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.
SOLUTION
Constraint:
y  r  e cos
(1)
Differentiate:
y  e  sin
(2)
Differentiate again:

y  e sin
e 2 cos
(3)
ymax occurs when cosθ =1 and ymin occurs when cosθ = -1
d max  ymax
ymin
d max  r  e
r
e
d max  2e
e
d max
2
(4)
yr
Substitute (4) into (1) to get Position
Max Velocity occurs when sin
d max
cos 
2
 1
vmax  e 
  vmax
e
(5)
y  vmax sin
Substitute (5) into (2) to get velocity and assume cw rotation.
Substitute (4) and (5) into (3)

y 
d max 
sin
2
2
d max 4vmax
cos
2
2
d max

y 
Acceleration:
d max 
sin
2
2
2vmax
cos 
d max
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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 and that the
distance AB is 900 m, determine (a) the acceleration
a, (b) the take-off velocity vB .
SOLUTION
Given:
xA  0 , xB  900 m , vA  0 , t  30 s
Uniform Acceleration:
xB  xA  v At 
1 2
at
2
Substitute known values:
900 m  0  0 
1
2
a  30 s 
2
a  2.0 m/s2 
(a) Solve for acceleration
Uniform Acceleration:
vB  vA  at
Substitute known values:
vB  0  2 m/s 2  30 s 


vB  60.0 m/s
(b) Velocity at takeoff (point B)
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37
PROBLEM 11.34
A motorist is traveling at 54 km/h when she
observes that a traffic light 240 m ahead of her
turns red. The traffic light is timed to stay red for
24 s. If the motorist wishes to pass the light
without stopping just as it turns green again,
determine (a) the required uniform deceleration of
the car, (b) the speed of the car as it passes the
light.
SOLUTION
Uniformly accelerated motion:
x0  0
(a)
v0  54 km/h  15 m/s
1
x  x0  v0t  a t 2
2
when t  24s, x  240 m:
240 m  0  (15 m/s)(24 s) 
1
a (24 s) 2
2
a  0.4167 m/s 2
(b)
a  0.417 m/s2 
v  v0  a t
when t  24s:
v  (15 m/s)  ( 0.4167 m/s)(24 s)
v  5.00 m/s
v  18.00 km/h
v  18.00 km/h 
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38
Problem 11.35
Steep safety ramps are built beside mountain highways to
enable vehicles with defective brakes to stop safely. A truck
enters a 225-m ramp at a high speed v0 and travels 160 m 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.
SOLUTION
Given:
xo  0 , xA  160 m , t A  6 s , vA 
Uniform Acceleration:
vA  vo  at A
Substitute known values:
1
vo  vo   a  6 
2
a
1
vo
12
Uniform Acceleration:
xA  xo  vot A 
1 2
at A
2
Substitute known values:
160  0  vo  6
1
2
vo  6
24
vo 
1
vo , Lramp  225 m
2
320
m/s and a 
9
(a)
vB  vo  atB
Substitute known values:
0
Solve for tB
tB  12 s
320
9
80
m/s 2
27
80
tB
27
tB t A  6.0 s 
Additional time to stop
(b)
xB  xo  votB 
Substitute known values:
xB  0 
Solve for xB
xB  213.3 m
1 2
atB
2
320
(12)
9
1 80
122
2 27
xB
Additional distance travelled by truck
xA  53.3 m 
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39
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 27 m 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
maximum altitude and assuming that g = 9.81 m /s2, determine (a) the speed v1 of the
rocket at the end of powered flight, (b) the maximum altitude reached by the rocket.
SOLUTION
1
y = y1 + v1t + at 2
2
(a) We have
At tland ,
y=0
Then
0 = 27 m + v1 (16 s)
1
+ ( -9.81 m/s2 )(16 s)2
2
or
v1 = 76.7925 m/s
(b) We have
v 2 = v12 + 2a( y - y1 )
v1 = 76.8 m/s !
At
y = ymax ,
v=0
Then
0 = (76.7925 m/s)2 + 2( -9.81 m/s2 )( ymax - 27) m
ymax = 327.57 m
or
ymax = 328 m !
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40
PROBLEM 11.37
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/s 2 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
Dwe have
v 2  v02  2a( x
x0 )
2
vBC
 0  2(4.8 m/s2 )(3 0) m
Then, at B
 28.8 m2 /s2
2
vD2  vBC
 2aCD ( xD
and at D
(vBC  5.3666 m/s)
xC )
d  xD
xC
(7.2 m/s) 2  (28.8 m 2 /s 2 )  2(4.8 m/s2 )d
or
d  2.40 m 
or
(b)
For A
Band C
D we have
v  v0  at
Then A
B
5.3666 m/s  0  (4.8 m/s 2 )t AB
t AB  1.118 04 s
or
and C
D 7.2 m/s  5.3666 m/s  (4.8 m/s 2 )tCD
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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41
PROBLEM 11.38
A sprinter in a 100
100-m
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
35 m, a  constant
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
for
0
x 35 m
1
35 m  a(5.4 s)2
2
a  2.4005 m/s 2
a  2.40 m/s 2 
(b)
First note that v  vmax for 35 m
Now
(c)
x 100 m.
v 2  0  2a ( x 0)
for
0
When x  35 m:
2
vmax
 2(2.4005 m/s 2 )(35 m)
or
vmax  12.9628 m/s
We have
When x  100 m:
x  x1  v0 (t
x 35 m
vmax  12.96 m/s 
for
t1 )
35 m
100 m  35 m  (12.9628 m/s)(t 2
x 100 m
5.4) s
t2  10.41 s 
or
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PROBLEM 11.39
Automobile A starts from O and
accelerates at the constant rate of
0.75 m/s2. 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.
SOLUTION
Place origin at 0.
Motion of auto.
 xA 0  0,  vA 0  0, aA  0.75 m/s2
x A   x A  0   v A 0 t 
1 2
1
aAt  0  0 
 0.75 t 2
2
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
0   xB  0
 6  20 
6t m
At t  20 s , x B  0.
Hence,
x B  120
 xB 0  120 m
6t
When the vehicles pass each other, xB  x A .
120
6t  0.375 t 2
0.375 t 2  6 t
t
t
6
6
120  0
 4  0.375  120 
 2  0.375 
(6)2
14.697
 11.596 s
0.75
and
27.6 s
t  11.60 s 
Reject the negative root.
Corresponding values of xAand xB.
xA   0.37511.596  50.4 m
2
xB  120
 6 11.596   50.4 m
x  50.4 m 
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43
PROBLEM 11.40
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 At
(v A ) 0  180 km/h  50 m/s
At t  8 s:
Then
v A  225 km/h  62.5 m/s
62.5 m/s  50 m/s  a A (8 s)
a A  1.563 m/s 2 
or
(b)
We have
1
1
xA  ( xA )0  (vA )0 t  a At 2  50 m  (50 m/s)(8 s)  (1.5625 m/s2 )(8 s)2  500 m
2
2
and
1
xB  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
aB  3.13 m/s2 
or
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PROBLEM 11.41
As relay runner A enters the 20-m-long exchange zone with a
speed of 12.9 m/s, he begins to slow down. He hands the baton to
runner B 1.82 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.
SOLUTION
(a)
For runner A:
At t = 1.82 s:
x A = 0 + (v A ) 0 t +
1
a At 2
2
20 m = (12.9 m/s)(1.82 s) +
1
a A (1.82 s) 2
2
a A = −2.10 m/s2 W
or
Also
At t = 1.82 s:
vA = (vA )0 + a At
(v A )1.82 = (12.9 m/s) + (−2.10 m/s 2 )(1.82 s)
= 9.078 m/s
For runner B:
vB2 = 0 + 2aB xB − 0
When
xB = 20 m, vB = v A : (9.078 m/s)2 = 2aB (20 m)
or
aB = 2.0603 m/s 2
aB = 2.06 m/s2 W
(b)
For runner B:
vB = 0 + aB (t − t B )
where t B is the time at which he begins to run.
At t = 1.82 s:
or
9.078 m/s = (2.0603 m/s2 )(1.82 − t B )s
t B = −2.59 s
Runner B should start to run 2.59 s before A reaches the exchange zone. W
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45
PROBLEM 11.42
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 0.54 and that B has
a constant deceleration of 0.36 m/s2 determine (a) when and
where A will overtake B, (b) the speed of each automobile at
that time.
SOLUTION
a A = +0.54 m/s2
aB = -0.36 m/s2
| v A |0 = 36 km/h = 10 m/s
| v B |0 = 54 km/h = 15 m/s
A overtakes B
Motion of auto A:
v A = ( v A )0 + a At = 10 + 0.54t
(1)
1
1
x A = ( x A )0 + ( v A )0 t + aAt 2 = 0 + 10t + (0.54)t 2
2
2
(2)
v B = ( v B )0 + aB t = 15 - 0.36t
(3)
1
1
x B = ( x B )0 + ( v B )0 t + aB t 2 = 22.5 + 15t + ( -0.36)t 2
2
2
(4)
Motion of auto B:
(a)
A overtakes B at t = t1 .
x A = x B : 10t1 + 0.27t12 = 22.5 + 15t1 - 0.18t12
0.45t12 - 5t1 - 22.5 = 0
Eq. (2):
t1 = -3.437 s and t1 = 14.548 s
t1 = 14.548 s !
x A = 10(14.548) + 0.27(14.548)2
x A = .203 m !
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46
PROBLEM 11.42 (Continued)
(b)
Velocities when
t1 = 14.548 s
Eq. (1):
v A = 10 + (0.54)(14.548)
v A = 17.856 m/s
Eq. (3):
v A = 64.3 km/h Æ !
v B = 15 - 0.36(14.548)
v B = 9.7627 m/s
v B = 35.1 km/h Æ !
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47
PROBLEM 11.43
Two automobiles A and B are approaching
each other in adjacent highway lanes.
At t = 0, A and B are 1 km apart, their
speeds are vA = 108 km / h and vB = 63
km / 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.
SOLUTION
(a)
1
x A = 0 + ( v A ) 0 t + a At 2
2
We have
At t = 40 s:
( v A )0 = 108 km/h = 30 m/s
1
1000 m = (30 m/s)(40 s) + aA ( 40 s)2
2
a A = -0.250 m/s2 !
or
1
x B = 0 + ( v B ) 0 t + aB t 2
2
Also,
At t = 42 s:
1
1000 m = (17.5 m/s)(42 s) + aB ( 42 s)2
2
aB = 0.30045 m/s2
or
(b)
( v B )0 = 63 km/h = 17.5 m/s
aB = 0.300 m/s2 !
When the cars pass each other
x A + x B = 1000 m
Then
or
Solving
1
2
(30 m/s)t AB + ( -0.250 m/s)t AB
+ (17.5 m/s)t AB
2
1
2
= 1000 m
+ (0.30045 m/s2 )t AB
2
2
2
+ 95t AB
- 2000 = 0
0.05045t AB
t = 20.822 s and t = -1904 s
t > 0 fi t AB = 20.8 s !
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48
PROBLEM 11.43 (Continued)
(c)
We have
v B = ( v B ) 0 + aB t
At t = t AB :
v B = 17.5 m/s + (0.30045 m/s2 )(20.822 s)
= 23.756 m/s
or
v B = 85.5 km/h !
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49
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.
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.
y B  ( y B ) 0  ( vB ) 0 t
1 2
gt
2
with ( yB )0  0, (vB )0  3 m/s, and g  9.81 m/s 2 .
yB  3t 4.905t 2
The elevator undergoes uniform motion.
yE  ( yE )0  vE t
with ( y E ) 0  10 m and v E  4 m/s.
(a)
Set y B  y E
Time of impact.
3t
4.905t 2  10  4t
4.905t 2  t 10  0
t  1.3295 and
(b)
t  1.330 s 
1.5334
Location of impact.
y B  (3)(1.3295) (4.905)(1.3295) 2  4.68 m
y E  10  (4)(1.3295)  4.68 m
(checks)
4.68 m below the man 
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50
PROBLEM 11.45
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.
2
Assuming a constant acceleration g 9.81 m/s , determine (a)
( the time t1,
(b) the velocity of B relative to A at the time of the explosion.
SOLUTION
Place origin at ground level. The motion of rockets A and B is
v A  (v A ) 0
Rocket A:
gt  100
y A  ( y A ) 0  (v A ) 0 t
vB  (vB )0
Rocket B:
9.81t
(1)
1 2
gt  100t 4.905t 2
2
(2)
t1 )  100 9.81(t
t1 )
(3)
1
g (t t1 ) 2
2
 100(t t1 ) 4.905(t t1 ) 2
(4)
g (t
y B  ( y B )0  (vB )0 (t
t1 )
Time of explosion of rockets A and B. y A  y B  300 ft
300  100t
From (2),
4.905t 2
4.905t 2
100t  300  0
t  16.732 s and 3.655 s
300  100(t t1 ) 4.905(t t12 )
From (4),
t
(a)
Time t1:
(b)
Relative velocity at explosion.
and
3.655 s
t  16.732 s
Since rocket A is falling,
Since rocket B is rising,
t1  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 
vB/A  128.3 m/s 
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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 90 km / h. At t = 0, A starts and
accelerates at a constant rate aA, 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 = 90 m and vA = vB,
determine (a) the acceleration aA,
(b) when the vehicles pass each other,
(c) the distance d between the vehicles
at t = 0.
SOLUTION
For t ≥ 0:
v A = 0 + a At
1
x A = 0 + 0 + a At 2
2
0 £ t < 5 s:
x B = 0 + ( v B )0 t ( v B )0 = 90 km/h = 25 m/s
At t = 5 s:
x B = (25m/s)(5 s) = 125 m
For t ≥ 5 s:
1
v B = ( v B )0 + aB (t - 5) aB = - aA
6
1
x B = ( x B )5 + ( v B )0 (t - 5) + aB (t - 5)2
2
Assume t > 5 s when the cars pass each other.
At that time (t AB ),
v A = vB :
aAt AB = (25 m /s) -
x A = 90 m:
90 m =
aA
(t AB - 5)
6
1
2
aAt AB
2
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52
PROBLEM 11.46 (Continued)
(
aA 73 t AB - 53
Then
2
t AB
Solving
With t AB > 5 s,
or
(b)
90
2
25t AB
- 210t AB + 150 = 0
or
(a)
) = 25
t AB = 0.788 s and t AB = 7.6117 s
90 m =
1
aA (7.6117)2
2
aA = 3.10677 m/s2
a A = 3.11 m/s2 !
t AB = 7.61 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
1 Ê -3.10677 ˆ
È
˘
(7.6117 - 5)2 ˙
= 90 m + Í125 m + 25 m/ s (7.6117 - 5) + Á
˜
¯
2Ë
6
Î
˚
d = 279 m !
or
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11110
PROBLEM 11.47
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, (dd) the relative velocity of the counterweight W with respect to the
elevator.
SOLUTION
Choose the positive direction downward.
(a)
Velocity of cableC.
yC  2 y E  constant
vC  2 v E  0
v E  4 m/s
But,
or
(b)
vC  2v E  8 m/s
v C  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 Cwith respect to E.
vC/E  vC
vE  ( 8 m/s) (4 m/s)  12 m/s
vC/E  12.00 m/s 
(d)
Relative velocity of W with respect to E.
vW /E  vW
vE  ( 4 m/s) (4 m/s)  8 m/s
vW /E  8.00 m/s 
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54
PROBLEM 11.48
The elevator shown starts from rest and moves upward with a constant
acceleration. If the counterweight W moves through 10 m in 5 s, determine (a)
the accelerations of the elevator and the cable C, (b) the velocity of the elevator
after 5 s.
SOLUTION
We choose Positive direction downward for motion of counterweight.
yW =
At t = 5 s,
1
aW t 2
2
yW = 10 m
10 m =
1
aW (5 s)2
2
aW = 0.8 m/s2
(a)
Accelerations of E and C.
Since
Thus:
Also,
Thus:
(b)
aW = 0.800 m/s2 Ø
yW + yE = constant vW + vE = 0, and aW + aE = 0
aE = - aW = -(0.8 m/s2 ),
a E = 0.800 m/s2 ≠ !
yC + 2 yE = constant, vC + 2vE = 0, and aC + 2aE = 0
aC = -2aE = -2( -0.8 m/s2 ) = 1.6 m/s2 ,
aC = 1.600 m/s2 Ø !
Velocity of elevator after 5 s.
vE = ( vE )0 + aE t = 0 + ( -0.8 m/s2 )(5 s) = - 4 m/s
( vE )5 = 4.00 m/s ≠ !
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55
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.
SOLUTION
From the diagram:
x A  4 y B  constant
(a)
v A  4vB  0
Differentiate
Sketch:
(1)
x A  0.5 m/s
Solve (1) for vB
vB  0.125 m/s
v B  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
or
v B / A  0.5
 0.5  m/s
0.125
m/s
vB/ A  0.5154 m/s
14 
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56
Problem 11.50
An athlete pulls handle A to the left with a constant
acceleration. Knowing that after the weight B has been
lifted 100 mm. its velocity is 0.6 m/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.
SOLUTION
(a)
From the diagram:
Sketch:
x A  3 y B  constant
v A  3v B  0
(1)
a A  3a B  0
(2)
Uniform Acceleration of weight B

v B 2   v B o +2aB yB   yB o
2
Substitute in known values

 0.6 m/s2 =02  2aB  0.1 m  0
aB  1.8 m/s 2
Using (2)
a A  3(1.8 m/s 2 )  0
aA  5.4 m/s2  
aB  1.8 m/s2  
(b) Uniform Acceleration of Handle A
vA   vA o  aAt


v A  0  5.4 m/s2  0.5 s 
vA  2.7 m/s  
or
Also
y A   y A o   v A o t 
y A   y A o  0 

1
a At 2
2

1
2
5.4 m/s2  0.5 s
2
yA  0.675 m  
or
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1
PROBLEM 11.51
Slider block B moves to the right
with a constant velocity of 300
mm/s. Determine (a)) the velocity
of slider block A, (b)) the velocity
of portion C of the cable, (c)
( the
velocity of portion D of the cable,
(d)) the relative velocity of portion
C of the cable with respect to slider
block A.
SOLUTION
x B  ( x B  x A )  2 x A  constant
From the diagram
Then
2 v B  3v A  0
(1)
and
2 a B  3a A  0
(2)
Also, we have
 x D  x A  constant
Then
vD  v A  0
(a)
Substituting into Eq. (1)
2(300 mm/s)  3v A  0
or
(b)
From the diagram
Then
Substituting
From the diagram
Then
Substituting
v A  200 mm/s

v C  600 mm/s

v D  200 mm/s

x B  ( x B  xC )  constant
2 v B  vC  0
2(300 mm/s)  vC  0
or
(c)
(3)
( xC  x A )  ( x D  x A )  constant
vC  2 v A  v D  0
600 mm/s  2(200 mm/s)  v D  0
or
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58
PROBLEM 11.51 (Continued)
(d)
We have
vC/A  vC  v A
 600 mm/s  200 mm/s
v C/A  400 mm/s
or
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
PROBLEM 11.52
At the instant shown, slider block
B is moving with a constant
acceleration, and its speed is 150
mm/s. Knowing that after slider
block A has moved 240 mm to the
right its velocity is 60 mm/s,
determine (a) the accelerations of
A and B, (b) the acceleration of
portion D of the cable, (c) the
velocity and change in position of
slider block B after 4 s.
SOLUTION
x B  ( x B  x A )  2 x A  constant
From the diagram
Then
2 v B  3v A  0
(1)
and
2 a B  3a A  0
(2)
(a)
First observe that if block A moves to the right, v A  and Eq. (1)
Eq. (1) at t  0
v B  . Then, using
2(150 mm/s)  3( v A ) 0  0
(v A ) 0  100 mm/s
or
Also, Eq. (2) and a B  constant
a A  constant
Then
v A2  (v A )02  2a A [ x A  ( xA )0 ]
When x A  ( x A ) 0  240 mm:
(60 mm/s) 2  (100 mm/s)2  2a A (240 mm)
or
aA  
40
mm/s2
3
a A  13.33 mm/s 2
or
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60

PROBLEM 11.52 (Continued)
Then, substituting into Eq. (2)
2aB  3 
40
mm/s2  0
3
aB  20 mm/s 2
or
(b)
a B  20.0 mm/s 2

a D  13.33 mm/s 2

v B  70.0 mm/s

x B  ( x B ) 0  440 mm

From the diagram,  x D  x A  constant
vD  v A  0
aD  a A  0
Then
Substituting
aD  
40
mm/s2  0
3
or
(c)
We have
vB  ( vB ) 0  a B t
At t  4 s:
vB  150 mm/s  (20.0 mm/s 2 )(4 s)
or
Also
At t  4 s:
x B  ( x B ) 0  (v B ) 0 t 
1
aB t 2
2
xB  ( xB )0  (150 mm/s)(4 s)

1
( 20.0 mm/s 2 )(4 s) 2
2
or
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PROBLEM 11.53
Slider block A moves to the left with a constant velocity of 6 m/s.
Determine (a) the velocity of block B, (b) the velocity of portion D of
the cable, (c) the relative velocity of portion C of the cable with respect
to portion D.
SOLUTION
From the diagram, we have
x A + 3 yB = constant
Then
v A + 3vB = 0
(1)
and
a A + 3aB = 0
(2)
(a)
Substituting into Eq. (1)
6 m/s + 3vB = 0
v B = 2.00 m/s �
or
(b)
From the diagram
yB + yD = constant
Then
vB + vD = 0
v D = 2.00 m/s �
(c)
From the diagram
x A + yC = constant
Then
v A + vC = 0
Now
vC/D = vC − vD = (−6 m/s) − (2 m/s) = −8 m/s
vC = −6 m/s
v C/D = 8.00 m/s �
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62
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.
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,
xM  3xB  constant
vM  3vB  0
xL  ( xL  xB )  constant
2vL  vB  0
vM  100 mm/s
with
v
vB   M  33.333 m/s
3
vB
vL 
 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. vB/L  vB  vL  33.333  (16.667)  16.667
v B/L  16.67 mm/s �
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63
PROBLEM 11.55
Collar A starts from rest and moves upward with a constant acceleration. Knowing that
after 8 s the relative velocity of collar B with respect to collar A is 0.6 m /s, determine
(a) the accelerations of A and B, (b) the velocity and the change in position of B after 6 s.
SOLUTION
From the diagram
2 y A + y B + ( y B - y A ) = constant
Then
v A + 2v B = 0
(1)
and
a A + 2 aB = 0
(2)
(a) Eq. (1) and ( vA )0 = 0 = ( v B )0
Also, Eq. (2) and a A is constant and negative fi a B is constant
and positive
Then
v A = 0 + a At
v B = 0 + aB t
Now
v B /A = v B - v A = ( aB - aA )t
From Eq. (2)
1
aB = - a A
2
So that
3
v B /A = - a A t
2
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64
PROBLEM 11.55 (Continued)
At t = 8 s:
3
0.6 m/s = - aA (8 s)
2
1
m/s2
20
or
aA = -
and then
1Ê 1
ˆ 1
aB = - Á - m/s2 ˜ =
m/s2
¯ 40
Ë
2 20
a B = 0.250 m/s2 Ø !
or
(b)
At t = 6 s:
ˆ
Ê 1
v B = Á m/s2 ˜ (6 s) = 0.15 m/s
¯
Ë 40
vB = 0.1500 m/s Ø !
or
Now
At t = 6 s:
a A = 0.0500 m/s2 ≠ !
1
y B = ( y B ) 0 + 0 + aB t 2
2
1 Ê 1 mˆ
2
y B - ( y B )0 = Á
˜ (6 s) = 0.45 m
2 Ë 40 s2 ¯
yB - ( yB )0 = 0.450 m Ø !
or
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or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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65
PROBLEM 11.56
Block A starts from rest at t = 0 and moves downward with a constant
acceleration of 150 mm/s2. Knowing that block B moves up with a constant
velocity of 75 mm/s, determine (a) the time when the velocity of block C is
zero, (b) the corresponding position of block C.
SOLUTION
The cable lengths are constant.
L1 = 2 yC + 2 yD + constant
L 2 = y A + yB + ( yB − yD ) + constant
Eliminate yD.
L1 + 2 L 2 = 2 yC + 2 yD + 2 y A + 2 yB + 2( yB − yD ) + constant
2( yC + y A + 2 yB ) = constant
Differentiate to obtain relationships for velocities and accelerations, positive downward.
vC + v A + 2vB = 0
(1)
aC + a A + 2aB = 0
(2)
Use units of mm and seconds.
Motion of block A:
v A = a At = 150t
yA =
Motion of block B:
1
1
a At 2 = (150)t 2 = 75t 2
2
2
v B = 75 mm/s
vB = −75 mm/s
yB = vB t = −75t
Motion of block C:
From (1),
vC = −v A − 2vB = −150t − 2( −75) = 150 − 150t
yC =
(a)
Time when vC is zero.
(b)
Corresponding position.
t
0
vC dt = 150t − 75t 2
150 − 150t = 0
t = 1.000 s �
yC = (150)(1) − (75)(1) 2 = 75 mm
yC = 75.0 mm �
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66
PROBLEM 11.57
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/s 2 . 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.
SOLUTION
From the diagram
3 y A  4 y B  xC  constant
Then
3v A  4 v B  vC  0
(1)
and
3 a A  4 a B  aC  0
(2)
( v B )  0,
Given:
a A  constent
(a C )  75 mm/s 2
At t  2 s,
v B  480 mm/s
v C  280 mm/s
(a)
Eq. (2) and a A  constant and aC  constant
a B  constant
Then
vB  0  aB t
At t  2 s:
480 mm/s  a B (2 s)
aB  240 mm/s 2
or
a B  240 mm/s 2 
or
a A  345 mm/s2 
Substituting into Eq. (2)
3a A  4(240 mm/s 2 )  (75 mm/s 2 )  0
a A   345 mm/s
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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
( v C ) 0  130.0 mm/s

Then, substituting into Eq. (1) at t  0
3( v A ) 0  4(0)  (130 mm/s)  0
v A  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
x C  ( x C ) 0  728 mm
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68

PROBLEM 11.58
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
accelera
of A and
C, (c) the change in position of block A after 5 s.
SOLUTION
From the diagram
3 y A  4 y B  xC  constant
Then
3v A  4 v B  vC  0
(1)
and
3 a A  4 a B  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)
( v C ) 0  10.00 mm/s
or
We have
1
xC  ( xC )0  (vC )0 t  aC t 2
2
At t  3 s:
1
57 mm  (10 mm/s)(3 s)  aC (3 s)2
2
aC  6 mm/s 2
Now
or
aC  6.00 mm/s 2
v B  constant  a B  0
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

PROBLEM 11.58 (Continued)
Then, substituting into Eq. (2)
3a A  4(0)  (6 mm/s 2 )  0
a A  2 mm/s 2
(c)
We have
At t  5 s:
or
a A  2.00 mm/s 2 
1
y A  ( y A ) 0  (v A )0 t  a A t 2
2
y A  ( y A )0  ( 30 mm/s)(5 s) 
1
( 2 mm/s 2 )(5 s) 2
2
 175 mm
y A  ( y A ) 0  175.0 mm 
or
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70
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/s 2 upward and the relative acceleration of block D with respect to block A is
110 mm/s 2 downward, determine (a)) the velocity of block C after 3 s, (b) the
change in position of block D after 5 s.
SOLUTION
From the diagram
Cable 1:
2 y A  2 y B  yC  constant
Then
2v A  2vB  vC  0
(1)
and
2 a A  2 a B  aC  0
(2)
Cable 2:
( y D  y A )  ( y D  y B )  constant
Then
 v A  v B  2vD  0
(3)
and
 a A  aB  2aD  0
(4)
Given: At t  0, v  0; all accelerations constant;
aC/B  60 mm/s2 , aD /A  110 mm/s 2
(a)
We have
aC /B  aC  a B   60 or
a B  aC  60
and
a D /A  a D  a A  110
a A  a D  110
or
Substituting into Eqs. (2) and (4)
Eq. (2):
2( a D  110)  2( aC  60)  aC  0
3aC  2 a D  100
or
Eq. (4):
 ( a D  110)  ( aC  60)  2 a D  0
 aC  a D   50
or
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(5)
(6)
PROBLEM 11.59 (Continued)
Solving Eqs. (5) and (6) for aC and a D
aC  40 mm/s 2
aD  10 mm/s 2
Now
vC  0  a C t
At t  3 s:
vC  (40 mm/s 2 )(3 s)
v C  120.0 mm/s 
or
(b)
We have
1
yD  ( yD )0  (0)t  aD t 2
2
At t  5 s:
1
yD  ( yD )0  (10 mm/s 2 )(5 s) 2
2
y D  ( y D ) 0  125.0 mm 
or
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72
PROBLEM 11.60*
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/s 2 , (b)) the change in position of
block D when the velocity of block C is 600 mm/s upward.
SOLUTION
From the diagram
Cable 1:
2 y A  2 y B  yC  constant
Then
2v A  2vB  vC  0
(1)
and
2 a A  2 a B  aC  0
(2)
Cable 2: ( y D  y A )  ( y D  y B )  constant
Then
 v A  vB  2vD  0
(3)
and
 a A  aB  2aD  0
(4)
t0
Given: At
v0
( y A )0  ( yB )0  ( yC )0
All accelerations constant.
At t  2 s
yC /A  280 mm
v B /A  80 mm/s
When
y A  ( y A ) 0  160 mm
y B  ( y B ) 0  320 mm
aB
10 mm/s 2
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PROBLEM 11.60* (Continued)
(a)
We have
1
y A  ( y A )0  (0)t  aAt 2
2
and
1
yC  ( yC )0  (0)t  aC t 2
2
Then
1
yC/A  yC  y A  (aC  a A )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 a B  ( a A  140)  0
or
1
aA  (140  2aB )
3
Now
vB  0  aB t
(6)
v A  0  a At
v B / A  v B  v A  ( a B  a A )t
Also
1
yB  ( yB )0  (0)t  aB t 2
2
When
v B /A  80 mm/s :
80  ( a B  a A )t
y A  160 mm :
1
160  aAt 2
2
y B  320 mm :
1
320  aB t 2
2
(7)
Then
1
160  (aB  aA )t 2
2
Using Eq. (7)
320  (80)t or t  4 s
Then
160 
1
a A (4)2
2
or
a A  20.0 mm/s 2 
and
320 
1
aB (4)2
2
or
a B  40.0 mm/s 2 
Note that Eq. (6) is not used; thus, the problem is over-determined.
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74
PROBLEM 11.60* (Continued)
(b)
Substituting into Eq. (5)
aC  20  140  120 mm/s 2
and into Eq. (4)
(20 mm/s 2 )  (40 mm/s 2 )  2aD  0
or
aD  30 mm/s 2
Now
vC  0  a C t
When vC  600 mm/s:
600 mm/s  ( 120 mm/s 2 )t
or
t 5s
Also
1
yD  ( yD )0  (0)t  aD t 2
2
At t  5 s:
1
yD  ( yD )0  (30 mm/s2 )(5 s)2
2
y D  ( y D ) 0  375 mm 
or
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PROBLEM11.61
A particle moves in a straight line with a constant acceleration
of 4 m/s 2 for 6 s, zero acceleration for the next 4 s, and a
constant acceleration of 4 m/s 2 for the next 4 s. Knowing
that the particle starts from the origin and that its velocity is
8 m/s during the zero acceleration time interval, (a) construct
the v  t and x  t curves for 0 t 14 s, (b) determine the
position and the velocity of the particle and the total distance
traveled when t 14 s.
SOLUTION
From the a  t curve
A1   24 m/s, A2  16 m/s
(a) Construct the v  t curve :
v6   8 m/s
v0  v6  A1
 8   24 
 16 m/s
v10   8 m/s
v14  v10  A2
 8  16

 8 m/s
From the v  t curve
A3  32 m, A4   8 m
A5   32 m, A6   8 m , A7  8 m
(b) Construct the x  t curve
x0  0
x4  x0  A3  32 m
x6  x4  A4  24 m
x10  x6  A5  8 m
x12  x10  A6   16 m
x14  x12  A7   8 m
Total Distance Traveled
0
4s
12 s
t
t
t

4 s,
d1  32  0  32 m
12 s,
d 2  16  32  48 m
14 s,
d3  8   16  8 m
d  32  48  8
d  88 m 
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76
Problem 11.62
A particle moves in a straight line with a constant acceleration
of 4 m/s 2 for 6 s, zero acceleration for the next 4 s, and a
constant acceleration of 4 m/s 2 for the next 4 s. Knowing that
the particle starts from the origin with v0  16 m/s,
(a) construct the v  t and x  t curves for 0 t 14 s,
(b) determine the amount of time during which the particle is
further than 16 m from the origin.
SOLUTION
From the a  t curve
A1   24 m/s, A2  16 m/s
(b) Construct the v  t curve :
v0  16 m/s
v6  v0  A1
 16   24
 8 m/s
v10  v6   8 m/s
v14  v10  A2
 8  16

 8 m/s
From the v  t curve
A3  32 m, A4   8 m
A5   32 m, A6   8 m , A7  8 m
Construct the x  t curve
x0  0
x4  x0  A3  32 m
x6  x4  A4  24 m
x10  x6  A5  8 m
x12  x10  A6   16 m
x14  x12  A7   8 m
(b)
Time for x

16 m.
From the x  t diagram, this is time interval t1 to t2 .
From 0
t
6 s,
dx
 v  16  4t
dt
Integrating, using limits x  0 when t  0 and x  16 m when t  t1
16
x 0  16t  2t 2
t
0
or
16  16t1  2t12
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PROBLEM 11.62 (Continued)
or
t12  8t1  8  0
Using the quadratic formula:
t1 
8
8    4 18 
4
 2 1
2
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  8t 2
Time Interval:
t  t2  t1
or
t2  7 s
t  5.83 s
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78
PROBLEM 11.63
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
20  60
 5 m/s 2
26  10
t  10 s to t  26 s:
a
t  26 s to t  41 s:
v  constant
t  41s to t  46 s:
a
t  46 s:
a0
a0
5  (20)
 3 m/s2
46  41
v  constant
a0
x2  x1  (area under v t curve from t1 to t 2 )
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
20  5
 17.5 m
2
x50  17.5  4(5)  2.5 m
x46  80  5
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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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80
PROBLEM 11.64
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.
SOLUTION
(a)
at  slope of v  t curve at time t
From t  0 to t  10 s:
v  constant
20  60
 5 m/s2
26  10
t  10 s to t  26 s:
a
t  26 s to t  41 s:
v  constant
t  41s to t  46 s:
a
t  46 s:
a0
a0
5  (20)
 3 m/s2
46  41
v  constant
a0
x2  x1  (area under v t curve from t1 to t 2 )
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
20  5
 17.5 m
2
x50  17.5  4(5)  2.5 m
x46  80  5
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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
We have
t1  33.3 s
10 s
t1
22 s
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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82
Problem 11.65
A particle moves in a straight line with the velocity shown in the
figure. Knowing that x = – 48 m 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
a0
10 s,
10 s < t
18 s,
a
18  6
 1.5 m/s 2
18  10
18 s < t
30 s,
a
18  18
  3 m/s 2
30  18
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 m
A2 
1
(6  18)(18  10)  96 m
2
A3 
1
(18)(24  18)  54 m
2
A4 
1
(18)(30  24)   54 m
2
A5  (  18)(40  30)   180 m
Value of x at specific times:
x0   48 m
x10  x0  A1  12 m
x18  x10  A2  108 m
x24  x18  A3  162 m
x30  x24  A4  108 m
x40  x30  A5   72 m
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
PROBLEM 11.65 (Continued)
(a) Maximum value of x occurs when: v  0, i.e. t  24 s.
xmax  162 m 
(b) Times when x  108 m.
From the x  t curve
t  18 s and t  30 s 
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84
PROBLEM 11.66
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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PROBLEM 11.66 (C
(Continued)
(a)
Now
x  area under v t curve for given time interva
interval
Then
(586  600) m  t1
55.555  13.888
m/s
2
→
평속
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 ⇒ 평속
t total  (0.4032  40.0346  4.3203) s
(b)
We have
ainitial 

t total  44.8 s 
vinitial
t1
[13.888  (55.555)] m/s
0.4032 s
 103.3 m/s 2
ainitial  103.3 m/s 2 
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86
PROBLEM 11.67
A commuter train traveling at 60 km / h is
4.5 km from a station. The train then decelerates so that its speed is 30 km / h when
it is 0.75 km 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 is travel the first
3.75 km, (b) the speed of the train as it
arrives at the station, (c) the final constant
deceleration of the train.
SOLUTION
Given: At t = 0, v = 60 km/h x = 0; when x = 3.75 km, v = 30 km/h;
at t = 7.5 min, x = 4.5 km; constant decelerations
The v - t curve is first drawn is shown.
(a)
A1 = 3.75 km
We have
or
1h
Ê 60 + 30 ˆ
km/h ¥
= 3.75 km
(t1 min) Á
Ë 2 ˜¯
60 min
t1 = 5 min !
or
(b)
A2 = 0.75 km
We have
or
1h
Ê 30 + v2 ˆ
km/h ¥
= 0.75 km
(7.5 - 5) min ¥ Á
˜
Ë 2 ¯
60 min
v2 = 6 km/h !
or
(c)
We have
afinal = a12
=
(6 - 30) km/h 1000 m 1 min
1h
¥
¥
¥
(7.5 - 5) min
km
60 s 3600 s
afinal = -0.0444 m/s2 !
or
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87
PROBLEM 11.68
A temperature sensor is attached to slider AB, which moves back
and forth through 1500 mm. The maximum velocities of the
slider are 300 mm /s to the right and 750 mm /s to the left. When
the slider is moving to the right, it accelerates and decelerates
at a constant rate of 150 mm /s2; when moving to the left, the
slider accelerates and decelerates at a constant rate of 500 mm /s2.
Determine the time required for the slider to complete a full cycle,
and construct the v − t and x - t curves of its motion.
SOLUTION
The v - t curve is first drawn as shown. Then
ta =
td =
vright
aright
=
300 mm/s
150 mm/s2
=2s
vleft 750 mm/s
=
aleft 500 mm/s2
= 1.5 s
A1 = 1500 mm
Now
or
[(t1 - 2) s](300 mm/s) = 1500 mm
or
t1 = 7 s
and
A2 = 1500 mm
or
{[(t2 - 7) - 1.5] s}(750 mm/s) = 1500 mm
t2 = 10.5 s
or
tcycle = t2
Now
tcycle = 10.5 s !
We have xii = xi + (area under v - t curve from ti to tii )
1
At
t = 2 s:
x2 = (2)(300) = 300 mm
2
t = 5 s:
x5 = 300 + (5 - 2)(300)
= 1200 mm
t = 7 s:
t = 8.5 s :
t = 9 s:
t = 10.5 s :
x7 = 1500 mm
1
x8.5 = 1500 - (1.5)(750)
2
= 937.5 mm
x9 = 937.5 mm - (0.5)(750)
= 562.5 mm
x10.5 = 0
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88
PROBLEM 11.69
In a water-tank
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/s 2 at t  0 to
2 m/s 2 at t  t1 and then remains equal to  2 m/s 2 until
t  1.4 s. Knowing that v  1.8 m/s when t  t1 , determine
(a) the value of t1 , (b) the velocity
ocity and the position of the
model at t  1.4 s.
SOLUTION
Given:
v0  6 m/s; for 0
for
t1
at
t  0 a  12 m/s 2 ;
at t  t1
a  2 m/s 2 , v  1.8 m/s 2
t
t1 ,
t 1.4 s a  2 m/s2 ;
The a t and v t curves are first drawn as shown. The time axis is not
drawn to scale.
(a)
vt1  v0  A1
We have
1.8 m/s  6 m/s  (t1 s)
12  2
m/s 2
2
t1  0.6 s 
(b)
v1.4  vt1  A2
We have
v1.4  1.8 m/s  (1.4  0.6) s 2 m/s 2
v1.4  0.20 m/s 
Now x1.4  A3  A4 , where A3 is most easily determined using
integration. Thus,
2  (12)
50
t  12  t  12
0.6
3
for 0 t t1 :
a
Now
dv
50
 a  t  12
dt
3
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PROBLEM 11.69 (Continued)
At t  0, v  6 m/s:
v
6
dv 
t
0
50
t  12 dt
3
25 2
t  12t
3
or
v 6
We have
dx
25
 v  6  12t  t 2
dt
3
Then
A3 
xt1
0
dx 
0.6
0
(6  12t 
25 2
t )dt
3
0.6
 6t  6t 2 
25 3
 2.04 m
t
9
0
Also
A4  (1.4  0.6)
1.8  0.2
 0.8 m
2
Then
x1.4  (2.04  0.8) m
x1.4  2.84 m 
or
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90
PROBLEM 11.70
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.
SOLUTION
Geometry of “bell-shaped”
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:
= x=6m
(a)
When t  20 s:
v20  60 m/s 
x20  (60 m/s) (20 s)  (shaded area)
 1200 m  6 m
(b)
From t  6 s to t  14 s:
x20  1194 m 
t 8s
x  (60 m/s)(14 s  6 s)  (shaded area)
 (60 m/s)(8 s)  6 m  480 m  6 m  474 m
vaverage 
x 474 m

t
8s
vaverage  59.25 m/s 
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91
PROBLEM 11.71
In a 400-m race, runner A reaches her maximum velocity v A 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 v B 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.
Runner A:
t1  4 s, t 2  25  4  21 s
A1 
1
(4)(vA )max  2(vA )max
2
A2  21(v A ) max
A1  A2 
x  200 m
23(v A ) max  200
Runner B:
or
t1  5 s,
A1 
(v A ) max  8.6957 m/s
t2  25.2  5  20.2 s
1
(5)(vB )max  2.5(vB )max
2
A2  20.2(v B ) max
A1  A2 
22.7(v B ) max  200
(vB ) max  8.8106 m/s
or
Sketch v t curve for second 200 m.
A3  vmaxt3 
t3 
Runner A:
vmax
x  200 m
v  | a |t3  0.1t3
1
vt3  200
2
0.05t32  vmaxt3  200  0
or
(vmax )2  (4)(0.05)(200)
(2)(0.05)
(vmax ) A  8.6957, (t3 ) A  146.64 s
Reject the larger root. Then total time

 10 vmax
and
(vmax )2  40

27.279 s
t A  25  27.279  52.279 s
t A  52.2 s 
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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
x  vave ( t ) 
x
1
(5.999  5.968)(52.279  51.965)
2
x  1.879 m 
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93
PROBLEM 11.72
A car and a truck are both traveling at the constant
speed of 50 km/h; the car is 12 m behind the
truck. The driver of the car wants to pass the
truck, i.e., he wishes to place his car at B, 12 m
in front of the truck, and then resume the speed of
50 km/h. The maximum acceleration of the car is
1.5 m/s2 and the maximum deceleration obtained
by applying the brakes is 6 m/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 75 km/h ? Draw the
v − t curve.
SOLUTION
Relative to truck, car must move a distance:
Allowable increase in speed:
D x = 4.8 + 12 + 15 + 12 = 43.8 m
125
Dvm = 75 - 50 = 25 km/h =
m/s
18
Acceleration Phase:
t1 =
125/18 125
=
s = 4.6296 s
1.5
27
1 Ê 125 ˆ Ê 125 ˆ
A1 = Á
˜ = 16.075 m
˜Á
2 Ë 18 ¯ Ë 27 ¯
Deceleration Phase:
t3 =
125 /18 125
=
s = 1.1574 s
6
108
1 Ê 125 ˆ Ê 125 ˆ
A3 = Á
˜ = 4.0188 m
˜Á
2 Ë 18 ¯ Ë 100 ¯
But: ∆ x = A1 + A2 + A3 :
43.8 = 16.075 + (125 /18)t2 + 4.0188
t2 = 3.414 s
t total = t1 + t2 + t3 = 4.6296 s + 3.414 s + 1.15745 s = 9.201 s
t F = 9.20 s !
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94
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 50 km / h in
the shortest possible time. What is the maximum
speed reached? Draw the v - t curve.
SOLUTION
Relative to truck, car must move a distance:
D x = 4.8 + 12 + 15 + 12 = 43.8 m
Dvm = 1.5t1 = 6t2 ;
D x = A1 + A2 :
t2 =
1
t1
4
1
43.8 m = ( Dvm )(t1 + t2 )
2
t ˆ
1
Ê
43.8 m = (1.5t1 ) Á t1 + 1 ˜
Ë
2
4¯
t12 = 46.72 t1 = 6.835 s t2 =
1
t1 = 1.709
4
t total = t1 + t2 = 6.835 + 1.709
t F = 8.54 s !
Dvm = 1.5t1 = 1.5(6.835) = 10.2525 m/s = 36.909 m/s
Speed vtruck = 50 km/h, vm = 50 km/h + 36.91 km/h
vm = 86.9 km/h !
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95
PROBLEM 11.74
Car A is traveling on a highway at a constant
speed ( v A )0 = 90 km / h and is 120 m from
the entrance of an access ramp when car B
enters the acceleration lane at that point at
a speed ( v B )0 = 25 km/h. Car B accelerates
uniformly and enters the main traffic lane
after traveling 60 m in 5 s. It then continues
to accelerate at the same rate until it reaches
a speed of 90 km/h, which it then maintains.
Determine the final distance between the two
cars.
SOLUTION
( v A )0 = 90 km / h, (v B )0 = 25 km / h; at t = 0,
Given:
( x A )0 = -120 m, ( x B )0 = 0; at t = 5 s,
x B = 60 m; for 25 km / h ≤ v B < 90 km/ h,
aB = constant; for v B = 90 km / h,
aB = 0
First note
90 km / h = 25 m /s
25 km / h = 125 /18 m /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 = 60 m:
È (125/18 + v B )5 ˘
(5 s) Í
˙ m /s = 60 m
2
Î
˚
( v B )5 =
or
307
m/s = 17.0556 m/s
18
Then, using similar triangles, we have
(25 - 125 /18) m/s (307 /18 - 125 /18) m/s
=
( = aB )
t1
5s
t1 = 8.9286 s
or
Finally, at t = t1
È
˘
Ê 25 + 125/18 ˆ
x B /A = x B - x A = Í(8.9286 s) Á
m/s ˙
˜
Ë
¯
2
Î
˚
- [-120 m + (8.9286 s) (25 m/s)]
or
x B /A = 39.4 m !
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96
PROBLEM 11.75
An elevator starts from rest and moves upward, accelerating at a rate of 1.2 m/s2
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 v E  0; For 0
vE
2
7.8 m/s, aE  1.2 m/s ;
For vE  7.8 m/s, a E  0;
At t  2 s, vB  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/s 2 , while the slope of the curve for the ball is  g (9.81 m/s 2 ).
The time t1 is the time when v E reaches 7.8 m/s.
Thus,
v E  (0)  a E t
or
7.8 m/s  (1.2 m/s 2 )t1
or
t1  6.5 s
The time ttop is the time at which the ball reaches the top of its trajectory.
Thus,
vB  ( vB ) 0  g (t  2)
or
0  20 m/s  (9.81 m/s2 )(ttop  2) s
or
ttop  4.0387 s
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97
PROBLEM 11.75 (C
(Continued)
Using the coordinate system shown, we have
0
t
t1 :
yE  12 m 
1
aE t 2 m
2
1
(4.0387  2) s (20 m/s)
2
 20.387 m
At t  t top :
yB 
and
yE  12 m 
At
t  [2  2(4.0387  2)] s  6.0774 s, y B  0
and at t  t1 ,
1
yE  12 m  (6.5 s) (7.8 m/s)  13.35 m
2
1
(1.2 m/s 2 )(4.0387 s) 2
2
 2.213 m
The ball hits the elevator ( y B  y E ) when ttop t
For t
t top :
t1.
yB  20.387 m 
1
g (t  ttop ) 2 m
2
Then,
yB  yE
when
1
(9.81 m/s 2 ) (t  4.0387) 2
2
1
 12 m  (1.2 m/s 2 ) (t s) 2
2
20.387 m 
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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98
PROBLEM 11.76
Car A is traveling at 60 km / h when it
enters a 40 km / h speed zone. The driver
of car A decelerates at a rate of 5 m /s2 until
reaching a speed of 40 km / h, which she
then maintains. When car B, which was
initially 20 m behind car A and traveling
at a constant speed of 70 km / h, enters
the speed zone, its driver decelerates at
a rate of 6 m /s2 until reaching a speed of
35 km / h. Knowing that the driver of car B
maintains a speed of 35 km / h, determine
(a) the closest that car B comes to car A,
(b) the time at which car A is 25 m in
front of car B.
SOLUTION
( v A )0 = 60 km/h; For 40 km/h < v A £ 60 km/h;
Given:
a A = -5 m/s2 ; For v A = 40 km/h, aA = 0;
( x A/B )0 = 20 m; ( v B )0 = 70 km/h;
when x B = 0, aB = - 6 m/s2 ; for v B = 35 km/h,
aB = 0
100
m/s
9
175
35 km/h =
m/s
18
60 km/h = 50 / 3 m/s, 40 km/h =
First note
70 km/h =
At t = 0
175
m/s
9
The v − t curves of the two cars are as shown.
At
t = 0:
car A enters the speed zone
t = (tB)1:
car B enters the speed zone
t = tA:
car A reaches its final speed
t = t min:
vA = vB
t = (tB) 2: car B reaches its final speed
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106
99
PROBLEM 11.76 (Continued)
(a)
aA =
We have
( v A )final - ( v A )0
tA
Ê 100 50 ˆ
- ˜ m/s
ÁË
9
3¯
2
-5 m/s =
tA
or
or
t A = 1.1111 s
Also
20 m = (t B )1 ( v B )0
or
ˆ
Ê 175
20 m = (t B )1 Á
m/s˜
¯
Ë 9
aB =
and
or
(t B )1 = 1.0286 s
( v B )final - ( v B )0
(t B )2 - (t B )1
Ê 175 175 ˆ
ÁË
˜ m/s
18
9 ¯
- 6 m/s2 =
[(t B )2 - 1.0286] s
or
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
100
m/s
9
v B = ( v B )0 + aB [t - (t B )1 ]
vA =
Then
or
Finally
at t = t min:
100 175
=
+ ( - 6 m/s2 )(t min - 1.0286) s
9
9
t min = 2.4175 s
( x A/ B ) min = ( x A )tmin - ( x B )tmin
¸
Ï È ( v ) + ( v A )final ˘
+ (t min - t A )( v A )final ˝
= Ìt A Í A 0
˙
2
˚
˛
Ó Î
Ï
È ( v ) + ( v A )final ˘ ¸
- Ì( x B )0 + (t B )1 ( v B )0 + [t min - (t B )1 ] Í B 0
˙˝
2
Î
˚˛
Ó
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100
PROBLEM 11.76 (Continued)
È
˘
Ê 50 100 ˆ
100
+
˙
m/s
2
4175
1
1111
s
m/s
+
(
.
.
)
¥
= Í(1.1111 s) Á 3
9 ˜
Í
˙
9
ÁË
˜¯
2
Î
˚
È
˘
Ê 175 100 ˆ
ˆ
Ê 175
+
˙
m/s˜ + (2.4175 - 1.0286) s ¥ Á 9
m/s
- Í-20 m + (1.0286 s) Á
˜
9
¯
Ë 9
Í
˙
ÁË
˜¯
2
Î
˚
= (15.432 + 14.516) m - ( -20 + 20.00 + 21.219) m
= 8.729 m
(b)
or (xA/B)min = 8.73 m !
Since ( x A/B ) ≤ 20 m for t ≤ t min , it follows that x A/B = 25 m 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 - t min )( v A )final ]
+ ( v B )final ˘
¸
Ï
È (v )
+ [t - (t B )2 ]( v B )final ˝
- Ì[(t B )2 - (t min )] Í A final
˙
2
Î
˚
˛
Ó
or
100
È
˘
25 m = 8.729 m + Í(t - 2.4175) s ¥
m/s ˙
9
Î
˚
È
˘
Ê 100 175 ˆ
Ê 175 ˆ
+
˙
m/s
(
t
.
)
s
m/s
+
2
6489
¥
- Í(2.6489 - 2.4175) s ¥ Á 9
˜
18
˜
ÁË
Í
˙
ÁË
˜¯
18 ¯
2
Î
˚
t = 14.25 s !
or
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101
PROBLEM 11.77
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 = 0.4 m 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, A2 , A3 and A4.
2
(4)(0.2) = 0.5333 m/s
3
A2 = (8)(0.2) = 1.6 m/s
A1 =
1
(8 + 4)(0.1) = 0.6 m/s
2
1
A4 = (4)(0.1) = 0.2 m/s
2
A3 =
Velocities: v0 = 0
v0.2 = v0 + A1 + A2
v0.2 = 2.13 m/s W
v0.3 = v0.2 + A3
v0.3 = 2.73 m/s W
v0.4 = v0.3 + A4
v0.4 = 2.93 m/s W
Sketch the v −t curve and divide its area into A5 , A6 , and A7 as shown.
∫
0.4
x
dx = 0.8 − x =
t
vdt
or
x = 0.4 −
2
(0.2)(0.1) = 0.01333 m,
3
With A5 + A6 =
∫
0.4
t
vdt
x0.3 = 0.1137 m W
x0.2 = 0.4 − ( A5 + A6 ) − A7
At t = 0.2 s,
and
0.4
x0.3 = 0.4 − A5 − (2.73)(0.1)
At t = 0.3 s,
With A5 =
∫
2
(0.8)(0.2) = 0.1067 m,
3
A7 = (2.13)(0.2) = 0.426 m
x0.2 = 0.4 − 0.1067 − 0.426
x0.2 = −0.1327 m W
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102
PROBLEM 11.78
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, (cc) 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  va  (area under at curve from ta to tb )
We have
Then
at
t  2 s:
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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103
PROBLEM 11.78 (Continued)
Now
x at t  18 s
x18  0 
(area under the v t curve from t  0 to t  18 s)
 (1 s)(15 m/s)  (1 s)
15  9
m/s
2
1
+ (2.5 s)(1.5 m/s)+ (2.5 s)(7.5 m/s)
3
2
(13.5 s)(13.5 m/s)
3
 [15  12  (3.75  6.25)  (20.25  121.50)] m
 (13.5 s)(1.5 m/s) 
 178.75 m
(c)
First note
or
x18  178.8 m 
x1  15 m
x18  178.75 m
Now
vave 
x (178.75  15) m

 9.6324 m/s
t
(18  1) s
vave  34.7 km/h 
or
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104
PROBLEM 11.79
An airport shuttle train travels between two terminals that are 2.5 km apart. To maintain passenger comfort,
the acceleration of the train is limited to ± 1.2 m / s2, and the jerk, or rate of change of acceleration, is limited to
± 0.24 m / s2 per second. If the shuttle has a maximum speed of 30 km / h, determine (a) the shortest time for the
shuttle to travel between the two terminals, (b) the corresponding average velocity of the shuttle.
SOLUTION
Given:
xmax = 2.5 km; | amax| = 1.2 m /s2
Ê da ˆ
= 0.24 m/s2 /s; vmax = 30 km/h
ÁË ˜¯
dt max
First note
(a)
25
m/s
3
2.5 km = 2500 m
30 km/h =
To obtain tmin, the train must accelerate and decelerate at the maximum rate to maximize the
time for which v = vmax . The time Dt required for the train to have an acceleration of 1.2 m / s2 is found
from
a
Ê da ˆ
= max
ÁË ˜¯
Dt
dt max
1.2 m/s2
or
Dt =
or
Dt = 5 s
0.24 m/s2 /s
Now,
da
1
ˆ
Ê
v5 = ( Dt )( amax ) Á since
= constant˜
¯
Ë
dt
2
1
or
v5 = (5 s)(1.2 m/s2 ) = 3 m/s
2
Then, since v5 < vmax , the train will continue to accelerate at 1.2 m / s2 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
curve is 0.24 m /s2/s.
after 5 s, the speed of the train is
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105
PROBLEM 11.79 (Continued)
Now
at t = (10 + Dt1 ) s, v = vmax :
25
È1
˘
2 Í (5 s)(1.2 m/s2 )˙ + ( Dt1 )(1.2 m/s2 ) =
m/s
3
Î2
˚
or
Dt1 = 1.9444 s
Then
at t = 5 s :
1
v = 0 + (5)(1.2) = 3 m/s
2
t = 6.9444 s : v = 3 + (1.9444)(1.2) = 5.33328 m/s
1
t = 11.9444 s : v = 5.33328 + (5)(1.2) = 8.33328 m/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 xmax , we have
È
˘
Ê 3 + 5.33328 ˆ
2 Í(1.9444 s) Á
m/s ˙
˜
¯
Ë
2
Î
˚
+ (10 + Dt2 ) s ¥ (8.33328 m/s) = 2500 m
or
Dt2 = 288.06 s
Then
t min = 4(5 s) + 2(1.9444 s) + 288.06 s
= 311.95 s
t min = 5 min11.95 s !
or
(b)
We have
vave =
Dx
2500 m
=
= 8.014 m/s = 28.850 km/h
Dt 311.95 s
vave = 28.9 km/h !
or
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106
PROBLEM 11.80
During a manufacturing process, a conveyor belt starts from rest and travels a total of 400 mm before temporarily
coming to rest. Knowing that the jerk, or rate of change of acceleration, is limited to ± 1.5 m/s2 per second,
determine (a) the shortest time required for the belt to move 400 mm, (b) the maximum and average values of
the velocity of the belt during that time.
SOLUTION
Given:
(a)
At
t = 0, x = 0, v = 0; xmax = 400 mm = 0.4 m
when
Ê da ˆ
= 1.5 m /s3
x = xmax , v = 0; Á ˜
Ë dt ¯ max
Observing that vmax must occur at t = 12 t min , the a - t curve must have the shape shown. Note that the
magnitude of the slope of each portion of the curve is 1.5 m/s2 / s.
We have
at t = Dt :
t = 2Dt :
1
1
v = 0 + ( Dt )( amax ) = amax Dt
2
2
vmax =
1
1
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 xmax, we have
(2Dt )( vmax ) = xmax
vmax = amax Dt fi 2amax Dt 2 = xmax
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107
PROBLEM 11.80 (Continued)
Now
or
Then
amax
= 1.5 m/s2 /s so that
Dt
2(1.5Dt m/s3 ) Dt 2 = 0.4 m
Dt = 0.5109 s
t min = 4Dt
t min = 2.04 s !
or
(b)
We have
vmax = amax D t
= (1.5 m/s2 /s ¥ D t ) D t
= 1.5 m/s2 /s ¥ (0.5109s)2
vmax = 0.392 m/s !
or
Also
vave =
Dx
0.4 m
=
Dt total 2.04 s
vave = 0.196 m/s !
or
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115
108
PROBLEM 11.81
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.
SOLUTION
at curve; at
Given:
1.
t  2 s, v  0
The at curve is first approximated with a series of rectangles, each of width t  0.25 s. The area
( t)(aave) of each rectangle is approximately equal to the change in velocity v for the specified
interval of time. Thus,
v aave t
v are given in columns 1 and 2, respectively, of the following table.
where the values of aave and
2.
v(2)  v0 
Now
and approximating the area
2
0
2
0
a dt  0
a dt under the at curve by
aave t
v, the initial velocity is then
equal to
v0  
v
Finally, using
v2  v1  v12
where v12 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
( t )(vave ) of each rectangle is approximately equal to the change in position
interval of time. Thus
x for the specified
x vave t
where vave and
x are given in columns 4 and 5, respectively, of the table.
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PROBLEM 11.81 ((Continued)
4.
With x0  0 and noting that
x2  x1  x12
where x12 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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110
PROBLEM 11.82
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.
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
t  2 s. The
area ( t )(aave ) of each rectangle is approximately equal to the change in velocity
specified interval of time. Thus,
v for the
v aave t
where the values of aave and
2.
v are given in columns 1 and 2, respectively, of the following table.
Noting that v0  0 and that
v2  v1  v12
where v12 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
t  2 s. The
area ( t )(vave ) of each rectangle is approximately equal to the change in position x for the
specified interval of time.
Thus,
x
vave t
where vave and x are given in columns 4 and 5, respectively, of the table.
4.
With x0  0 and noting that
x2  x1  x12
where x12 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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111
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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112
PROBLEM 11.83
A training airplane has a velocity of 38 m /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.
SOLUTION
Given: a-v curve:
v0 = 38 m/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
v -v
Dt = 2 1
a
Dx =
or
For the five regions shown above, we have
-4
D x, m
18.5
Dt, s
0.5
30
-10
19.8
0.6
30
24
-13.5
12
0.4444
4
24
12
-16
13.5
0.75
5
12
0
-17
4.235
0.7059
68.035
3.0003
Region
v1, m /s
v2, m/s
a, m /s2
1
38
36
2
36
3
S
(a)
From the table, when v = 0
t = 3.00 s !
(b)
From the table and assuming x0 = 0, when v = 0
x = 68.0 m !
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113
PROBLEM 11.84
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 = 250 mm, (b) when v = 2000 mm/s.
SOLUTION
Given: v - x curve
First note that the slope of the above curve is dv
. Now
dx
a=v
(a)
dv
dx
When
x = 250 mm, v = 1375 mm/s
Then
Ê 1000 m/s ˆ
a = (1375 mm/s ) Á
Ë 337.5 mm ˜¯
a = 4070 mm/s2 !
or
(b)
When v = 2000 mm/s, we have
Ê 1000 mm/s ˆ
a = 2000 mm/s Á
Ë 700 mm ˜¯
a = 2860 mm/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 cm = 50 cm, 1 cm = 50 cm /s). In the above
solution, Dv and D x were measured directly, so different scales could be used.
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114
PROBLEM11.85
An elevator starts from rest and rises 40 m to its maximum
velocity inT 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
First, find the areas A1 and A2 on the a
area method.
t curve using the moment-area
t curve
A1 
1
T
0.6
 0.1T m/s
2
3
A2 
1
2T
0.6
 0.2T m/s
2
3
(a) Apply the moment-area
area formula: x
T
0
x0  v0t
x  v0t
40  0
t adt
2 T
3 3
A1 T
7 2
T
90
T
A2 T
T
3
1 2T
3 3
8 2
T
90
15 2
T
90
1
 T2
6

T 2  40 6
 240 s 2
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
A1  0.1T
A4 
Apply the moment-area
area formula: x
A3 
t curve
1
T
0.6
 0.05T
2
6
1
T
 0.0375T
0.45
2
6
x0  v0t
T 2
0
T
2
t adt
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PROBLEM 11.85 (Continued)
x  v0
0
T
2
A1
T
2
2T
9
0.1T
5T
18
 0.035417 15.49
A3
2 T
3 6
A4
1 T
3 6
0.05T
T
9
0.0375T
T
 0.035417T 2
18
2
x  8.50 m 
v  v0
A1
A3
A4  0.1875T
v  2.90 m/s 
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PROBLEM 11.86
The acceleration of an object subjected to the pressure wave of a
large explosion is defined approximately by the curve shown. The
object is initially at rest and is again at rest at time t1. Using the
method of section 11.8, determine (a) the time t1, (b) the distance
through which the object is moved by the pressure wave.
SOLUTION
(a)
Since v = 0 when t = 0 and when t = t1 the change in v between t = 0 and t = t1 is zero.
Thus, area under a−t curve is zero
A1 + A2 + A3 = 0
1
1
1
(30)(0.6) + (−10)(0.2) + (−10)(t1 − 0.8) = 0
2
2
2
9 − 1 − 5t1 + 4 = 0
(b)
t1 = 2.40 s W
Position when t = t1 = 2.4 s
⎛2⎞
x = x0 + v0 t1 + A1 (t1 − 0.2) + A2 (t1 − 0.733) + A3 ⎜ ⎟ (t1 − 0.8)
⎝3⎠
⎡1
⎤2
= 0 + 0 + (9)(2.4 − 0.2) + (−1)(2.4 − 0.733) + ⎢ (−10)(2.4 − 0.8) ⎥ (2.4 − 0.8)
⎣2
⎦3
= 19.8 m − 1.667 m − 8.533 m
x = 9.60 m W
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124
117
PROBLEM11.87
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 andx–t
and
curves
for 0 t 4 s, (b)) determine the position of the particle at t 3 s.
(Hint: Use table inside the front cover.)
SOLUTION
t  0, x  0, v  8 m/s
Given
At
(a) We have
v2  v1 (area under a t curve from t1 to t2)
x2  x1
and
(area under v t curve from t1 to t2)
Then, using the formula for the area of a parabolic spandrel, we have
1
(2)(12)  0
3
1
t  4 s: v  0
(2)(12)  8 m/s
3
at t  2 s: v  8
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.
Now at t  2 s: x  0
t  4 s: x  4
(2)(8)
4m
3 1
(2)(8)
0
3 1
The x t curve is then drawn as shown.
(b) We have

At t  3 s: a  3(3 2) 2  3 m/s 2
1
v0
(1)(3)  1 m/s
3
(1)(1)
x4
3 1
x3  3.75 m 
or
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118
PROBLEM11.88
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:
A1 
0.75 8  6 m/s
A2  2 4  8 m/s
A3  6 6  36 m/s
Find the values of velocity at times indicated:
v0  2 m/s
(a) Sketch v
v8  v0
A1  8 m/s
v12  v8
A2  0
v18  v12
A3  36 m/s
t curve using straight line portions over the constant acceleration periods.
Find the indicated areas under the v t curve:
1
2 8 8  40 m
2
1
A5 
8 4  16 m
2
1
A6 
36 6  108 m
2
A4 
Find the values of position at times indicated:
x0  0
Sketch the x
x8  x0
A4  40 m
x12  x8
A5  56 m
x18  x12
A6  52 m
t curve using the positions and times calculated:
(b) Position at t=18s
x18  52 m 
Velocity at t-18s
v18  36 m/s 
Total Distance Traveled:  56
d  164 m 
108
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PROBLEM 11.89
A ball is thrown so that the motion is defined by
the equations x  5t and y  2 6t 4.9t 2 , where
x and y are expressed in meters and t is expressed
in seconds. Determine (a) the velocity at t  1 s,
(b) the horizontal distance the ball travels before
hitting the ground.
SOLUTION
Units are meters and seconds.
Horizontal motion:
vx 
dx
5
dt
Vertical motion:
vy 
dy
 6 9.8t
dt
(a)
vx  5
Velocity at t 1 s.
v y  6 9.8  3.8
v  vx2
3.82  6.28 m/s
3.8
5
 37.2
Horizontal distance:
( y  0)
tan 
(b)
vy
v 2y  52
vx

v  6.28 m/s
37.2 
y  2 6t 4.9t 2
t  1.4971 s
x  (5)(1.4971)  7.4856 m
x  7.49 m 
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120
PROBLEM 11.90
The motion of a vibrating particle is defined by the position vector
r  10(1 e 3t )i (4e 2t 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.
SOLUTION
r  10(1 e 3t )i (4e 2t sin15t ) j
Then
v
and
a
(a)
dr
 30e 3t i [60e 2t cos15t 8e 2t sin15t ]j
dt
dv
 90e 3t i [ 120e 2t cos15t 900e 2t sin15t 120e 2t cos15t 16e 2t sin15t ]j
dt
 90e 3t i [ 240e 2t cos15t 884e 2t sin15t ]j
When t  0:
v  30i 60 j mm/s
v  67.1 mm/s
63.4 
a  90i 240 j mm/s 2
a  256 mm/s 2
69.4 
v  8.29 mm/s
36.2 
a  336 mm/s 2
86.6 
When t  0.5 s:
v  30e 1.5i [60e 1 cos 7.5 8e 1 sin 7.5]
 6.694i 4.8906 j mm/s
a  90e 1.5i [ 240e 1 cos 7.5 884e 1 sin 7.5 j]
 20.08i 335.65 j mm/s 2
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121
PROBLEM 11.91
The motion of a vibrating particle is defined by the position vector
r = (4sin t )i − (cos 2 t ) j, where r is expressed in m and t in
seconds. (a) Determine the velocity and acceleration when t = 1 s.
(b) Show that the path of the particle is parabolic.
SOLUTION
r = (4sin t )i − (cos 2 t ) j
v = (4 cos t )i + (2 sin 2 t ) j
a = −(4 2 sin t )i + (4 2 cos 2 t ) j
(a)
(b)
When t = 1 s:
v = (4 cos )i + (2 sin 2 ) j
v = −(4 m/s)i W
a = −(4 2 sin )i − (4 2 cos ) j
a = −(4 2m/s 2 ) j W
Path of particle:
Since r = xi + y j ;
x = 4sin t ,
Recall that cos 2 = 1 − 2sin 2
y = − cos 2 t
and write
y = − cos 2 t = −(1 − 2sin 2 t )
But since x = 4sin t or sin t =
(1)
1
x, Eq.(1) yields
4
2
⎡
⎛1 ⎞ ⎤
y = − ⎢1 − 2 ⎜ x ⎟ ⎥
⎝ 4 ⎠ ⎦⎥
⎣⎢
y=
1 2
x − 1 (Parabola) W
8
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122
PROBLEM 11.92
The motion of a particle is defined by the equations x = 100t - 50 sin t and y = 100 - 50 cos t, where x and
y are expressed in mm and t is expressed in seconds. Sketch the path of the particle, 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
x = 4t - 2 sin t
We have
(a)
t, s
0
x, mm
0
y, mm
50
p
2
107.08
100
p
314.25
150
3p
2
521.24
100
2p
628.32
50
y = 4 - 2 cos t
We have
x = 100t - 50 sin t = 25( 4t - 2 sin t )
Then
vx =
Now
v 2 = v x2 + v 2y = 625( 4 - 2 cos t )2 + 2500 sin 2 t
dx
= 25 ( 4 - 2 cos t )
dt
y = 100 - 50 cos t = 25 ( 4 - 2 cos t )
vy =
dy
= 50 sin t
dt
= 12500 - 10000 t
By observation,
for vmin, cos t = 1, so that
2
vmin
= 2500
or
vmin = 50 mm/s !
2
vmax
= 22500
or
vmax = 150 mm /s !
cos t = 1
or
t = 2np s !
for vmax, cos t = -1, so that
(b)
When v = vmin:
where n = 0, 1, 2, . . .
Then
x = 25 {4(2np ) - 2 sin (2np )}
or
x = 200 np mm !
y = 25( 4 - 2(1)) = 50
or
y = 50.0 mm !
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135
123
PROBLEM 11.92 (Continued)
Also,
v x = 25 {4 - 2(1)} = 50 mm/s
v y = 50 sin (2np ) = 0
When v = vmax :
q v min = 0 Æ !
cos t = -1
or t = (2n + 1)p s !
where n = 0, 1, 2,
Then
x = 25 {4(2n + 1)p - 2 sin (2n + 1)p }
y = 25 {4 - 2( -1)}
Also,
or
x = 100(2n + 1)p mm !
or
y = 150 mm !
v x = 25{4 - 2( -1)} = 150 mm/s
v y = 50 sin (2n + 1)p = 0
q vmax = 0 Æ !
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124
PROBLEM 11.93
The damped motion of a vibrating particle is defined by the
position vector r  x1[1 1/(t 1)]i ( y1e t/2 cos 2 t ) j, where
t is expressed in seconds. For x1  30 mm and y1  20 mm,
determine the position, the velocity, and the acceleration of the
particle when (a) t  0, (b) t  1.5 s.
SOLUTION
We have
r  30 1
Then
v
1
i 20
(t 1) 2
 30
1
i 20
(t 1) 2

At t  0:
t/ 2
i 20(e
cos 2 t ) j
dr
dt
2
e
t/2
e
t/2
t/ 2
cos 2 t
2 e
sin 2 t j
1
cos 2 t
2
2 sin 2 t
j
dv
dt
 30
(a)
t 1
 30
a
and
1
2
i 20
(t 1)3
60
i 10 2 e
(t 1)3
r  30 1
2
t/2
e
t/2
1
cos 2 t
2
2 sin 2 t
e
t/2
(
sin 2 t
4 cos 2 t ) j
(4 sin 2 t 7.5 cos 2 t ) j
1
i 20(1) j
1
r  20 mm 
or
v  30
1
i 20
1
(1)
1
0
2
or
a
j
v  43.4 mm/s
46.3° 
a  743 mm/s2
85.4° 
60
i 10 2 (1)(0 7.5) j
(1)
or
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125
PROBLEM 11.93 (Continued)
(b)
At t  1.5 s:
1
i 20e 0.75 (cos 3 ) j
2.5
 (18 mm)i ( 1.8956 mm) j
r  30 1
or
r  18.10 mm
6.01° 
v  5.65 mm/s
31.8° 
a  70.3 mm/s2
86.9° 
30
1
i 20 e 0.75
cos 3
0 j
2
2
(2.5)
 (4.80 mm/s)i (2.9778 mm/s) j
v
or
a
60
i 10 2 e 0.75 (0 7.5 cos 3 ) j
3
(2.5)
 ( 3.84 mm/s 2 )i (70.1582 mm/s2 ) j
or
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126
Problem 11.94
A girl operates a radio-controlled
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  10i 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
dy 2
where:
dx  4tdt and dy  3t 2 dt
Substituting
ds  16t 2 9t 4 dt
Integrating:
(c) Velocity can be found by
s
0
ds 
2
0
t 16 9t 2 dt
s
3/2 2
1
16 9t 2 |
0
27
v
dr
dt
s  11.52 m 
v  4ti 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  4i 6tj m/s 2
At t=2 s
a  4i 12 j m/s 2
a  12.65 m/s 2 
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127
PROBLEM 11.95
The three-dimensional motion of a particle is defined by the position vector r (Rt cos nt)i ctj
(Rt sin nt)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
Then
v
and
a
n t )i
ctj ( Rt sin
dr
 R(cos
dt
nt
n t sin
n sin
nt
n sin
R( n cos
nt
n cos
nt
2
n t cos
dv
dt
 R(
 R ( 2 n sin
Now
v 2  vx2
v 2y
vz2
 [ R (cos
nt
 R2
cos 2
sin 2
 R2 1
n t )i
2 nt sin
nt
2 n t sin
2 2
nt
c2
cj R(sin
2
n t cos
nt
n t )i
n t )k
nt
2
n t sin
n t )k
R (2 n cos
n t cos
nt
n t cos
2 2
2
n t sin
nt
2 2
2
n t cos
nt
a 2  ax2
 R
 R2
a 2y
2
c2
nt
2 2
nt
c2 
az2
2 n sin
nt
nt
nt
nt
2
n t sin
4 n2 sin 2
nt
4 n3t sin
4 n2 cos 2
nt
2
2
n t cos
R 2 n cos
 R 2 4 n2
n t )]
nt
v  R2 1
or
Also,
n t cos
n t )k
(c) 2 [ R(sin
n t cos
nt
n t )i
2
n t sin
nt
2
n t )]
n t sin
nt
n t )k
4 n3t sin
(0)2
2
n t cos
n t cos
nt
nt
4 2
2
n t cos
4 2
2
n t sin
nt
nt
4 2
nt
aR n 4
or
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128
2 2
nt 
PROBLEM 11.96*
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
described by the particle lies on the hyperboloid (y/A)2 - (x/A)2 (z /B)2 = 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.
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 sin t
sin t =
2
Ê yˆ
t2 = Á ˜ -1
Ë A¯
z
Bt
Then
cos t =
Now
Ê zˆ
Ê xˆ
cos2 t + sin 2 t = 1 fi Á ˜ + Á ˜ = 1
Ë Bt ¯
Ë At ¯
2
2
2
2
2
Ê zˆ
Ê xˆ
t2 = Á ˜ + Á ˜
Ë B¯
Ë A¯
or
2
Then
Ê zˆ
Ê xˆ
Ê yˆ
ÁË ˜¯ - 1 = ÁË ˜¯ + ÁË ˜¯
A
A
B
or
Ê zˆ
Ê xˆ
Ê yˆ
ÁË ˜¯ - ÁË ˜¯ - ÁË ˜¯ = 1
A
A
B
2
(a)
2
2
2
Q.E.D.
!
With A = 3 and B = 1, we have
dr
t
= 3(cos t - t sin t )i + 3
j + (sin t + t cos t )k
dt
t2 +1
Ê t ˆ
2
ÁË t 2 + 1 ˜¯
+
1
t
t
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
1
= -3(2 sin t + t cos t )i + 3 2
j + (2 cos t - t sin t )k
(t + 1)3/2
v=
and
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129
PROBLEM 11.96* (Continued)
At t = 0:
v = 3(1 - 0)i + (0) j + (0)k
v = v x2 + v 2y + v z2
v = 3 m/s !
or
and
a = -3(0)i + 3(1) j + (2 - 0)k
Then
a2 = (0)2 + (3)2 + (2)2 = 13
a = 3.61 m/s2 !
or
(b)
If r and v are perpendicular, r ◊ v = 0
(
)
t ˆ
Ê
j + (sin t + t cos t )k ] = 0
[(3t cos t )i + 3 t 2 + 1 j + (t sin t )k ] ◊ [3(cos t - t sin t )i + 3
ÁË
˜
t 2 + 1¯
t ˆ
Ê
or
+ (t sin t )(sin t + t cos t ) = 0
(3t cos t )[3(cos t - t sin t )] + (3 t 2 + 1) 3
ÁË
˜
2
t + 1¯
Expanding
(9t cos2 t - 9t 2 sin t cos t ) + (9t ) + (t sin 2 t + t 2 sin t cos t ) = 0
10 + 8 cos2 t - 8t sin t cos t = 0
or ( with t π 0)
7 + 2 cos 2t - 2t sin 2t = 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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130
PROBLEM 11.97
An airplane used to drop water on brushfires is flying horizontally in a straight line at 315 km / h at an altitude of 80 m.
Determine the distance d at which the pilot should release the
water so that it will hit the fire at B.
SOLUTION
First note
v0 = 315 km/h
= 87.5 m/s
Vertical motion. (Uniformly accelerated motion)
y = 0 + (0)t -
1 2
gt
2
At B:
1
- 80 m = - (9.81 m/s2 )t 2
2
or
t B = 4.03855 s
Horizontal motion. (Uniform)
x = 0 + ( v x )0 t
At B:
d = (87.5 m/s)( 4.03855 s)
d = 353 m !
or
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131
PROBLEM 11.98
A ski jumper starts with a horizontal take-off velocity of 25
m/s and lands on a straight landing hill inclined at 30o.
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.
SOLUTION
y  x tan 30
(a) At the landing point,
Horizontal motion:
Vertical motion:
y  y0
from which
t2 
d 
(b) Landing distance:
(c) Vertical distance:
or
v x 0 t  v0t
vy t
1 2
gt 
2
0
1 2
gt
2
2y
2 x tan 30
2v t tan 30

 0
g
g
g
t 
Rejecting the t  0 solution gives
x  x0
2 25 tan 30
2v0 tan 30

g
9.81
25 2.94
x
v0t


cos 30
cos 30
cos 30
h  x tan 30
y
h  v0t tan 30
1 2
gt
2
t  2.94 s 
d  84.9 m 
Differentiating and setting equal to zero,
dh
 v0 tan 30
dt
Then,
hmax 
gt  0
or
v0 v0 tan 30 tan 30
g
2
t 
vo tan 30
g
1
v tan 30
g 0
2
g
25 tan 30
v 2 tan 2 30
 0

2g
2 9.81
2
2
hmax  10.62 m 
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PROBLEM 11.99
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
corresponding to
h 788 mm and h 1068 mm.
SOLUTION
(a)
y0  1.5 m, (v y )0  0
Vertical motion:
or
t 
2( y0 y)
g
yh
or
tB 
2( y0 h)
g
When h  788 mm  0.788 m,
tB 
(2)(1.5 0.788)
 0.3810 s
9.81
When h  1068 mm  1.068 m,
tB 
(2)(1.5 1.068)
 0.2968 s
9.81
y  y0
(v y )0 t
At Point B,
Horizontal motion:
1 2
gt
2
x0  0, (vx )0  v0 ,
x  v0t
or
v0 
x
x
 B
t
tB
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133
PROBLEM 11.99 (Continued)
With xB  12.2 m,
we get
and
32.02 m/s
(b)
v0
41.11 m/s
v0 
12.2
 32.02 m/s
0.3810
v0 
12.2
 41.11 m/s
0.2968
115.3 km/h
or
Vertical motion:
v y  (v y )0
Horizontal motion:
vx  v0
dy

dx
(v y ) B
v0
148.0 km/h 
gt  gt

For h  0.788 m,
tan

(9.81)(0.3810)
 0.11673,
32.02
 6.66 
For h  1.068 m,
tan

(9.81)(0.2968)
 0.07082,
41.11
 4.05 
(vx ) B

gtB
v0
tan
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134
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.
SOLUTION
Vertical motion. (Uniformly accelerated motion)
y = 0 + (0)t -
1 2
gt
2
Horizontal motion. (Uniform)
x = 0 + ( v x )0 t = v0 t
At B:
y:
t B = 0.45152 s
or
Then
x:
y:
or
1
- 0.6 m = - (9.81 m/s2 )t 2
2
tC = 0.34975 s
or
Then
(2.1) m = ( v0 ) B (0.45152 s)
( v0 ) B = 4.651 m/s
or
At C:
1
-1 m = - (9.81 m/s2 )t 2
2
x:
3.7 m = ( v0 )C (0.34975 s)
( v0 )C = 10.579 m/s
4.65 m/s ≤ v0 ≤ 10.58 m/s !
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135
PROBLEM 11.101
Water flows from a drain spout with an initial velocity of
0.75 m /s at an angle of 15° with the horizontal. Determine the
range of values of the distance d for which the water will enter
the trough BC.
SOLUTION
( v x )0 = (0.75 m/s) cos 15∞ = 0.72444 m/s
First note
( v y )0 = -(0.75 m/s) sin 15∞ = - 0.19411 m/s
Vertical motion. (Uniformly accelerated motion)
y = 0 + ( v y )0 t -
1 2
gt
2
At the top of the trough, y = -3 + 0.36 = -2.64 m
1
- 2.64 m = ( - 0.19411 m/s) t - (9.81 m/s2 ) t 2
2
t BC = 0.71412 s (the other root is negative)
or
Horizontal motion. (Uniform)
x = 0 + ( v x )0 t
In time t BC
x BC = (0.72444 m/s)(0.71412 s) = 0.5173 m
Thus, the trough must be placed so that
x B £ 0.5173 m
xC ≥ 0.5173 m
0 £ d £ 0.5173 m !
Since the trough is 0.6 m wide, it then follows that
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136
PROBLEM 11.102
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 v 0 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.
SOLUTION
Vertical motion:
v0  13 m/s,
 33 , x0  0, y0  0.6 m
v y  v0 sin
gt
y  y0
v0 sin
t
1 2
gt
2
At maximum height,
vy  0
or
t 
v0 sin
g
(a)
t 
ymax  0.6
13sin 33
13sin 33
 0.7217 s
9.81
0.7217
1
2
9.81 0.7217
2
1.8 m
3.16 m
Horizontal motion:
x  x0
v0 cos
At x  15.2 m,
t 
(b) Corresponding value of y :
ymax  3.16 m 
yes 
3.7 m
t
or
t 
x x0
v0 cos
15.2 0
 1.3941 s
13cos 33
y  0.6
13sin 33
1.3941
1
2
9.81 1.3941
2
y  0.937 m 
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137
PROBLEM 11.103
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.
SOLUTION
First note
(vx )0  (13.40 m/s) cos 20  12.5919 m/s
(v y )0  (13.40 m/s) sin 20  4.5831 m/s
(a)
Horizontal motion.. (Uniform)
x  0 (vx )0 t
At C
9 m  (12.5919 m/s) t or tC  0.71475 s
Vertical motion.. (Uniformly accelerated motion)
y  y0
At C:
(v y ) 0 t
1 2
gt
2
yC  2.1 m (4.5831 m/s)(0.71475 s)
1
(9.81 m/s 2 )(0.71475 s)2
2
 2.87 m
yC
(b)
2.43 m (height of net)
ball clears net 
1
(9.81 m/s2 )t 2
2
At B, y  0:
0  2.1 m (4.5831 m/s)t
Solving
tB  1.271175 s (the other root is negative)
Then
d  (vx )0 t B  (12.5919 m/s)(1.271175 s)
 16.01 m
b  (16.01 9.00) m  7.01 m from the net 
The ball lands
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PROBLEM 11.104
A golfer hits a golf ball with an
initial velocity of 50 m /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
( v x )0 = (50 m/s) cos 25∞
First note
( v y )0 = (50 m/s)sin 25∞
x B = d cos 5∞
and at B
Now
y B = - d sin 5∞
Horizontal motion. (Uniform)
x = 0 + ( v x )0 t
d cos 5∞ = (50 cos 25∞)t
At B
or t B =
cos 5∞
d
50 cos 25∞
Vertical motion. (Uniformly accelerated motion)
y = 0 + ( v y )0 t -
1 2
gt
2
( g = 9.81 m/s2 )
1 2
gt B
2
At B:
- d sin 5∞ = (50 sin 25∞)t B -
Substituting for t B
Ê cos 5∞ ˆ
1 Ê cos 5∞ ˆ 2
- d sin 5∞ = (50 sin 25∞) Á
d - gÁ
d
˜
2 Ë 50 cos 25∞ ˜¯
Ë 50 cos 25∞ ¯
2
or
or
2
(50 cos 25∞)2 (tan 5∞ + tan 25∞)
9.81 cos 5∞
= 232.733 m
d=
d = 233 m !
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139
PROBLEM 11.105
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.
SOLUTION
First note
( v x )0 = v0 cos 40∞
( v y )0 = v0 sin 40∞
Horizontal motion. (Uniform)
x = 0 + ( v x )0 t
4.2 = ( v0 cos 40∞) t
At B:
or t B =
4.2
v0 cos 40∞
Vertical motion. (Uniformly accelerated motion)
y = 0 + ( v y )0 t -
1 2
gt
2
0.45 = ( v0 sin 40∞) t B -
At B:
( g = 9.81 m/s2 )
1 2
gt B
2
Substituting for t B
Ê
4.2 ˆ 1 Ê
4.2 ˆ
0.45 = ( v0 sin 40∞) Á
- gÁ
˜
v
cos
40
∞
2
v
cos
40∞ ˜¯
Ë 0
¯
Ë 0
1
or
v02 = 2
2
(9.81)(17.64) / cos2 40∞
-0.45 + 4.2 tan 40∞
v0 = 6.93 m/s !
or
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140
PROBLEM 11.106
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.
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 x0  0, y0  2m. The components of initial velocity are (vx )0  v0 cos 40 m/s
and (v y )0  v0 sin 40 .
Horizontal motion:
Vertical motion:
x  x0
(vx )0 t  (v0 cos 40 )t
1 2
gt
2
1
 2 (v0 sin 40 ) 
(9.81)t 2
2
y  y0
(v y ) 0 t 
v0t 
Then
y  2 x tan 40
t2 
(2)
x
cos 40
From (1),
Point B:
(1)
2 x tan 40
4.905
(3)
4.905t 2
y
(4)
x  8 10 cos35  16.1915 m
y  1.5 10sin 35  7.2358 m
16.1915
 21.1365 m
cos 40
2 16.1915 tan 40 7.2358
t2 
4.905
21.1365
v0 
1.3048
v0t 
t  1.3048 s
v0  16.199 m/s
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PROBLEM 11.106 (Continued)
Point C:
x  8 (10 7) cos35  21.9256 m
y  1.5 (10 7)sin 35  11.2508 m
v0t 
21.9256
 28.622 m
cos 40
t2 
2 21.9256 tan 40
4.905
v0 
28.622
1.3656
11.2508
t  1.3656 s
v0  20.96 m/s
16.20 m/s v0
Range of values of v0.
21.0 m/s 
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PROBLEM 11.107
A basketball player shoots when she is 5 m 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) 225 mm, (b) 425 mm.
SOLUTION
First note
( v x )0 = v0 cos 30∞
( v y )0 = v0 sin 30∞
Horizontal motion. (Uniform)
x = 0 + ( v x )0 t
(5 - d ) = ( v0 cos 30∞) t
At B:
or t B =
5-d
v0 cos 30∞
Vertical motion. (Uniformly accelerated motion)
1 2
gt ( g = 9.81 m/s2 )
2
1
0.9 = ( v0 sin 30∞) t B - gt B2
2
y = 0 + ( v y )0 t -
At B:
Substituting for tB
Ê 5-d ˆ 1 Ê 5-d ˆ
0.9 = ( v0 sin 30∞) Á
- g
Ë v cos 30∞ ˜¯ 2 ÁË v cos 30∞ ˜¯
0
or
v02 =
2 g (5 - d ) 2
È 1
˘
3 Í (5 - d ) - 0.9˙
Î 3
˚
2
0
d~m
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143
PROBLEM 11.107 (Continued)
2(9.81) (5 - 0.225)
2
(a)
d = 225 mm = 0.225 m
v02 =
or
3 È 13 (5 - 0.225) - 0.9˘
Î
˚
2(9.81) (5 - 0.425)
2
(b)
d = 425 mm = 0.425 m
or
v02 =
3 È 13 (5 - 0.425) - 0.9˘
Î
˚
v0 = 8.96 m/s !
v0 = 8.87 m/s !
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144
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.
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 x0  0, y0  h  2.5 m. The
components of initial velocity are (vx )0  v0 cos5 and (v y )0  v0 sin 5 .
Horizontal motion:
Vertical motion:
x  x0
(vx )0 t  (v0 cos5 )t
(1)
1 2
gt
2
1
 2.5 (v0 sin 5 )t 
(9.81)t 2
2
y  y0
(v y ) 0 t 
x
cos5
From (1),
v0t 
Then
y  2.5 x tan 5
t2 
(3)
2.5 x tan 5
4.905
4.905t 2
y
(4)
At the minimum speed the ball just clears the net.
x  12.2 m,
(2)
y  0.914 m
12.2
 12.2466 m
cos 5
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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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
y0
18.6
v0t 
 18.6710 m
cos 5
2.5 18.6 tan 5 0
t2 
t  0.42181 s
4.905
18.6710
v0 
v0  44.26 m/s
0.42181
37.7 m/s v0
Range for v0.
44.3 m/s 
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PROBLEM 11.109
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.
C
SOLUTION
First note
(vx )0  v0 cos 6
(v y )0  v0 sin 6
Horizontal motion. (Uniform)
x  x0
(vx )0 t
Vertical motion.. (Uniformly accelerated motion)
y  y0
(v y ) 0 t
1 2
gt
2
( g  9.81 m/s2 )
x  (0.175 m) sin 10
At Point B:
y  (0.175 m) cos 10
x : 0.175 sin 10  0.020 (v0 cos 6 )t
tB 
or
0.050388
v0 cos 6
y : 0.175 cos 10  0.205 ( v0 sin 6 )tB
1 2
gtB
2
Substituting for tB
0.050388
0.032659  ( v0 sin 6 )
v0 cos 6
1
0.050388
(9.81)
2
v0 cos 6
2
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PROBLEM 11.109 (Continued)
v02 
or
1
(9.81)(0.050388) 2
2
cos 2 6 (0.032659 0.050388 tan 6 )
(v0 ) B  0.678 m/s
or
x  (0.175 m) cos 30
At Point C:
y  (0.175 m) sin 30
x : 0.175 cos 30  0.020 (v0 cos 6 )t
tC 
or
0.171554
v0 cos 6
y : 0.175 sin 30  0.205 ( v0 sin 6 )tC
1 2
gtC
2
Substituting for tC
0.171554
0.117500  ( v0 sin 6 )
v0 cos 6
or
or
v02 
1
0.171554
(9.81)
2
v0 cos 6
2
1
(9.81)(0.171554) 2
2
cos 2 6 (0.117500 0.171554 tan 6 )
(v0 )C  1.211 m/s
0.678 m/s v0 1.211 m/s 
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148
PROBLEM 11.110
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.
SOLUTION
First note
(vx )0  v0 cos 65
(v y )0  v0 sin 65
Horizontal motion. (Uniform)
x  0 (vx )0 t
At either B or C, x  5 m
s  (v0 cos 65 )tB,C
t B ,C 
or
5
(v0 cos 65 )
Vertical motion.. (Uniformly accelerated motion)
1 2
gt
2
y  0 (v y ) 0 t
( g  9.81 m/s 2 )
At the tree limbs, t  tB ,C
yB ,C  (v0 sin 65 )
5
v0 cos 65
1
5
g
2
v0 cos 65
2
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149
PROBLEM 11.110 (Continued)
v02 
or
1
(9.81)(25)
2
cos 2 65 (5 tan 65
yB , C )

686.566
5 tan 65 yB , C
At Point B:
v02 
686.566
5 tan 65 5
or
(v0 ) B  10.95 m/s
At Point C:
v02 
686.566
5 tan 65 5.9
or
(v0 )C  11.93 m/s
10.95 m/s v0 11.93 m/s 
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150
PROBLEM 11.111
The pitcher in a softball game throws a ball with an initial velocity v0 of 72 km/h at an angle
horizontal. If the height of the ball at Point B is 0.68 m, determine (a) the angle , (b)
( the angle
velocity of the ballat Point B forms with the horizontal.
with the
that the
SOLUTION
First note
v0  72 km/h  20 m/s
and
(a)
(vx )0  v0 cos
 (20 m/s) cos
(v y )0  v0 sin
 (20 m/s) sin
Horizontal motion.. (Uniform)
x  0 (vx )0 t  (20 cos ) t
14  (20 cos )t or t B 
At Point B:
7
10 cos
Vertical motion.. (Uniformly accelerated motion)
y  0 (v y )0 t
At Point B:
1 2
gt  (20 sin )t
2
0.08  (20 sin )t B
1 2
gt
2
( g  9.81 m/s2 )
1 2
gt B
2
Substituting for tB
7
0.08  (20 sin )
10 cos
Now
1
cos 2
 sec 2
2
1
49
g
2 cos 2
8  1400 tan
or
1
7
g
2 10 cos
 1 tan 2
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151
PROBLEM 11.111 (Continued)
or
24.5 g (1 tan 2 )
8  1400 tan
Then
240.345 tan 2
248.345  0
1400 tan
 10.3786
Solving
 79.949
and
Rejecting the second root because it is not physically reasonable, we have
 10.38 
(b)
We have
vx  (vx )0  20 cos
and
v y  (v y ) 0
gt  20 sin
(v y ) B  20 sin
gtB
At Point B:
 20 sin
Noting that at Point B, v y
gt
7g
10 cos
0, we have
tan



|(v y ) B |
vx
7g
10 cos
20 sin
20 cos
7
9.81
200 cos 10.3786
sin 10.3786
cos 10.3786
 9.74 
or
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152
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
that v0 forms with the vertical, (b) the maximum height above Point A
reached by the rocket, and (c) the duration of the flight.
SOLUTION
Set the origin at Point A.
x0  0, y0  0
Horizontal motion:
x  v0t sin
sin
Vertical motion:
y  v0t cos
1 2
gt
2
sin 2
cos

1
y
v0t
cos2

1
x2
(v0t )2
1 24
g t
4
At Point B,
x2

x
v0t
(1)
1 2
gt
2
(2)
1 2
gt
2
y
x2
y2
gyt 2
v02
gy t 2
( x2
2
1
1 24
g t  v02t 2
4
y2 )  0
y 2  100 m, x  100 cos 30 m
y  100 sin 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 s 2
and 1.6458 s 2
t  15.8916 s
and 1.2829 s
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153
(3)
PROBLEM 11.112 (Continued)
0
Restrictions on :
tan
120

x
y
1
gt 2
2

100 cos 30
 0.0729
50 (4.905)(15.8916)2
 4.1669
100 cos 30
 2.0655
50 (4.905)(1.2829) 2
 115.8331
and
Use
 4.1669 corresponding to the steeper possible trajectory.
(a)
Angle .
(b)
Maximum height.
 4.17 
v y  0 at
y  ymax
v y  v0 cos
gt  0
t
v0 cos
g
ymax  v0t cos

(c)
Duration of the flight.
v 2 cos 2
1
gt  0
2
2g
(75) 2 cos 2 4.1669
(2)(9.81)
ymax  285 m 
(time to reach B)
t  15.89 s 
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154
PROBLEM 11.113
The initial velocity v0 of a hockey
puck is 160 km / 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
First note
and
400
m/s
9
ˆ
Ê 400
m/s˜ cos a
( v x )0 = v0 cos a = Á
¯
Ë 9
v0 = 160 km/h =
ˆ
Ê 400
m/s˜ sin a
( v y )0 = v0 sin a = Á
¯
Ë 9
(a)
Horizontal motion. (Uniform)
ˆ
Ê 400
x = 0 + ( v x )0 t = Á
cos a ˜ t
¯
Ë 9
At the front of the net,
Then
or
x=5m
ˆ
Ê 400
cos a ˜ t
5=Á
¯
Ë 9
9
tenter =
80 cos a
Vertical motion. (Uniformly accelerated motion)
y = 0 + ( v y )0 t -
1 2
gt
2
1
ˆ
Ê 400
=Á
sin a ˜ t - gt 2
¯
Ë 9
2
( g = 9.81 m/s2 )
At the front of the net,
1 2
ˆ
Ê 400
yfront = Á
sin a ˜ tenter - gtenter
¯
Ë 9
2
ˆ 1 Ê
ˆ
9
9
ˆÊ
Ê 400
=Á
- g
sin a ˜ Á
¯ Ë 80 cos a ˜¯ 2 ÁË 80 cos a ˜¯
Ë 9
= 5 tan a Now
2
81g
12800 cos2 a
1
= sec2 a = 1 + tan 2 a
cos2 a
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167
155
PROBLEM 11.113 (Continued)
Then
yfront = 5 tan a -
or
tan 2 a -
81g
(1 + tan 2 a )
12800
ˆ
Ê 12800
64000
yfront ˜ = 0
tan a + Á1 +
¯
Ë
81g
81g
(
È 64000
64000
81g ± Í - 81g
Î
)
2
(
)
1/ 2
˘
- 4 1 + 12800
81g yfront ˙
˚
2
Then
tan a =
or
2
ÈÊ
Ê
ˆ˘
64000
64000 ˆ
12800
tan a =
± ÍÁ - Á1 +
yfroont ˜ ˙
˜
162 ¥ 9.81 ÍË 162 ¥ 9.81¯
81 ¥ 9.81
Ë
¯˙
Î
˚
or
tan a = 40.2713 ± [( 40.2713)2 - (1 + 16.1085 yfront )]1/ 2
1/ 2
Now 0 £ yfront £ 1.2 m 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 = 1.2 m
(b)
Then
tan a = 40.2713 - [( 40.2713)2 - (1 + 16.1085 + 1.2)]1/ 2
or
a max = 14.2093∞
9
tenter =
80 cos a
We had found
=
a max = 14.66∞ !
9
80 cos 14.2093∞
tenter = 0.1161 s !
or
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168
156
PROBLEM 11.114
A worker uses high-pressure
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 .
SOLUTION
First note
By observation, dmax occurs when
(vx )0  v0 cos
 (11.5 m/s) cos
(v y )0  v0 sin
 (11.5 m/s) sin
ymax  1.1 m.
Vertical motion.. (Uniformly accelerated motion)
v y  (v y )0
 (11.5 sin )
y  ymax
When
Then
at
1 2
gt
2
1 2
 (11.5 sin ) t
gt
2
y  0 (v y )0 t
gt
gt
B , (v y ) B  0
(v y ) B  0  (11.5 sin ) gt
11.5 sin
g
( g  9.81 m/s 2 )
or
tB 
and
yB  (11.5 sin ) t B
1 2
gtB
2
Substituting for tB and noting yB  1.1 m
11.5 sin
1.1  (11.5 sin )
g
or
sin 2

1
(11.5)2 sin 2
2g

2.2 9.81
11.52
1 11.5 sin
g
2
g
2
 23.8265
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157
PROBLEM 11.114 (Continued)
(a)
Horizontal motion. (Uniform)
x  0 (vx )0 t  (11.5 cos ) t
At Point B:
x  d max
where
tB 
Then
and t  t B
11.5
sin 23.8265  0.47356 s
9.81
d max  (11.5)(cos 23.8265 )(0.47356)
d max  4.98 m 
or
(b)
 23.8 
From above
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158
PROBLEM 11.115
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 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
 (8 m/s) cos
(v y )0  v0 sin
 (8 m/s) sin
Horizontal motion. (Uniform)
x  0 (vx )0 t  (8 cos ) t
At Point B:
or
x  d : d  (8 cos ) t
d
tB 
8 cos
Vertical motion.. (Uniformly accelerated motion)
1 2
gt
2
1 2
 (8 sin ) t
gt ( g  9.81 m/s 2 )
2
1 2
0  (8 sin ) t B
gt B
2
y  0 (v y ) 0 t
At Point B:
Simplifying and substituting for tB
0  8 sin
d
or
(a)
1
d
g
2 8 cos
64
sin 2
g
(1)
When h  0, the water can follow any physically possible trajectory. It then follows from
Eq. (1) that d is maximum when 2  90
 45 
or
Then
d
64
sin (2 45 )
9.81
d max  6.52 m 
or
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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 increases
in value from 0 to 45° and then d decreases as is further increased. Thus, dmax occurs for the value
of closest to 45° and for which the water just passes over the first row of corn plants. At this row,
xcom  1.5 m
tcorn 
so that
1.5
8 cos
Also, with ycorn  h, we have
1 2
gtcorn
2
h  (8 sin ) tcorn
Substituting for tcorn and noting h  1.8 m,
1.5
1.8  (8 sin )
8 cos
or
Now
Then
or
 sec 2
 1 tan 2
1.8  1.5 tan
0.172441 tan 2
Solving
2
2.25 g
128 cos 2
1.8  1.5 tan
1
cos 2
1
1.5
g
2 8 cos
2.25(9.81)
(1 tan 2 )
128
1.5 tan
1.972441  0
 58.229
and
 81.965
From the above discussion, it follows that d  d max when
 58.2 
Finally, using Eq. (1)
d
64
sin (2 58.229 )
9.81
d max  5.84 m 
or
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160
PROBLEM 11.116
A ball is dropped onto a step at Point A and rebounds with a
velocity v0 at an angle of 15° with the vertical. Determine the
value of v0, knowing that just before the ball bounces at Point B,
its velocity vB forms an angle of 12° with the vertical.
SOLUTION
First note
( v x )0 = v0 sin 15∞
( v y )0 = v0 cos 15∞
Horizontal motion. (Uniform)
v x = ( v x )0 = v0 sin 15∞
Vertical motion. (Uniformly accelerated motion)
v y = ( v y )0 - gt
At Point B, v y < 0
= v0 cos 15∞ - gt
tan 12∞ =
Then
tB =
or
1 2
gt
2
1
= ( v0 cos 15∞) t - gt 2
2
y = 0 + ( v y )0 t -
(vx ) B
v0 sin 15∞
=
| ( v y ) B | gt B - v0 cos 15∞
ˆ
v0 Ê sin 15∞
+ cos 15∞˜
Á
g Ë tan 12∞
¯
g = 9.81 m /s2
= 0.22259 v0
Noting that
We have
or
y B = -0.2 m,
1
- 0.2 = ( v0 cos 15∞)(0.22259 v0 ) - (9.81)(0.22259 v0 )2
2
v0 = 2.67 m /s !
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161
PROBLEM 11.117
The velocities of skiers A and B are
as shown. Determine the velocity
of A with respect to B.
SOLUTION
We have
vA = vB + vA/B
The graphical representation of this equation is then as shown.
Then
v 2A/B = 102 + 142 - 2(10)(14) cos 15∞
or
vA/B = 5.05379 m/s
and
10
5.05379
=
sin a
sin 15∞
or
a = 30.8°
vA/B = 5.05 m/s
55.8° !
Alternative solution.
vA/B = v A - vB
= 10 cos 10i - 10 sin 10 j - (14 cos 25i - 14 sin 25 j)
= - 2.84i + 4.1
14j
= 5.05 m/s
55.8°
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162
PROBLEM 11.118
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.
SOLUTION
From the diagram
yD  constant
Cable 1:
yA
Then
vA vD  0
Cable 2:
( yB
yD ) ( yC
vB
Then
vC
(1)
yD )  constant
2vD  0
(2)
Combining Eqs. (1) and (2) to eliminate vD ,
vC  0
(3)
Now
vA/C  vA vC  300 mm/s
(4)
and
vB/A  vB
(5)
Then
(3) (4) (5)
2vA
vB
vA  200 mm/s
(2v A vB vC ) (vA vC ) (vB vA )  ( 300) (200)
v A  125 mm/s 
or
vB
and using Eq. (5)
( 125)  200
v B  75 mm/s 
or
125 vC  300
Eq. (4)
vC  175 mm/s 
or
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PROBLEM 11.119
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.
SOLUTION
50
m/s
3
v B  90 km/h  25 m/s
vA  60 km/h 
Law of cosines
2
50
+ (25)2
3
v B / A  34.46 m/s
vB2 / A 
vB  v A
2
50
(25) cos110
3
Law of sines
vB/A
sin
sin110

50
34.46
3
 90
 90
(a)
Relative velocity:
(b)
Change in position for t  4 s.
 27.03
27.03  62.96
vB/A = 34.5 m/s
DrrB/A = vB/ADt = (34.46 m/s)
(c)
63.0° 
vB/A = 137.8 m
63.0° 
Distance between autos 2 seconds after auto A has passed intersection.
Auto A travels for 2 s.
vA 
rA  v At 
50
m/s
3
20°
50
100
m/s (2s ) 
m/s
3
3
rA = 33.3 m
20°
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PROBLEM 11.119 (Continued)
Auto B
vB = 25 m/s
rBvB = (25 m/s)(5s) = 125 m
rB  rA
rB/A
Law of cosines
2
100
(125)2
3
rA  139.95 m
rB2/ A 
2
100
(125) cos 110°
3
Distance between autos 140m
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165
PROBLEM 11.120
Shore-based
based radar indicates that a ferry leaves its slip with a
velocity v  18 km/h 70°, while instruments aboard the
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
vF  vR
v F/R
or v F  v F/R
vR
The graphical representation of the second equation is then as shown.
We have
vR2  182 18.42
or
vR  3.1974 km/h
and
18
3.1974

sin
sin 10
or
 77.84
2(18)(18.4) cos 10
Noting that
v R  3.20 km/h
17.8° 
Alternatively one could use vector algebra.
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PROBLEM 11.121
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 ofB with respect to A,
(b) the velocity of A if ground-based
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
15
interval.
SOLUTION
(a)
We have
vC  v A
vC/A
and
vC  v B
vC/B
Then
vA
vC/A  v B
vC/B
or
vB
v A  vC/A
vC/B
Now
vB
v A  v B/A
so that
v B/A  vC/A
vC/B
or
vC/A  vC/B
v B/A
The graphical representation of the last equation is then as shown.
We have
vB2/A  3502
or
vB/A  405.175 km/h
and
400
405.175

sin
sin 65
or
 63.474
75
4002
2(350)(400) cos 65
 11.526
v B/A  405 km/h
11.53° 
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167
PROBLEM 11.121 (Continued)
(b)
We have
vC  v A
or
v A  (30 km/h) j (350 km/h)( cos 75 i sin 75 j )
vC/A
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
rC/B  rC
Now
v Bt
rC  (rC )0
vC t
rB  [(rC )0
(rB )0 ] ( vC
v B )t
 [(rC )0 (rB )0 ] vC/B t
rC/B  (rC/B )t2
Then
For
t  15 min:
(rC/B )t1  vC/B (t2
rC/B  (400 km/h)
t1 )  vC/B t
1
h  100 km
4
rC/B  100 km
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40° 
PROBLEM 11.122
Instruments in an airplane which is in level flight indicate that the velocity relative
to the air (airspeed) is 120 km/h and the direction of the relative velocity vector
(heading) is 70° east of north. Instruments on the ground indicate that the velocity of
the airplane (ground speed) is 110 km/h and the direction of flight (course) is 60°
east of north. Determine the wind speed and direction.
SOLUTION
Let i and j be unit vectors in directions east and north respectively.
Velocity of plane relative to air.
v P/ A  120 km/h cos 20 i
sin 20 j
v P  110 km/h cos 30 i
sin 30 j
Velocity of plane.
But
vP  v A
v P/ A
Velocity of air.
v A  vP
v P/ A
v A  110cos 30

17.50 km/h i
vA 
tan
120 cos 20 i
17.50

17.50
13.96
2
110sin 30
120sin 20 j
13.96 km/h j
13.96
2
 22.4 km/h
 51.4
v A  22.4 km/h at 51.4 west of north 
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169
PROBLEM 11.123
Knowing that the velocity of block B with respect to block A is
vB/A = 5.6 m /s
70°, determine the velocities of A and B.
SOLUTION
From the diagram
2 x A + 3x B = constant
2v A + 3v B = 0
2
| vB | = v A
3
v B = v A + v B /A
Then
or
Now
and noting that vA and vB must be parallel to surfaces A and B, respectively, the graphical representation of this
equation is then as shown. Note: Assuming that vA is directed up the incline leads to a velocity diagram that
does not “close.”
First note
a = 180∞ - ( 40∞ + 30∞ + q B )
= 110∞ - q B
2
vA
vA
5.6
= 3
=
sin (110∞ - q B ) sin 40∞ sin (30∞ + q B )
Then
v A sin 40∞ =
or
2
v A sin (110∞ - q B )
3
or
sin (110∞ - q B ) = 0.96418
or
q B = 35.3817∞
and
q B = 4.6183∞
For q B = 35.3817∞ :
vB =
or
v A = 5.94 m/s v B = 3.96 m/s
5.6 sin 40∞
2
vA =
3
sin (30∞ + 35.3817∞)
v A = 5.94 m/s
30° !
v B = 3.96 m/s
35.4° !
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170
PROBLEM 11.123 (Continued)
For q B = 4.6183∞ :
or
vB =
5.6 sin 40∞
2
vA =
3
sin (30∞ + 4.6183∞)
v A = 9.50 m/s
v B = 6.34 m/s
v A = 9.50 m/s
30° !
v B = 6.34 m/s
4.62° !
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171
PROBLEM 11.124
Knowing that at the instant shown block A has a velocity of 200 mm /s
and an acceleration of 150 mm /s2 both directed down the incline,
determine (a) the velocity of block B, (b) the acceleration of block B.
SOLUTION
From the diagram
2x A + x B /A = constant
Then
2 v A + v B /A = 0
| v B /A | = 400 mm/s
or
2 a A + a B /A = 0
and
| aB /A | = 300 mm/s2
or
Note that vB/A and aB/A must be parallel to the top surface of block A.
(a)
We have
v B = v A + v B /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 = 2002 + 4002 - 2(200)( 400) cos 15∞
or
v B = 213.194 mm/s
and
200 213.194
=
sin a
sin 15∞
or
a = 14.05∞
v B = 213.194 mm/s
(b)
54.1° !
The same technique that was used to determine vB can be used to determine aB. An alternative method is
as follows.
We have
a B = a A + a B /A
= (150i ) + 300( - cos 15∞i + sin 15∞ j)*
= - (139.778 mm/s2 )i + (77.646 mm/s2 ) j
a B = 159.9 mm/s2
or
54.1° !
* Note the orientation of the coordinate axes on the sketch of the system.
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172
PROBLEM 11.125
A boat is moving to the right with a constant
deceleration of 0.3 m/s2 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.
SOLUTION
Horizontal motion of the ball:
vx  (vx )0 ,
xball  (vx )0 t
Vertical motion of the ball:
v y  (v y )0
gt
yB  (v y )0 t
At maximum height,
1 2
gt , (v y ) 2 (v y )02  2 gy
2
vy  0
and
y  ymax
(v y )2  2 gymax  (2)(9.81)(8)  156.96 m2 /s2
(v y )0  12.528 m/s
At time of catch,
y  0  12.528
tcatch  2.554 s
or
Motion of the deck:
vx  (vx )0
Motion of the ball relative to the deck:
(vB/D ) x  (vx )0
xB/D  (vx )0 t
(vB/D ) y  (v y )0
(a)
(b)
At time of catch,
1
(9.81)t 2
2
and
v y  12.528 m/s
aDt,
xdeck  (vx )0 t
a D t ]  aD t
[(vx )0
(vx )0 t
gt ,
1
aDt 2
2
1
aD t 2 
2
1
aD t 2
2
yB/D  yB
1
( 0.3)(2.554)2
2
(vB/D ) x  ( 0.3)(2.554)  0.766 m/s
d  0.979 m 
d  xD/B 
(vB/D ) y  12.528 m/s
or 0.766 m/s
v B/D  12.55 m/s
86.5 
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173
PROBLEM 11.126
The assembly of rod A and wedge B starts from rest and moves
to the right with a constant acceleration of 2 mm/s2. Determine
(a) the acceleration of wedge C, (b)) the velocity of wedge C
when t  10 s.
SOLUTION
(a)
We have
aC  a B
aC/B
The graphical representation of this equation is then as shown.
First note
Then
 180
 55
(20
105 )
aC
2

sin 20
sin 55
aC  0.83506 mm/s2
(b)
aC  0.835 mm/s 2
75° 
vC  8.35 mm/s
75° 
For uniformly accelerated motion
vC  0 aC t
At t  10 s:
vC  (0.83506 mm/s 2 )(10 s)
 8.3506 mm/s
or
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174
PROBLEM 11.127
Determine the required velocity of the belt B if the relative velocity with
which the sand hits belt B is to be (a) vertical, (b) as small as possible.
SOLUTION
A grain of sand will undergo projectile motion.
vsx = vsx = constant = -2.5 m /s
0
vsy = 2 gh = (2)(9.81 m /s2 )(1.5 m) = 5.4249 m /s Ø
y-direction.
v S /B = v S - v B
Relative velocity.
(a)
(1)
If vS /B is vertical,
- vS /B j = -2.5i - 5.4249 j - ( - v B cos 15i + v B sin 15 j)
= -2.5i - 5.4249 j + v B cos 15i - v B sin 15 j
Equate components.
i: 0 = -2.5 + v B cos 15∞
vB =
2.5
= 2.5882 m/s
cos 15∞
v B = 2.59 m/s
(b)
15° !
vS /C is as small as possible, so make vS /B ^ to v B into (1).
- vS /B sin 15∞i - vS /B cos 15∞ j = - 2.5i - 5.4249 j + v B cos 15∞i - v B sin 15∞ j
Equate components and transpose terms.
(sin 15∞) vS /B + (cos 15∞) v B = 2.5
(cos 15∞) vS /B - (sin 15∞) v B = 5.4249
Solving,
vS /B = 5.8871 m /s
v B = 1.0107 m /s
v B = 1.011 m/s
15° !
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175
PROBLEM 11.128
Conveyor belt A, which forms a 20°
angle with the horizontal, moves at a
constant speed of 1.2 m /s and is used
to load an airplane. Knowing that a
worker tosses duffel bag B with an
initial velocity of 0.75 m /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.
SOLUTION
First determine the velocity of the bag as it lands on the belt. Now
[( v B ) x ]0 = ( v B )0 cos 30∞
= (0.75 m/s) cos 30∞
[( v B ) y ]0 = ( v B )0 sin 30∞
= (0.75 m/s)sin 30∞
Horizontal motion. (Uniform)
x = 0 + [( v B ) x ]0 t
( v B ) x = [( v B ) x ]0
= (0.75 cos 30∞) t
= 0.75 cos 30∞
Vertical motion. (Uniformly accelerated motion)
y = y0 + [( v B ) y ]0 t -
1 2
gt
2
= 0.45 + (0.75 sin 30∞)t -
( v B ) y = [( v B ) y ]0 - gt
1 2
gt
2
= 0.75 sin 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
0.45 + (0.75 sin 30∞) t -
1 2
gt = [(0.75 cos 30∞) t ]tan 20∞
2
or
1
(9.81) t 2 + 0.75(cos 30∞ tan 20∞ - sin 30∞) t - 0.45 = 0
2
4.905t 2 - 0.13859t - 0.45 = 0
Solving
t = 0.31735 s and t = - 0.28909 s (Reject)
or
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190
176
PROBLEM 11.128 (Continued)
The velocity v B of the bag as it lands on the belt is then
v B = (0.75 cos 30∞)i + [0.75 sin 30∞ - 9.81(0.31735)]j
= (0.64952 m/s)i - (2.7382 m/s) j
Finally
v B = v A + v B /A
or
v B /A = (0.64952i - 2.7382 j) - 1.2(cos 20∞i + sin 20∞ j)
= - (0.47811 m/s)i - (3.1486 m/s) j
v B /A = 3.18 m/s
or
81.4° !
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177
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
vT  15 km/h
Case:
Case:
vrain/train
vT  24 km/h
;
;
vR /T
vR / T
30°
45°
Case:
(vR ) y tan 30  15 (vR ) x
(1)
Case:
(vR ) y tan 45  24 (vR ) x
(2)
Substract (1) from (2)
(vR ) y (tan 45
tan 30 )  9
(vR ) y  21.294 km/h
Eq. (2):
21.294 tan 45  25 (vR ) x
(vR ) x  2.706 km/h
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178
PROBLEM 11.129 (Continued)
3.706
21.294
 7.24
21.294
vR 
 21.47 km/h  5.96 m/s
cos 7.24
tan

vR  5.96 m/s
82.8° 
Alternate solution
Alternate, vector equation
v R  vT
For first case,
v R  15i vR /T 1 ( sin 30 i cos30 j)
For second case,
v R  24i vR /T 2 ( sin 45 i cos 45 j)
v R /T
Set equal
15i vR /T 1 ( sin 30 i cos30 j)  24i 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
j:
vR /T 2 sin 45  9
(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,
vR /T 1  24.5885 km/h
vR /T 2  30.1146 km/h
Substitute v R /T 2 back into equation for v R .
v R  24i 30.1146( sin 45 i cos 45 j)
v R  2.71i 21.29 j
 tan 1
21.29
 82.7585
2.71
v R  21.4654 km/hr  5.96 m/s
v R  5.96 m/s
82.8° 
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PROBLEM 11.130
Instruments in airplane A indicate that with respect to the air the plane is
headed 30° north of east with an airspeed of 480 km/h. At the same time
radar on ship B indicates that the relative velocity of the plane with
respect to the ship is 416 km/h in the direction 33° north of east.
Knowing that the ship is steaming due south at 20 km/h, determine (a)
the velocity of the airplane, (b) the wind speed and direction.
SOLUTION
Let the x-axis be directed east, and the y-axis be directed north.
Airspeed:
Plane relative to ship:
v A/W = 480 km/h
30° = ( 480 km/h )(cos 30°i + sin 30° j)
v A/B = ( 416 km/h )(cos 33°i + sin 33° j) km/h
v B = 20 km/h
Ship:
= −20 j
(a) Velocity of airplane.
v A = v B + v A/B = −20 j + 416 (cos 33°i + sin 33° j)
= (348.89 km/h ) i + ( 206.57 km/h ) j
v A = 405 km/h
30.6° t
vW = 74.7 km/h
26.6° t
(b) Wind velocity.
vW = v A + vW / A = v A − v A/W
= 348.89i + 206.57 j − 480 ( cos 30°i + sin 30° j)
= − (66.80 km/h ) i − (33.43 km/h ) j
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180
PROBLEM 11.131
When a small boat travels north at 5 km/h, a flag mounted on
its stern forms an angle  50 with the centerline of the boat
as shown. A short time later, when the boat travels east at 20
km/h, angle
is again 50°. Determine the speed and the
direction of the wind.
SOLUTION
We have
vW  v B
vW/B
Using this equation,
quation, the two cases are then graphically represented as shown.
With vW now defined, the above diagram is redrawn for the two cases for clarity.
Noting that
 180
 40
We have
(50
vW
5

sin 50
sin (40
90
)
)
 180
 130
vW
20

sin 50
sin (130
(50
)
)
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PROBLEM 11.131 (Continued)
5
sin (40
Therefore
or
or
or
Then
sin 130 cos
)
cos 130 sin
tan

20
sin (130
)
 4(sin 40 cos

sin 130
cos 130
cos 40 sin )
4 sin 40
4 cos 40
 25.964
vW 
5 sin 50
 15.79 km/h
sin (40 25.964 )
vW  15.79 km/h
26.0° 
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182
PROBLEM 11.132
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 1 m /s, determine the speed
and the direction of the train.
SOLUTION
v D = v B + v D /B
We have
v D = v C + v D /C
The graphical representations of these equations are then as shown.
Then
vD
1
=
sin 8∞ sin (22∞ + a )
Equating the expressions for
vD
1
=
sin 7∞ sin (23∞ - a )
vD
1
sin 8∞
sin 7∞
=
sin (22∞ + a ) sin (23∞ - a )
or
or
sin 8∞ (sin 23∞ cos a - cos 23∞ sin a ) = sin 7∞ (sin 22∞ cos a
+ cos 22∞ sin a )
tan a =
sin 8∞ sin 23∞ - sin 7∞ sin 22∞
sin 8∞ cos 23∞ + sin 7∞ cos 22∞
or
a = 2.0728∞
Then
vD =
(1) sin 8∞
= 0.3412 m/s
sin (22∞ + 2.0728∞)
v D = 0.341 m/s
2.07° !
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183
PROBLEM 11.132 (Continued)
Alternate solution using components.
v B = (1 m/s)
30∞ = (0.866 m/s)i + (0.5 m/s) j
vC = (1 m/s)
30∞ = (0.866 m/s)i - (0.5 m/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
0.866i + 0.5 j - (u1 cos 22∞)i - (u1 sin 22∞) j = 0.866i - 0.5 j - (u2 cos 23∞)i + (u2 sin 23∞) j
Separate into components, and transpose and change
u1 cos 22∞ - u2 cos 23∞ = 0
u1 sin 22∞ + u1 sin 23∞ = 1
Solving for u1 and u2 ,
u1 = 1.3018 m/s
u2 = 1.3112 m/s
v D = 0.866i + 0.5 j - (1.3018 cos 22∞)i - (1.3112 sin 22∞) j
= -(0.341 m/s))i + (0.008158 m/s) j
v D = 0.341 m/s
2.07° !
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184
PROBLEM 11.133
Determine the smallest radius that should be used for a
highway if the normal component of the acceleration of a car
traveling at 72 km/h is not to exceed 0.8 m/s2 .
SOLUTION
an 
v2
an  0.8 m/s 2
v  72 km/h  20 m/s
0.8 m/s 2 
(20 m/s) 2
 500 m 
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PROBLEM 11.134
Determine the maximum speed that the cars of the roller-coaster
can reach along the circular portion AB of the track if is 25 m
and the normal component of their acceleration cannot exceed 3 g.
SOLUTION
We have
an 
v2
Then
(vmax ) 2AB  (3 9.81 m/s2 )(25 m)
or
(vmax ) AB  27.124 m/s
(vmax ) AB  97.6 km/h 
or
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186
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 3g forward acceleration.
Knowing that the astronaut sits 5 m from the axis of rotation
and experiences 3 g’s inward, determine her velocity.
SOLUTION
Given:
an  3 g ,
an 
Rearrange
Substitute in known values
5 m
v2
v
an
v
3g
v  5*3*9.81 m/s
vA  12.13 m/s 
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187
PROBLEM 11.136
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 20 mm/s2, determine the
magnitude of its total acceleration when (a) t = 0, (b) t = 2 s.
SOLUTION
(a)
At t = 0, v A = 0, which implies ( aA )n = 0
aA = ( aA )t
a A = 20 mm/s2 !
or
(b)
We have uniformly accelerated motion
v A = 0 + ( a A )t t
At t = 2 s:
Now
Finally,
v A = (20 mm/s2 )(2 s) = 40 mm/s
( aA )n =
v 2A ( 40 mm/s)2
=
= 17.7778 mm/s2
rA
90 mm
a2A = ( aA )t2 + ( a A )2n
= (20)2 + (17.7778)2
aA = 26.8 mm/s2 !
or
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200
188
PROBLEM 11.137
A monorail train starts from rest on a curve of radius 400 m and accelerates at the constant rate at . 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 at .
SOLUTION
When v = 72 km/h = 20 m/s and r = 400 m,
an =
v 2 (20)2
=
= 1.000 m/s2
r
400
a = an2 + at2
But
at = a2 - 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 + 2at ( x1 - x0 ) = 2at 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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189
PROBLEM 11.138
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
increases at a constant rate of 10 mm/s2, determine (a) the
magnitude of the acceleration when t 4 s, (b) the time for the
magnitude of the acceleration to be 80 mm/s2.
SOLUTION
Tangential acceleration:
at  10 mm/s 2
Speed:
v  at t
Normal acceleration:
an 

at2t 2
 0.8 m  800 mm
where
(a)
v2
When t  4 s
v  (10)(4)  40 mm/s
an 
Acceleration:
(40)2
 2 mm/s 2
800
a  at2
an2  (10)2
(2)2
a  10.20 mm/s 2 
(b)
Time when a  80 mm/s2
a 2  an2
(80)2 
at2
(10) 2 t 2
800
2
102
t 4  403200 s 4
t  25.2 s 
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190
PROBLEM 11.139
A monorail train starts from rest on a curve of radius 400 m and accelerates at the constant rate at . If the
2
maximum total acceleration of the train must not exceed 1.5 m/s , determine (a) the shortest distance in
which the train can reach a speed of 72 km/h, (b) the corresponding constant rate of acceleration at .
SOLUTION
When v  72 km/h  20 m/s and
 400 m,
an 
But
v2

(20)2
 1.000 m/s 2
400
a  an2
at2
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
x1 
(b)
2at ( x1
x0 )  2at x1
v12
(20)2

2at (2)(1.11803)
x1  178.9 m 
Corresponding tangential acceleration.
at  1.118 m/s 2 
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191
PROBLEM 11.140
A motorist starts from rest at Point A on a circular entrance ramp
whent 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.
SOLUTION
v0  0
Speeds:
v1  100 km/h  27.78 m/s
s
Distance:
2
(150)
v12  v02
Tangential component of acceleration:
at 
vB2  0
v12
v02
2s
vB2  v02
At Point B,
100  335.6 m

2at s
(27.78) 2 0
 1.1495 m/s 2
(2)(335.6)
2at sB
where
sB 
2
(150)  235.6 m
(2)(1.1495)(235.6)  541.69 m2 /s2
vB  23.27 m/s
(a)
At t  20 s,
Since v
(b)
v  v0
at t  0
vB  83.8 km/h 
(1.1495)(20)  22.99 m/s
 150 m
vB , the car is still on the curve.
Normal component of acceleration:
an 
Magnitude of total acceleration:
|a| 
v2

at2
(22.99) 2
 3.524 m/s 2
150
an2  (1.1495)2
(3.524)2
| a |  3.71 m/s 2 
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192
PROBLEM 11.141
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 10
m/s2, and the speed of B is decreasing at the rate of 6 m/s2. For
the position shown, determine (a)) the velocity of B relative to
A, (b) the acceleration of B relative to A.
A
SOLUTION
v A  240 km/h  66.67 m/s
Speeds:
vB  200 km/h  55.56 m/s
vA  66.67 m/s
Velocities:
vB  55.56 m/s
(a)
vB/A  vB
Relative velocity:
50°
vA
vB/A  (55.56 cos 50 )
 30.96
55.56 sin 50
66.67
42.56
 52.63 m/s
53.96°
vB /A  189.5 km/h
Tangential accelerations:
(aA )t  10 m/s 2
(aB )t  6 m/s 2
an 
Normal accelerations:
(  )
Car B:
(  300 m)
Acceleration of B relative to A:
50°
v2
Car A:
(aB ) n 
(b)
54.0° 
(aA )n  0
(55.56)2
 10.288
300
aB/A  aB
40°
aA
a B/A  (a B )t
6
(aB ) n  10.288 m/s 2
(a B )n
50
 (6 cos 50
(a A )t
10.288
40
10.288cos 40
(6sin 50
 21.738
(a A )n
2.017
10
0
10)
10.288sin 40 )
a B/A  21.8 m/s2
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193
5.3° 
PROBLEM 11.142
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/s 2 . 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
m radius. Knowing that at the given instant the
2
speed of B is being decreased at the rate of 3 m/s , 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 vB  540 km/h  150 m/s
(a)
vB  v A
We have
v B/A
The graphical representation of this equation is then as shown.
We have
vB2 /A  4502
5402
2(450)(540) cos 60°
vB/A  501.10 km/h
and
540 501.10

sin
sin 60
 68.9
v B/A  501 km/h
(b)
First note
Now
a A  8 m/s 2
( aB ) n 
v 2B
B

(a B )t  3 m/s 2
60
(150 m/s) 2
300 m
(a B )n  75 m/s 2
Then
68.9 
30
a B  (a B )t + (a B )n
 3( cos 60 i sin 60 j) + 75( cos 30 i sin 30 j)
= (66.452 m/s 2 )i (34.902 m/s 2 ) j
Finally
aB  a A
a B/A
a B/A  ( 66.452i 34.902 j) (8i )
 (74.452 m/s 2 )i (34.902 m/s 2 ) j
a B/A  82.2 m/s 2
25.1 
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194
PROBLEM 11.143
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 with a speed of 120 km/h that is increasing at 5 m/s2.
Three seconds later, determine (a) the total acceleration of the
car in xy components, (b) the linear velocity of the car in xy
components.
SOLUTION
Speeds:
vto  120 km/h  33.33 m/s
vt  vto
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R
s 122.5
 
 1.75 rad
R
70
s  so vto t
 100.3
Tangential Acceleration:
at  5 m/s 2
Normal Acceleration:
an 
an 
(a)
Accelerations:
v2
48.332
 33.37 m/s 2
70
ax  5cos(10.3 ) 33.37sin(10.3 )
ax  1.047 m/s 2 
a y  5sin(10.3 ) 33.37 cos(10.3 )
a y  33.726 m/s 2 
(b)
Velocities:
vx  48.33cos(10.3 )
vx  47.55 m/s 
v y  48.33sin(10.3 )
v y  8.64 m/s 
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195
PROBLEM 11.144
An airplane flying at a constant speed of 240 m/s makes a banked horizontal turn.
What is the minimum allowable radius of the turn if the structural specifications
require that the acceleration of the airplane shall never exceed 4g?
SOLUTION
Given:
v  240 m/s, amax  4 g
Acceleration:
an 
Rearrange:

v2
an
Substitute in known values:

2402
m
4*9.81
v2
 1467.9 m 
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196
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
(a A ) n 
v 2A
A
or
A 
(50 m/s)2
(9.81 m/s 2 )cos 25
A  281 m 
or
(b)
We have
( aB ) n 
vB2
B
where Point B is the highest point of the trajectory, so that
vB  (vA ) x  v A cos 25
Then
B 
[(50 m/s) cos 25°]2
9.81 m/s2
B  209 m 
or
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197
PROBLEM 11.146
Three children are throwing
snowballs at each other. Child A
throws a snowball with a horizontal
velocity v0. If the snowball just
passes over the head of child B and
hits child C, determine the radius of
curvature
of
the
trajectory
described by the snowball (a) at
Point B, (b) at Point C.
SOLUTION
The motion is projectile motion. Place the origin at Point A.
Horizontal motion:
v x  v0
x  v0t
Vertical motion:
y0  0,
(v y )  0
v y  gt
y
2h
,
g
t
1 2
gt
2
whereh is the vertical distance fallen.
| v y|  2 gh
v 2  vx2
Speed:
v 2y  v02
2 gh
Direction of velocity.
cos 
v0
v
Direction of normal acceleration.
an  g cos 
Radius of curvature:
At Point B,

gv0 v 2

v
v3
gv0
hB  1 m; xB  7 m
tB 
(2)(1 m)
 0.45152 s
9.81 m/s 2
xB  v0 t B
v0 
vB2  (15.504)2
xB
7m

 15.504 m/s
t B 0.45152 s
(2)(9.81)(1)  259.97 m 2 /s 2
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198
PROBLEM 11.146 (Continued)
(a)
Radius of curvature at Point B.
B 
At Point C
(259.97 m2 /s2 )3/ 2
(9.81 m/s2 )(15.504 m/s)
hC  1 m 2 m  3 m
vC2  (15.504)2
(b)
B  27.6 m 
(2)(9.81)(3)  299.23 m 2 /s2
Radius of curvature at Point C.
C 
(299.23 m2 /s2 )3/2
(9.81 m/s 2 )(15.504 m/s)
C  34.0 m 
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199
PROBLEM 11.147
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) At Point A. a A  g
 9.81 m/s 2
Sketch tangential and normal components of acceleration at A.
(a A )n  g cos50
A 
v A2
(2)2

(a A ) n
9.81cos50
A  0.634 m 
(b) At Point B, 1 meter below Point A.
Horizontal motion: (vB ) x  (vA ) x  2cos50  1.286 m/s
Vertical motion:
(vB ) 2y  (v A ) 2y
2a y ( yB
 (2 cos 40 )2
yA )
(2)( 9.81)( 1)
2 2
 21.97 m /s
(vB ) y  4.687 m/s

tan
(vB ) y
(vB ) x

4.687
,
1.286
or
 74.6
aB  g cos74.6
B 

(vB ) 2x (vB )2y
vB 2

(aB )n
g cos 74.6
(1.286)2 21.97
9.81cos 74.6
B  9.07 m 
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200
PROBLEM 11.148
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.
SOLUTION
(a)
We have
(a A )n 
v A2
A
4
(9.81 m/s 2 ) (25 m)
5
or
v 2A 
or
vA  14.0071 m/s
vA  14.01 m/s
(b)
We have
Where
Then
( aB ) n 
vB2
B
v B  (v A ) x 
B 
36.9° 
4
5
4
vA
5
14.0071 m/s
2
9.81 m/s2
 B  12.80 m 
or
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201
PROBLEM 11.149
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.
SOLUTION
Horizontal motion.
vx  v0 cos
Vertical motion.
v y  v0 sin
Eliminate t.
x  v0 t cos
 gt
y  y0  v0 t sin

y  y0  x tan

1 2
gt
2
gx2
2 v02 cos2
(1)
Solving (1) for v0 and applying result at point B
v0 
gx 2
2 y0  x tan  y cos 2

2
2 1.5  6 tan 3  0.97 cos 2 3
v0  14.48 m/s 
(a)
Magnitude of initial velocity.
(b)
Minimum radius of curvature of trajectory.
an  g 
9.81 6
v2


v2
v2

an g cos
(2)
where is the slope angle of the trajectory.
The minimum value of  occurs at the highest point of the trajectory where cos   1
and v  vx  v0 cos
Then
 min 
v02 cos 2
g
2

14.48 cos 2 3
9.81
min  21.3 m 
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202
PROBLEM 11.150
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  the angle
formed by the trajectory and the
horizontal at a given
gi
Point C, show
that the radius of curvature of the
trajectory at C is    min /cos3  .
SOLUTION
For the arbitrary Point C, we have
vC2
C
(aC )n 
C 
or
vC2
g cos 
Noting that the horizontal motion is uniform, we have
(vA ) x  (vC ) x
(vC ) x  vC cos 
(vA ) x  v0 cos
where
v0 cos
Then
 vC cos 
or
vC 
cos
v0
cos
so that
C 
1
cos
v0
g cos  cos
(a)

v02 cos2
g cos3 
In the expression for C , v0 , , and g are constants, so that C is minimum where cos is
maximum. By observation, this occurs at Point B where   0.
min   B 
(b)
2
v02 cos 2
g
C 
v02 cos2
1
g
cos3 
C 
 min
cos3 
Q.E.D.

Q.E.D.

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203
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 nt)ictj (Rt sin nt)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
dr
 R(cos
dt
nt 
and
a
dv
R 
dt
n sin
nt 
R
n cos
nt 
n R [(2 sin
nt 
n t cos
n t )i  (2cos
nt 
n t sin
nt ) k ]
2
 c 2  R 2 (sin
nt 
n t cos
nt )
or
a
Now
v 2  R 2 (cos
nt 
 R2 1 
2 2
nt
 c2
Then
v  R2 1 
2 2
nt
 c2
and
dv

dt
n t )i  cj  R(sin
n t sin
n t sin
n sin
n cos
nt )
nt 
nt 
nt 
2
n t cos
nt
2
n t sin
nt
n t cos
n t )k
i
k
2
1/ 2
R 2 n2 t
R2 1 
2 2
nt
 c2
dv
dt
2
1/ 2
v2

2
Now
a 2  at2  an2 
At t  0:
dv
0
dt
a  n R (2 k ) or a  2 n R

v 2  R2  c 2
Then, with
dv
 0,
dt
we have
a
or
2 nR 
v2

R 2  c2


R2  c2

2 nR
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204
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.
SOLUTION
With
A = 3,
B =1
We have
r = (3t cos t ) i + 3 t 2 + 1 j + (t sin t )k
Now
v=
and
a=
(
)
Ê 3t ˆ
dr
= 3(cos t - t sin t )i + Á 2
˜ j + (sin t + t cos t )k
dt
Ë t + 1¯
È
Ê t ˆ˘
2
dv
˜
Á
= 3( - sin t - sin t - t cos t )i + 3 Í t + 1 - t Ë t 2 + 1 ¯ ˙ j
Í
˙
dt
t2 + 1
ÍÎ
˙˚
+ (cos t + cos t - t sin t )k
= - 3(2 sin t + t cos t )i + 3
1
j
(t + 1)1/ 2
2
+ (2 cos t - t sin t )k
Then
t2
v 2 = 9 (cos t - t sin t )2 + 9 2
+ (sin t + t cos t )2
t +1
Expanding and simplifying yields
v 2 = t 4 + 19t 2 + 1 + 8(cos2 t + t 4 sin 2 t ) - 8(t 3 + t )sin 2t
Then
v = [t 4 + 19t 2 + 1 + 8(cos2 t + t 4 sin 2 t ) - 8(t 3 + t )sin 2t ]1/ 2
and
dv 4t 3 + 38t + 8( -2 cos t sin t + 4t 3sin 2 t + 2t 4 sin t cos t ) - 8[(3t 2 + 1)siin 2t + 2(t 3 + t ) cos 2t ]
=
dt
2[t 4 + 19t 2 + 1 + 8(cos2 t + t 4 sin 2 t ) - 8(t 3 + t )sin 2t]1/ 2
Now
2
Ê v2 ˆ
Ê dv ˆ
a2 = at2 + an2 = Á ˜ + Á ˜
Ë dt ¯
Ë r¯
2
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223
205
PROBLEM 11.152* (Continued)
At t = 0:
a = 3j + 2k
or
a = 13 m/s2
dv
=0
dt
v 2 = 9 (m/s)2
Then, with
dv
= 0,
dt
we have
a=
or
r=
v2
r
9 m2 /s2
13 m/s2
r = 2.50 m !
or
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224
206
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/s 2 ,
determine the radius of the orbit of the indicated planet around the sun assuming that the orbit is circular.
Earth: (mean )orbit  107 Mm/h.
SOLUTION
g  274 m/s 2 ,
For the sun,
R
and
Given that an 
1
1
D
(1.39 109 )  0.695 109 m
2
2
gR 2
v2
and
that
for
a
circular
orbit
a

n
r
r2
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
(29.72 103 ) 2
r  149.8 Gm 
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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/s 2 ,
determine the radius of the orbit of the indicated planet around the sun assuming that the orbit is circular.
Saturn: (mean )orbit  34.7 Mm/h.
SOLUTION
g  274 m/s 2
For the sun,
R
and
Given that an 
1
1
D
(1.39 109 )  0.695 109 m
2
2
gR 2
v2
.and that for a circular orbit: an 
2
r
r
gR 2
v2
Eliminating an and solving for r,
r 
For the planet Saturn,
v  34.7
Then,
r 
106 m/h  9.639
(274)(0.695 109 ) 2
 1.425 1012 m
3 2
(9.639 10 )
103 m/s
r  1425 Gm 
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PROBLEM 11.155
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. Determine the speed of a
satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit 160 km above
the surface of the planet.
Venus: g = 8.53 m/s2 , R = 6161 km.
SOLUTION
an = g
We have
R2
r2
and an =
v2
r
R2 v 2
=
r
r2
Then
g
or
vcirc. = R
g
r
where r = R + h
For the given data
vcirc. = 6161 km
8.53 m/s2
3
(6161 + 160) ¥ 10 m
¥
3600 s
1h
vcirc. = 25.8 ¥ 103 km/h !
or
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209
PROBLEM 11.156
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. Determine the speed of a
satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit 160 km above
the surface of the planet.
Mars: g = 3.83 m/s2 , R = 3332 km.
SOLUTION
an = g
We have
R2
=
2
r2
and an =
v2
r
v2
r
Then
g
or
vcirc. = R
r
R2
g
r
where r = R + h
For the given data
vcirc. = 3332 km
3.83 m/s2
3
(3332 + 160) ¥ 10 m
¥
3600 s
1h
vcirc. = 12.56 ¥ 103 km/h !
or
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228
210
PROBLEM 11.157
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. Determine the speed of a
satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit 160 km above
the surface of the planet.
Jupiter: g = 26.0 m/s2 , R = 69, 893 km.
SOLUTION
an = g
We have
R2
r2
and an =
v2
r
R2 v 2
=
r
r2
Then
g
or
vcirc. = R
g
r
where r = R + h
For the given data
vcirc. = 69, 893
26.0 m/s2
3
(69, 893 + 160) ¥ 10 m
¥
3600 s
1h
vcirc. = 153.3 ¥ 103 km/h !
or
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211
PROBLEM 11.158
A satellite will travel indefinitely in a circular orbit around the earth if the normal component of its
2
acceleration is equal to g R / r , where g  9.81 m/s 2 , 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
gR 2
r2
Normal acceleration:
an 
Solve for v2:
v 2  ran 
Data:
g  9.81 m/s 2 ,
and
an 
v2 v2


r
gR 2
r
R  6370 km = 6.370 106 m
r  384 103 km = 384 106 m
v2 
9.81 6.370 106
384 106
v = 1.018 m/s
2
 1.0366 106 m 2 /s2
v  3670 km/h 
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212
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
11.153–11.155.)
SOLUTION
an  g
We have
Then
g
and an 
v2
r
where
r Rh
R 2 v2

r
r2
vR
or
R2
r2
g
r
The circumference s of the circular orbit is equal to
s2 r
Assuming that the speed of the telescope is constant, we have
s  vtorbit
Substituting for s and v
2 rR
or
torbit 

g
torbit
r
2 r 3/2
R g
2
[(6370  590) km]3/2
6370 km [9.81 103 km/s 2 ]1/2
1h
3600 s
torbit  1.606 h 
or
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213
PROBLEM 11.160
Satellites A and B are traveling in the same plane in circular orbits around the
earth at altitudes of 180 and 300 km, respectively. If at t = 0 the satellites are
aligned as shown and knowing that the radius of the earth is R = 6370 km,
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
and an =
R2 v 2
=
r
r2
or
v2
r
v=R
g
r
r = R+h
where
The circumference s of a circular orbit is
equal to
s = 2p r
Assuming that the speeds of the satellites are constant, we have
s = vT
Substituting for s and v
2p r = R
g
T
r
2p r 3/ 2 2p ( R + h)3/ 2
=
R g
R
g
or
T=
Now
hB > hA fi (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
È 2p ( R + hA )3/ 2 ˘
È 2p ( R + hB )3/ 2 ˘
NÍ
˙
˙ = ( N + 1) Í
g
g
˙˚
ÍÎ R
˙˚
ÍÎ R
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232
214
PROBLEM 11.160 (Continued)
or
N=
=
Then
( R + hA )3/ 2
( R + hB )
(
3/ 2
1
- ( R + hA )
)
6370 + 300 3 / 2
-1
6370 + 180
TC = N (T ) B = N
3/ 2
=
1
( )
R + hB 3 / 2
-1
R + hA
= 36.223 orbits
2p ( R + hB )3/ 2
R
g
2p [(6370 + 300) km ]
1h
= 36.223
¥
/
1
2
6370 m 9.81 m/s2 ¥ 1 km
3600 s
3/ 2
(
1000 m
)
TC = 54.6 h !
or
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215
PROBLEM 11.161
The oscillation of rod OA about O is defined by the relation q = (2 /p) (sin pt), where
q and t are expressed in radians and seconds, respectively. Collar B slides along the
rod so that its distance from O is r = 625/(t + 4) where r and t are expressed in mm and
seconds, respectively. When t = 1 s, determine (a) the velocity of the collar, (b) the total
acceleration of the collar, (c) the acceleration of the collar relative to the rod.
SOLUTION
625
t+4
q=
2
sin p t
p
625
(t + 4) 2
q = 2 cos p t
1250
( t + 4 )3
q = -2p sin p t
We have
r=
Then
r = -
and

r=
At t = 1 s
r = 125 mm
q =0
r = -25 mm/s
q = -2 rad/s

r = 10 mm/s2
q = 0
(a)
We have
v B = re r + rq eq = ( -1)e r + (5)( -2)eq
v A = -(25 mm/s)e r - (250 mm/s)eq !
or
(b)
We have
r - rq 2 )e r + ( rq + 2rq )eq
a B = ( 
= [10 - (125)(2)2 ]e r + [0 + 2( -25)( -2)]eq
a B = -( 490 mm/s2 )e r + (100 mm/s2 )eq !
or
(c)
We have
a B /OA = 
r
a B /OA = (10 mm/s2 )e r !
so that
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234
216
PROBLEM 11.162
The path of a particle P is a limaçon. The motion of the particle is
defined by the relations r  b 2  cos t and   t , where t and
 are expressed in seconds and radians, respectively. Determine
(a) the velocity and the acceleration of the particle when t  2 s,
(b) the value of  for which the magnitude of the velocity is
maximum.
SOLUTION
Differentiate the expressions for r and  with respect to time.
r  b 2  cos t ,
r   b sin t ,
2
r  
b cos t
  t,
  ,
  0
(a) At t  2 s,
sin t  0,
cos t  1
r  3b,
r  0,
vr  r  0 ,
v  r  3 b,
2
r  
b,
 
  2 rad,
rad/s
v  3 be 
a r  r  r 2  
2
b  3b
2
 4
2
b
a  r  2r  0,
a   4 2ber 
(b) Values of  for which v is maximum.
vr  r   b sin t
v  r   b 2  cos t

v 2  vr2  v 2 

2 2

2 2
2 2
b sin 2 t  2  cos t
2
b sin 2 t  4  4 cos t  cos2 t
b 5  4 cos t
v
2
But
is maximum when
cos t  1
  t,
or
hence
t  0,
2 ,
4 ,
6 , etc
  2N , N  0, 1, 2,  
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217
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 30º and is increasing at a
constant rate of 2º/s. Determine the velocity and
acceleration of the parasailer at this instant.
SOLUTION
Given:
v B  30i km/hr  8.333i m/s
aB  0
eˆr
r  200 m, r  0, r  0
  30
  2 / s  0.0349 rad/s
:
ê
r
바
P

B
=0
Relative Motion relations:
vP  vB  vP/ B
aP  aB  aP / B
Using Radial and Transverse components:
v P / B  rer  re
a  
r  r2 e  r  2r e
P/ B
Substitute in known values:
r

v P / B  6.981e m/s
a P / B  0.2437er m/s 2
?
Change to rectangular coordinates:
v P / B  6.981*sin 30e
i  6.981*cos 30D
j m/s
~
=3.491i  6.046 j m/s
~
a P / B  0.2437*  cos 30 i  0.2437 *sin 30 j m/s 2
…
과
 0.2111i  0.1219 j m/s 2
Substitute into Relative Motion relations:
apIB 성분
v P  8.33i  3.491i  6.046 j m/s
=11.824i  6.046 j m/s
어디에 존재 ?
ㄳ
a P  0  0.2111i  0.1219 j m/s 2
 0.2111i  0.1219 j m/s 2
Velocity:
vP  13.280 m/s
27.08 
Acceleration:
aP  0.2437 m/s
30.00 
2
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218
PROBLEM 11.164*
Pin P is attached to BC and slides freely in the slot of OA.
Determine the rate of change q& of the angle q, knowing that
BC moves at a constant speed v0. Express your answer in terms
of v0, h, b, and q.
SOLUTION
From the diagram
r
h
=
sin (90∞ - b ) sin ( b + q )
r (sin b cos q + cos b sin q ) = h cos b
or
h
tan b cos q + sin q
or
r=
Also,
vq = v0 sin ( b + q ) where vq = rq&
hq&
= v0 (sin b cos q + cos b sin q )
tan b cos q + sin q
Then
v cos b
q& = 0
(tan b cos q + sin q )2 !
h
or
Alternative solution.
h
tan b cos q + sin q
From above
r=
Then
r& = h
Now
v02 = vr2 + vq2 = ( r&)2 + ( rq& )2
or
È
v02 = Íhq&
Î
tan b sin q - cos q &
q
(tan b cos q + sin q )2
tan b sin q - cos q ˘
˙
(tan b cos q + sin q )2 ˚
2
2
Ê
ˆ
hq&
+Á
Ë tan b cos b + sinn q ˜¯
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219
PROBLEM 11.164* (Continued)
or
v0 = ±
hq&
tan b cos q + sin q
1/ 2
È (tan b sin q - cos q )2 ˘
+Í
+ 1˙
2
Î (tan b cos q + sin q )
˚
È
˘
tan 2 b + 1
hq&
=±
Í
2˙
tan b cos q + sin q Î (tan b cos q + sin q ) ˚
1/ 2
Note that as q increases, member BC moves in the indicated direction. Thus, the positive root is chosen.
v cos b
q& = 0
(tan b cos q + sin q )2 !
h
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220
PROBLEM 11.165
As rod OA rotates, pin P moves along the parabola BCD. Knowing that the
equation of this parabola is r  2b /(1  cos  ) and that   kt , determine the
velocity and acceleration of P when (a)   0, (b)   90.
SOLUTION
2b
  kt
1  cos kt
2bk sin kt
r 
  k   0
(1  cos kt )2
2bk

r
[(1  cos kt ) 2 k cos kt  (sin kt )2(1  cos kt )(k sin kt )]
(1  cos kt )4
r
(a)
When kt 0:
2bk
1
[(2) 2 k (1)  0]  bk 2
2
(2) 4
r b
r  0
r 
 0
  k
  0
v  r  bk
vr  r  0
v  bk e 
1
1
ar  
r  r 2  bk 2  bk 2   bk 2
2
2


a  r  2r  b(0)  2(0)  0
(b)
1
a   bk 2e r 
2
When kt 90°:
r  2b
r  2bk
  90
  k
2bk
[0  2k ]  4bk 2
19
  0
r 
v  r  2bk
v  2bk er  2bk e 
ar  
r  r2  4bk 2  2bk 2  2bk 2
a  r  2r  2b(0)  2(2bk )k  4bk 2
a  2bk 2 e r  4bk 2 e 
vr  r  2bk

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PROBLEM 11.166
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  , (a) show that the
acceleration of pin B is of constant magnitude,
(b) determine the direction of the acceleration of pin B.
SOLUTION
From the sketch:
r  2b cos 
r  2b sin  
Since   constant,   0

r  2b cos   2
ar  
r  r2  2b cos  2  (2b cos  )2
a  4b cos   2
r
a  r  2r  (2b cos  )(0)  2(2b sin  ) 2
a  4b sin   2

a  ar2  a2  4b2 ( cos  )2  ( sin  )2
a  4b 2
Since both b and  are constant, we find that
a  constant 
 tan 1
a
4b sin   2
 tan 1
ar
4b cos   2
 tan 1 (tan  )

Thus, a is directed toward A
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222
PROBLEM 11.167
To study the performance of a racecar, a high-speed
high
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,  , and , (b)) the magnitude of the acceleration in
terms of b,  , , and .
SOLUTION
(a)
We have
r
b
cos 
Then
r 
b sin 
cos 2 
We have
v 2  vr2  v2  (r)2  (r)2
or
2
b sin 
cos 2

b 22 sin 2
b22
1


cos 2 cos2
cos4
v

b
cos 
2

b
cos2
For the position of the car shown,  is decreasing; thus, the negative root is chosen.
v
b

cos2
Alternative solution.
From the diagram
or
r  v sin 
b sin 
 v sin 
cos2
v
or
b

cos2
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223
PROBLEM 11.167 (Continued)
(b)
For rectilinear motion
a
dv
dt
Using the answer from Part a
v
Then
b
cos2
d
b

dt
cos 2
 cos 2  (2 cos  sin  )
 b
cos 4
a
a
or
b
(  22 tan  ) 
cos2
Alternative solution
Then
b
b sin 
r 
cos
cos2 
( sin    2 cos  )(cos 2 )  ( sin  )(2 cos  sin  )

r b
cos 4
 sin   2 (1  sin 2  )
b

cos 2
cos3
Now
a 2  ar2  a2
where
 sin   2 (1  sin 2  )
b2
ar  
r  r 2  b


cos 
cos 2
cos 2
From above
r

ar 
and
b sin  
(  22 tan  )
2
cos 
a  r  2r 

Then
2 2
b
 sin   2 sin 

cos 
cos 2
a
b
b 2 sin 
2
cos 
cos 2
b cos  
(  2 tan  )
cos 2
b
(  22 tan  )[(sin  )2  (cos  )2 ]1/2
2
cos 
For the position of the car shown,  is negative; for a to be positive, the negative root is chosen.
b
a
(  2 2 tan  ) 
cos 2 
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224
PROBLEM 11.168
After taking off, a helicopter climbs
in a straight line at a constant angle
. Its flight is tracked by radar from
Point A.. Determine the speed of the
helicopter in terms
t
of d, ,  , and .
SOLUTION
From the diagram
r
d

sin (180  ) sin (   )
or
 r (sin cos   cos sin  )
d sin
or
rd
Then
r  d tan
tan
tan
cos   sin 
( tan sin   cos  ) 

(tan cos   sin  ) 2
tan sin   cos 
 d tan
(tan cos   sin  )2
From the diagram
vr  v cos (   )
where
vr  r
Then
d tan
tan sin   cos 
 v(cos
(tan cos   sin  )2
 v cos
cos   sin
(tan
sin  )
sin   cos  )
v
or
d tan sec

(tan cos  sin  )2
Alternative solution.
We have
v 2  vr2  v2  (r)2  (r)2
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225
PROBLEM 11.168 (Continued)
Using the expressions for r and r from above
v  d tan
or
tan sin   cos
(tan cos   sin  ) 2
d tan
(tan
v
(tan cos   sin  ) (tan
2
sin   cos  ) 2
1
cos   sin  ) 2
d tan
tan 2  1

(tan cos   sin  ) (tan cos   sin  ) 2
1/ 2
1/2
Note that as  increases, the helicopter moves in the indicated direction. Thus, the positive root is chosen.
v
d tan sec

(tan cos   sin  )2
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226
PROBLEM 11.169
At the bottom of a loop in the vertical plane, an
airplane has a horizontal velocity of 150 m /s and
is speeding up at a rate of 25 m /s2. The radius of
curvature of the loop is 2000 m. The plane is being
tracked by radar at O. What are the recorded values
of r, 
r , q and q for this instant?
SOLUTION
Geometry. The polar coordinates are
Ê 600 ˆ
= 36.87∞
v = (800)2 + (600)2 = 1000 m q = tan -1 Á
Ë 800 ˜¯
Velocity Analysis.
v = 150 m /s Æ
vr = 150 cos q = 120 m/s
vq = -150 sin q = -90 m/s
vr = r
r = 120 m/s !
v
90
vq = rq q = q = r
1000
q = -0.0900 rad/s !
Acceleration analysis.
at = 25 m/s2
an =
v 2 (150)2
=
= 11.25 m/s2
r
2000
a = 25 m /s2 Æ + 11.25 m /s2 ≠ = 27.41 m /s2
24.23°
b = 24.23∞
q - b = 12.64∞
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or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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227
PROBLEM 11.169 (Continued)
ar = a cos (q - b ) = 27.41 cos 12.64∞ = 26.74 m/s2
aq = - a sin (q - b ) = -27.41 sin 12.64∞ = -6.00 m/s2
a = 
r - rq 2 
r = a + rq 2
r
r

r = 26.74 + (1000)(0.0900)2
a = rq + 2rq

r = 34.8 m/s2 !
q
a
2rq
q = q r
r
=
-6.00 (2)(120)( -0.0900)
1000
1000
q = -0.0156 rad/s2 !
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228
PROBLEM 11.170
Pin C is attached to rod BC and slides freely in the slot of rod OA
which rotates at the constant rate . At the instant when
 60,


determine (a) r and  , (b) 
r and  . Express your answers in terms
of d and
SOLUTION
.
as polar coordinates with d  0,
Looking at d and
v d d ,
vd  d  0
a  d   2d   0,
(a)
Velocity analysis:
ad  d  d  2  d
2
r  d 3 for angles shown.
Geometry analysis:
Sketch the directions of v, er and e.
vr  r  v er  d cos120
1
r   d
2

1
 
2

v  r   v e  d cos 30
(b)
Acceleration analysis:
3
d cos 30 d 2
 

r
d 3
Sketch the directions of a, er and e.
ar  a e r  a cos150  
3

r  r2  
d
2
r  
3
d
2
2
3
d
2
2
3
 r 2  
d
2
a  a e  d
2
2
2
d 3
1
cos120   d
2
1
2
2

r  
3
d
4
1
2
  0 
2

2
a  r  2r
1
  (a  2r) 
r
1
1
 d
2
3d
2
1
 (2)  d
2
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PROBLEM 11.171
For the racecar of Problem 11.167, it was found that it took 0.5
s for the car to travel from the position   60 to the position
  35. Knowing that b  25 m, determine the average speed
of the car during the 0.5-s interval.
PROBLEM 11.167 To study the performance of a racecar,
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.
BC Determine
(a) the speed of the car in terms of b,  , and , (b) the
magnitude of the acceleration in terms of b,  , , and .
SOLUTION
From the diagram:
r12  25 tan 60  25 tan 35
 25.796 m
Now
vave 

r12
t12
25.796 m
0.5 s
 51.592 m/s
vave  185.7 km/h 
or
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230
PROBLEM 11.172
For the helicopter of Problem 11.169,
it was found that when the helicopter
was at B, the distance and the angle
of elevation of the helicopter were
r = 1000 m and q = 20∞, respectively.
Four seconds later, the radar station
sighted the helicopter at r = 1100 m and
q = 23.1∞. Determine the average speed
and the angle of climb b of the helicopter
during the 4-s interval.
SOLUTION
We have
r0 = 1000 m q 0 = 20∞
r4 = 1100 m q 4 = 23.1∞
From the diagram:
Dr 2 = 10002 + 11002
- 2(1000)(1100) cos (23.1∞ - 20∞)
or
Dr = 114.975 m
Now
vave =
Dr
Dt
114.975 m
=
4s
= 28.74375 m/s
vave = 103.5 km/h !
or
Also,
or
Dr cos b = r4 cos q 4 - r0 cos q 0
cos b =
1100 cos 23.1∞ - 1000 cos 20∞
114.975
b = 51.2∞ !
or
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231
PROBLEM 11.173
A particle moves along the spiral shown. Determine the magnitude
of the velocity of the particle in terms of b,  , and .
SOLUTION
1 2

Given:
r  be 2
Take time derivative
r  be 2 
Radial and Transverse Components
vr  r  be 2 
1 2

1 2

1 2

v  r  be 2 
Magnitude of Velocity
v 2  vr2  v2

1 2

be 2
2
 2  1 2
1 2

v  be 2
2  1
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1/2
 
PROBLEM 11.174
A particle moves along the spiral shown. Determine the magnitude of the
velocity of the particle in terms of b,  , and .
SOLUTION
b
2
Given:
r 
Take time derivative
r  
Radial and Transverse Components
vr  r  
2b 

3
2b 

3
b
v  r  2 

Magnitude of Velocity
v 2  vr2  v2

4b 2  2 b 2  2
  4
6


b2
4   2  2
6
v
12
b
4   2  
3

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PROBLEM 11.175
A particle moves along the spiral shown. Knowing that  is
, determine the
constant and denoting this constant by
magnitude of the acceleration of the particle in terms of b,
 , and  .
SOLUTION
1 2

r  be 2
 
Given:
Take time derivatives
1 2

r  be 2 
1 2

r  be 2
Radial Component of Acceleration
2
   2  
ar  r  r 2
1 2

Transverse Component of Acceleration
  0
and
 be 2
2
   2    2
1 2

 be 2
2
  
a  r  2r
1 2

1 2

1 2

 be 2
  2 2
 be 2   2be 2  2
Substitute Given Values
 
1 2

ar  be 2
Magnitude of Acceleration
  0
and
2

1 2

a  be 2
and
2
2
a 2  ar2  a2

1 2

be 2
2
 4  4 2
4
1 2

1
a  be 2   2  4 2
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2

PROBLEM 11.176
A particle moves along the spiral shown. Knowing that  is constant and
denoting this constant by , determine the magnitude of the acceleration of
the particle in terms of b,  , and  .
SOLUTION
b
,  
2
Given:
r 
Take time derivatives
r  
Radial Component of Acceleration
ar  r  r 2
and   0
2b 

3
2b
6b
r   3   4 2


2b  6b  2
b
  4   2  2
3



b
 4 2  6 2   2 2


Transverse Component of Acceleration
a  r  2r
b 
2b
  2  3  2
2

b
 3   4 2

  and
&&&

0

Substitute Given Values
ar 
Magnitude of Acceleration
b
6 2
4

2
a  
and
4b
3
2
a 2  ar2  a2

b2
36  12 2   4
8


b2
36  4 2   4
8

4

16b 2
6
4
4
a
12
b
36  4 2   4
4

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2

PROBLEM 11.177
The motion of a particle on the surface of a right circular cylinder is
defined by the relations RA,   2 t , and z  B sin 2 nt , where A andB
are constants and n is an integer. Determine the magnitudes of the
velocity and acceleration of the particle at any time t.
SOLUTION
RA
R  0
 2 t
  2
z  2 n B cos 2 nt
  0
R
  0

z  4 2 n 2 B sin 2 nt
z  B sin 2 nt
Velocity (Eq. 11.49)
v  R e R  Re  zk
v
 A(2 )e  2 n B cos 2 nt k
A 2  n 2 B 2 cos 2 2 nt 
v2
Acceleration (Eq. 11.50)
 - R2 )e  ( R  2 R)e  
a  (R
zk
R

a  4 2 Aek  4 2 n2 B sin 2 nt k
a4
2
A 2  n 4 B 2 sin 2 2 nt 
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236
PROBLEM 11.178
Show that r  h  sin  knowing that at the instant shown, step
AB of the step exerciser is rotating counterclockwise at a
constant rate .
SOLUTION
From the diagram
r 2  d 2  h 2  2dh cos
Then
2rr  2dh  sin
Now
r
sin
or

d
sin 
r
d sin
sin 
Substituting for r in the expression for r
d sin
sin 
r  dh  sin
r  h  sin 
or
Q.E.D.

Alternative solution.
First note
 180  (   )
v  v r  v  rer  re
Now
With B as the origin
vP  d 
(d  constant
d  0)
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PROBLEM 11.178 (Continued)
With O as the origin
(vP )r  r
where
(vP )r  vP sin
Then
r  d  sin
Now
h
sin
or
d sin  h sin 
substituting

d
sin 
r  h  sin 
Q.E.D.
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238
PROBLEM 11.179
The three-dimensional motion of a particle is defined by the relations R  A(1  et ),   2 t , and
z  B(1  e t ). Determine the magnitudes of the velocity and acceleration when (a) t 0, (b) t .
SOLUTION
R  A(1  et )
R  Aet
 2 t
  2
z  B(1  e t )
   Aet
R
  0

z   Be t
z  Be t
Velocity (Eq. 11.49)
v  R e R  Re  zk
v  Ae t e R  2 A(1  et )e  Be t k
(a)
When
t  0: et  e0  1;
v  Ae R  Bk
(b)
When
t  : et  e   0
v  2 Ae
v
A2  B 2 
v  2 A
Acceleration (Eq. 11.50)
  R1 )e  ( R  2 R)e  
a  (R
zk
R

 [ Aet  A(1  et )4 2 ]eR  [0  2 Aet (2 )]e  Bet k
(a)
When
t  0: et  e0  1
a   Ae R  4 Ae  B k
a  A2  (4 A)2  B2
(b)
When
t
a  (1  16 2 ) A2  B 2 
: e t  e  0
a  4 2A
a  4 2 Ae R
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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
dimensional motion of a particle is defined by the position vector r
(Rt cos nt)ictj (Rt sin nt)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
n t )i  ctj  ( Rt sin
Then
v
dr
 R(cos
dt
and
a
dv
dt
nt 
n t sin
nt 
n sin
n cos
nt 
n cos
n R[ (2sin
nt 
n t cos
 R  n sin
R

n t )k
n t )i  cj  R(sin
nt 
nt 
2
n t cos
nt
2
n t sin
nt
n t )i  (2 cos
nt 
n t cos
n t )k
i
k
nt 
n t sin
n t )k ]
It then follows that the vector ( v a) is perpendicular to the osculating plane.
i
( v a) 

k
nt 
n t sin
nt )
c R(sin
nt 
n t cos
nt )
(2sin
nt 
n t cos
nt )
0 (2 cos
nt 
n t sin
nt )
n R{c (2 cos
nt 
n t sin
n t )i  R[ (sin
nt 
n t sin
n t )(2 cos
c(2cos
nt 
n t sin
 (cos

j
n R R (cos
nR
nt 
n t )i  R
nt 
n t sin
2
n t cos
n t )(2sin
n t )] j  c (2sin
2 2
nt
j  c(2sin
nt 
nt 
nt 
n t cos
n t cos
n t )k
n t cos
nt)
n t )k
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240
PROBLEM 11.180* (Continued)
The angle formed by the vector ( v a) and the y axis is found from
cos

( v a) j
| ( v a) || j |
|j| 1
Where
( v a) j   n R 2 2 
|( v a) | 
nR
c 2 (2 cos
 c 2 (2sin
cos

c2 4 
c2 4 
2 2
nt
2 2
nt
2 2
nt
nt)
nt )
2
 R2 2 
2 2
nt
2
2 1/2
 R2 2 
2 2
nt
2
1/ 2
2 2
nt
 R2 2 
2 2
nt
2
1/2
2 2
n t
 R2 2 
2 2
nt
2
1/ 2
that the osculating plane forms with y axis (see the above diagram) is equal to

cos
 sin
 90
 cos (  90)  sin
R 2 

c2 4 
Then
n t cos
R 2 

Then
n t sin
 n R2 2 
nR
The angle
nt 
nt 
2
nR c 4 

Then
2 2
nt
tan

2 2
nt
R 2
2 2
nt
c 4
2 2
nt
2 2
nt
 R2 2 
2 2
nt
2 1/ 2
 tan
or
1
R 2
2 2
nt
c 4
2 2
nt
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241

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 - m, t - s;
A = 3, B - 1
First note that eb 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
t2 +1
j + (sin t + t cos t )k
( )
t
2
dv
t 2 +1
a=
= 3( - sin t - sin t - t cos t )i + 3 t + 12- t
j
dt
t +1
+ (cos t + cos t - t sin t )k
3
= -3(2 sin t + t cos t )i + 2
j + (2 cos t - t sin t )k
(t + 1)3/ 2
and
(a)
At t = 0:
v = (3 m/s)i
a = (3 m/s2 ) j + (2 m/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 q x = 0
cos q y = cos q z =
13
13
eb =
q x = 90∞
q y = 123.7∞
q z = 33.7∞
!
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261
242
PROBLEM 11.181* (Continued)
(b)
At t =
p
s:
2
ˆ
ˆ Ê 3p
Ê 3p
v = -Á
m/s˜ i + Á
m/s˜ j + (1 m/s)k
¯ Ë p2 + 4
Ë 2
¯
È
˘ Êp
24
ˆ
m/s2 ˙ j - Á m/s2 ˜ k
a = -(6 m/s2 )i + Í 2
3/ 2
¯
Ë
2
(
p
+
4
)
Î
˚
Then
v¥a = -
i
j
3p
2
3p
(p + 4)1/ 2
24
2
(p + 4)3/ 2
-6
2
k
1
-
p
2
È
˘ Ê
3p 2
24
3p 2 ˆ
+
= -Í
+
i
j
6
˙
Á
2
1/ 2
4 ˜¯
(p 2 + 4)3/ 2 ˚ Ë
Î 2(p + 4)
È
˘
18p
36p
+ Í- 2
k
+ 2
3/ 2
1/ 2 ˙
(p + 4) ˚
Î (p + 4)
= -4.43984i - 13.40220 + 12.99459k
and
Then
| v ¥ a | = [( -4.43984)2 + ( -13.40220)2 + (12.99459)2 ]1/ 2
= 19.18829
eb =
1
( -4.43984i - 13.40220 j + 12.99459k )
19.1829
cos q x = or
13.40220
4.43984
cos q y = 19.18829
19.18829
q x = 103.4∞
q y = 134.3∞
cos q z =
12.99459
19.18829
q z = 47.4∞
!
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262
243
PROBLEM 11.182
3
2
The motion of a particle is defined by the relation x  2t  15t  24t  4, where x and t are expressed in
meters and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total
distance traveled when the acceleration is zero.
SOLUTION
x  2t 3  15t 2  24t  4
dx
 6t 2  30t  24
dt
dv
 12t  30
a
dt
v
so
(a)
0  6t 2  30t  24  6 (t 2  5t  4)
Times when v 0.
(t  4)(t  1)  0
(b)
t  1.00 s, t  4.00 s 
Position and distance traveled when a 0.
a  12t  30  0
t  2.5 s
x2  2(2.5)3  15(2.5) 2  24(2.5)  4
so
x  1.50 m 
Final position
For
0 t 1 s, v
For
1s t
2.5 s, v
At
t  0,
x0  4 m.
At
t  1 s, x1  (2)(1)3  (15)(1) 2  (24)(1)  4  15 m
0.
0.
Distance traveled over interval: x1  x0  11 m
For
1s t
2.5 s,
v 0
Distance traveled over interval
| x2  x1 |  |1.5  15 |  13.5 m
Total distance:
d  11  13.5
d  24.5 m 
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244
PROBLEM 11.183
A drag car starts from rest and starts down the racetrack with and acceleration defined by a = 50 – 10 t ,
where a and t are in m/s2 and s, respectively. After reaching a speed of 125 m/s, a parachute is deployed to
help slow down the dragster. Knowing that this deceleration is defined by the relationship a= - 0.02v2, 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  10t m/s 2 for 0 v 125 m/s
a23  0.02v 2 after v  125 m/s
Find v(t) as car speeds up
dv
dt
a
dv  adt
v t
t
dv 
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
dx  vdt
Find distance to reach-125 m/s:
x2
t2
50t  5t 2 dt
dx 
0
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
t3
dv
dt
dt 
dv
a
10
dv
v2
125
dt  50
5
t3  5 
50 10
|
v 125
t3  9.6 s 
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PROBLEM 11.183 (Continued)
(b) Find total distance to slow down to 10 m/s
adx  vdv
x3
x2
dx 
vdv
a
v3
dv
v
v2
dx  50
10
x3  416.7  50ln v|
125
x3  543.0 m 
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246
PROBLEM 11.184
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
Compute areas under a  t curve.
A1  (0.75)(8)  6 m/s
A2  (2)(4)  8 m/s
A3  (6)(6)  36 m/s
v0  2 m/s
v8  v0  A1  8 m/s
v12  v8  A2  0
v18  v12  A3
v18  36 m/s 
Sketch v t curve using straight line portions over the constant
acceleration periods.
Compute areas under the v t curve.
1
(2  8)(8)  40 m
2
1
A5  (8)(4)  16 m
2
1
A6  (36)(6)  108 m
2
A4 
x0  0
x8  x0  A4  40 m
x12  x8  A5  56 m
x18  x12  A6
Total distance traveled  56  108
x18  52 m 
d  164 m 
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247
PROBLEM 11.185
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.
SOLUTION
(a)
We have
v B  v A  v B/A
The graphical representation of this equation is then as shown.
Then
vB2 /A  662  482  2(66)(48) cos 155
or
vB/A  111.366 km/h
and
48
111.366

sin
sin 155
or
 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.
7
 5.6 km southwest of the crossing.
at t  3 min, B is (48 km/h) 60
Now
rB  rA  rB/A
Then at t  3 min, we have
rB2/A  3.32  5.62  2(3.3)(5.6) cos 25
rB/A  2.96 km 
or
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248
PROBLEM 11.186
Block B starts from rest and moves downward with a constant acceleration.
Knowing that after slider block A has moved 400 mm its velocity is 4 m / s,
determine (a) the accelerations of A and B, (b) the velocity and the change
in position of B after 2 s.
SOLUTION
From the diagram, we have
x A + 3 y B = constant
Then
v A + 3v B = 0
(1)
and
a A + 3aB = 0
(2)
(a)
Eq. (2): aA + 3aB = 0 and a B is constant and
positive fi a A is constant and negative
Also, Eq. (1) and ( v B )0 = 0 fi ( v A )0 = 0
Then
v 2A = 0 + 2a A [ x A - ( x A )0 ]
When |∆ x A | = 0.4 m:
(4 m/s)2 = 2aA (0.4 m)
a A = 20 m/s2 Æ !
or
Then, substituting into Eq. (2):
-20 m/s2 + 3aB = 0
or
aB =
20
m/s2
3
a B = 6.67 m/s2 Ø !
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249
PROBLEM 11.1 86 (Continued)
(b)
We have
v B = 0 + aB t
At t = 2 s:
ˆ
Ê 20
m/s2 ˜ (2 s)
vB = Á
¯
Ë 3
vB = 13.33 m/s Ø !
or
Also
At t = 2 s:
1
y B = ( y B ) 0 + 0 + aB t 2
2
1 Ê 20
ˆ
y B - ( y B )0 = Á
m/s2 ˜ (2 s)2
¯
2Ë 3
yB - ( yB )0 = 13.33 m Ø !
or
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63
250
PROBLEM 11.187
Collar A starts from rest at t = 0 and moves downward with a constant
acceleration of 175 mm /s2. Collar B moves upward with a constant acceleration,
and its initial velocity is 200 mm/s. Knowing that collar B moves through
500 mm between t = 0 and t = 2 s, determine (a) the accelerations of collar
B and block C, (b) the time at which the velocity of block C is zero, (c) the
distance through which block C will have moved at that time.
SOLUTION
From the diagram
- y A + ( yC - y A ) + 2 yC + ( yC - y B ) = constant
Then
-2v A - v B + 4vC = 0
(1)
and
-2aA - aB + 4aC = 0
(2)
( v A )0 = 0
Given:
(a A ) = 175 mm/s2 Ø
( v B )0 = 200 mm/s ≠
a B = constant ≠
At t = 2 s
(a)
y - (y B )0 = 500 mm ≠
We have
At t = 2 s:
1
y B = ( y B ) 0 + ( v B ) 0 t + aB t 2
2
1
-500 mm = ( -200 mm/s)(2 s) + aB (2 s)2
2
aB = -50 mm/s2
or
a B = 50.0 mm/s2 ≠ !
Then, substituting into Eq. (2)
-2(175 mm/s2 ) - ( -50 mm/s2 ) + 4aC = 0
aC = 75 mm/s2
or
aC = 75.0 mm/s2 Ø !
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251
PROBLEM 11.187 (Continued)
(b)
Substituting into Eq. (1) at t = 0
-2(0) - ( -200 mm/s) + 4( vC )0 = 0 or (vC )0 = -50 mm/s
(c)
Now
vC = ( vC )0 + aC t
When vC = 0:
0 = ( -50 mm/s) + (75 mm/s2 )t
or
t=
t = 0.667 s !
1
yC = ( yC )0 + ( vC )0 t + aC t 2
2
We have
At t = 23 s:
2
s
3
Ê2 ˆ
Ê2 ˆ 1
yC - ( yC )0 = ( -50 mm/s) Á s˜ + (75 mm/s2 ) Á s˜
Ë3 ¯
Ë3 ¯ 2
=-
50
mm = −16.667 mm
3
or
2
yC - ( yC )0 = 16.67 mm ≠ !
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252
PROBLEM 11.188
A golfer hits a ball with an initial velocity of
magnitude v0 at an angle
with the
horizontal. Knowing that the ball must clear
the tops of two trees and land as close as
possible to the flag, determine v0 and the
distance d when the golfer uses (a) a six-iron
with  31, (b) a five-iron with  27.
SOLUTION
The horizontal and vertical motions are
x
t cos
1
1
y  (v0 sin )t  gt 2  x tan  gt 2
2
2
x  (v0 cos )t
(1)
2( x tan  y)
g
or
t2 
At the landing Point C:
yC  0,
And
xC  (v0 cos )t 
(a)
v0 
or
(2)
t 
2v0 sin
g
2v02 sin cos
g
(3)
 31
To clear treeA:
xA  30 m, y A  12 m
From (2),
t A2 
From (1),
(v0 ) A 
To clear tree B:
2(30 tan 31  12)
 1.22851 s2 ,
9.81
t A  1.1084 s
30
 31.58 m/s
1.1084cos31
xB  100 m,
yB  14 m
From (2),
(t B )2 
2(100 tan 31  14)
 9.3957 s2 ,
9.81
From (1),
(v0 ) B 
100
 38.06 m/s
3.0652cos31
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 cos 31
 130.38 m
9.81
d  xC  110
d  20.4 m 
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253
PROBLEM 11.188 (Continued)
(b)
 27
By a similar calculation,
t A  0.81846 s,
(v0 ) A  41.138 m/s,
tB  2.7447 s,
(v0 )B  40.890 m/s,
v0  41.138 m/s
v0  41.1 m/s 
xC  139.56 m,
d  29.6 m 
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PROBLEM 11.189
As the truck shown begins to back up with a constant acceleration
of 1.2 m/s 2 , the outer section B of its boom starts to retract with
a constant acceleration of 0.48 m/s 2 relative to the truck.
Determine (a) the acceleration of section B, (b) the velocity of
section B when t = 2 s.
SOLUTION
For the truck,
a A = 1.2 m/s 2
For the boom,
a B/ A = 0.48 m/s 2
50°
(a) a B = a A + a B/ A
Sketch the vector addition.
By law of cosines:
aB2 = a A2 + aB2/ A − 2a AaB/ A cos 50°
= (1.2) 2 + (0.48) 2 − 2(1.2)(0.48)cos50° = 0.9299
aB = 0.9643 m/s 2
Law of sines:
sin α =
aB/ A sin 50°
aB
α = 22.4°,
=
0.48sin 50°
= 0.38131
0.9643
a B = 0.964 m/s 2
22.4° �
v B = (vB )0 + aBt = 0 + (0.9643)(2)
(b)
v B = 1.929 m/s 2
22.4° �
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255
PROBLEM 11.190
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
at= (11.46 – 0.01878 v2) m/s2, what is her acceleration
at point B?
SOLUTION
at  11.46 – 0.01878v 2 m/s2
Given:
Distance traveled between points A and B:
sB  18.5  r m
 18.5  28 *
2
m
 62.48 m
Find velocity at point B:
ads  vdv
62.48
ds 
0
ds 
vdv
a
vB
vdv
11.46 – 0.01878v 2
0
vB
1
62.48 
ln 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

 11.46 – 0.01878vB2 et 
23.49
28
2
en
a B  1.097et  19.71en m/s2 
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PROBLEM 11.191
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 α = 25° 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.
SOLUTION
The motion is projectile motion. Place the origin at Point A. Then x0 = 0 and y0 = 0.
The coordinates of Point B are xB = 9 m and yB = −5.4 m
Horizontal motion:
Vertical motion:
vx = v0 cos 25°
(1)
x = v0t cos 25°
(2)
v y = v0 sin 25° − gt
(3)
y = v0 t sin 25° −
1 2
gt
2
(4)
At Point B, Eq. (2) gives
v0t B =
xB
9
=
= 9.9304 m
cos 25° cos 25°
Substituting into Eq. (4),
9
1
(sin 25°) − (9.81)t B2
cos 25°
2
t B = 1.3988 s
−5.4 =
(a)
Speed of the belt.
v0 =
v0t B 9.9304
=
= 7.099 m/s
tB
1.3988
v0 = 7.10 m/s �
Eqs. (1) and (3) give
vx = 7.099cos 25° = 6.434 m/s
v y = (7.099)sin 25° − (9.81)(1.3988) = −10.722 m/s
tan
−v y
vx
= 1.6665
= 59.03°
v = 12.504 m/s
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257
PROBLEM 11.191 (Continued)
Components of acceleration.
a = 9.81 m/s2
at = 9.81sin
an = 9.81cos = 9.81cos(59.03°) = 5.048 m/s2
(b)
an =
Radius of curvature at B.
=
v2
v 2 (12.504) 2
=
= 30.97
an
5.048
= 31.0 m �
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258
PROBLEM 11.192
The end Point B of a boom is originally 5 m from fixed Point
Awhen the driver starts to retract the boom with a constant radial
r  1.0 m/s2 and lower it with a constant
acceleration of 
angular acceleration   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.
SOLUTION
r0  5 m, r0  0, 
r  1.0 m/s 2
Radial motion.
1 2

rt  5  0  0.5t 2
2
  0  1.0t
r  r0  rt
r  r0  r0t 
At t  2 s,
r  5  (0.5)(2) 2  3 m
r  (1.0)(2)  2 m/s
Angular motion.
0  60 
3
rad, 0  0,   0.5 rad/s 2
1
  0  0  t 2   0  0.25t 2
2
3



  0   t  0  0.5t
 0  (0.25)(2) 2  0.047198 rad  2.70
3
  (0.5)(2)  1.0 rad/s

At t  2 s,
Unit vectors er and e .
(a)
Velocity of Point B at t 2 s.
v B  rer  re
 (2 m/s)er  (3 m)(1.0 rad/s)e
v B  (2.00 m/s)er  (3.00 m/s)e 
tan
v
3.0
  
 1.5
vr 2.0
 56.31
v  vr2  v2  (2)2  (3)2  3.6055 m/s
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259
PROBLEM 11.192 (Continued)
Direction of velocity.
v 2er  3e

 0.55470er  0.83205e
v
3.6055
et 

 2.70  56.31  59.01
v B  3.61 m/s
(b)
59.0°
Acceleration of Point B at t 2 s.
a B  (
r  r2 )er  (r  2r)e
 [1.0  (3)(1)2 ]er  [(3)(0.5)  (2)(1.0)(0.5)]e
a B  (4.00 m/s 2 )er  (2.50 m/s2 )e 

tan
a
2.50

 0.625
ar 4.00
 32.00
a  ar2  a2  (4) 2  (2.5)2  4.7170 m/s 2

 2.70  32.00  29.30
a B  4.72 m/s2
Tangential component:
29.3°
at  (a et )et
at  (4er  2.5e ) ( 0.55470er  0.83205e )et
 [(4)( 0.55470)  (2.5)(0.83205)]et
 (0.138675 m/s 2 )et  0.1389 m/s2
Normal component:
59.0°
an  a  at
a n  4e r  2.5e  (0.138675)(0.55470er  0.83205e )
 (3.9231 m/s 2 )e r  (2.6154 m/s 2 )e
an  (3.9231)2  (2.6154)2  4.7149 m/s2
(c)
Radius of curvature of the path.
an 
v2


v 2 (3.6055 m/s)2

an
4.7149 m/s
  2.76 m 
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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 system r = 150 m, r = −31.5 m/s,
r = −3 m/s 2 ,
= 25°, = 0.07 rad/s, = 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.
SOLUTION
(a)
Velocity of the skier.
(r = 150 m,
= 25°)
v = vr e r + v e = re r + r e
= ( −31.5 m/s)e r + (150 m)(0.07 rad/s)e
v = ( −31.5 m/s)er + (10.5 m/s)e �
Direction of velocity:
v = ( −31.5cos 25° − 10.5cos65°)i + (10.5sin 65° − 31.5sin 25°) j
= ( −32.986 m/s)i + ( −3.796 m/s) j
vy
−3.796
tan α =
=
α = 6.565°
vx −32.986
v = ( −31.5) 2 + (10.5) 2 = 33.204 m/s
v = 33.2 m/s
6.57° �
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261
PROBLEM 11.193 (Continued)
(b)
Acceleration of the skier.
a = ar e r + a e = (
r − r 2 )e r + (r + 2r )e
ar = −3 − (150)(0.07) 2 = −3.735 m/s 2
a = (150)(0.06) + 2( −31.5)(0.07) = 4.59 m/s 2
a = ( −3.74 m/s 2 )er + (4.59 m/s 2 )e �
a = ( −3.735)(i cos 25° + j sin 25°) + (4.59)(− i cos65° + j sin 65°)
= ( −5.325 m/s 2 )i + (2.581 m/s 2 ) j
ay
2.581
=
=
= −25.9°
ax −5.325
tan
a = ( −3.735) 2 + (4.59) 2 = 5.918 m/s 2
a = 5.92 m/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
x0 = 32.986 m/s
(from Part a)
x = x0 + x0t = 32.986t
Vertical motion:
(Uniformly accelerated motion)
y0 = 0
y 0 = 3.796 m/s
(from Part a)
y = 9.81 m/s 2
y = y0 + y0t +
1 2
yt = 3.796t + 4.905t 2
2
At the landing point,
x = d cos30°
y = 3 + d sin 30° or
(1)
y − 3 = d sin 30°
(2)
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262
PROBLEM 11.193 (Continued)
Multiply Eq. (1) by sin 30° and Eq. (2) by cos 30° and subtract
x sin 30° − ( y − 3)cos30° = 0
(32.986t )sin 30° − (3.796t + 4.905t 2 − 3)cos30° = 0
−4.2479t 2 + 13.206t + 2.5980 = 0
t = −0.18565 s and 3.2944 s
Reject the negative root.
x = (32.986)(3.2944) = 108.67 m
d=
x
cos30°
d = 125.5 m �
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PROBLEM 11.CQ1
A bus travels the 100 km between A and B at 50 km/h and then another 100
miles between B and C at 70 km/h. The average speed of the bus for the
entire 200-km trip is:
(a) more than 60 km/h
(b) equal to 60 km/h
(c) less than 60 km/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 km/h.
Answer: (c) 
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PROBLEM 11CQ2
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)?
(a) At time t2 both cars have traveled the same
distance
(b) At time t1 both cars have the same speed
(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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PROBLEM 11.CQ3
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
(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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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?
(a) The velocity and acceleration are both zero.
(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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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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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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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 of the acceleration of block B?
SOLUTION
Since aB  a A  aB/A we get
Answer: (d ) 
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PROBLEM 11.CQ8
The Ferris wheel is rotating with a constant angular velocity . What is
the direction of the acceleration of Point A?
(a)
(b)
(c)
(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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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
(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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PROBLEM 11.CQ10
A child walks across merry
merry-go-round A with a constant speed u relative to
A.. The merry
merry-go-round
round undergoes fixed axis rotation about its center with a
constant angular velocity
counterclockwise.When the child is at the
center of A, as shown, what is the direction of his acceleration when viewed
from above.
(a)
(b)
(c)
(d )
(e) The acceleration is zero.
SOLUTION
Polar coordinates are most natural for this problem, that is,
a  (
r  r2 )er  (r  2r)e
(1)
From the information given, we know 
r  0,   0, r  0,   , r  u. When we substitute
these values into (1), we will only have a term in the  direction.
Answer: (d
( )
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