PROBLEM 11.1
The motion of a particle is defined by the relation x = t 4 − 10t 2 + 8t + 12 , where x and t are expressed in
inches and seconds, respectively. Determine the position, the velocity, and the acceleration of the particle
when t = 1 s.
SOLUTION
x = t 4 − 10t 2 + 8t + 12
At t = 1 s,
v=
dx
= 4t 3 − 20t + 8
dt
a=
dv
= 12t 2 − 20
dt
x = 1 − 10 + 8 + 12 = 11
x = 11.00 in. 
v = 4 − 20 + 8 = −8
v = −8.00 in./s 
a = 12 − 20 = −8 
a = −8.00 in./s 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 feet
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 
x1 = 15.00 ft 
a1 = −6.00 ft/s 2 
At t = 2 s,
x2 = 2(2)3 − 9(2) 2 + 12(2) + 10 = 14
t = 2.00 s 
x2 = 14.00 ft 
a2 = (12)(2) − 18 = 6
a2 = 6.00 ft/s 2 
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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
= 20 cos 2t − 30sin 2t
dt
a=
dv
= −40sin 2t − 60 cos 2t
dt
For trigonometric functions set calculator to radians:
(a) At t = 1 s.
x1 = 10sin 2 + 15cos 2 + 100 = 102.9
x1 = 102.9 mm 
v1 = 20cos 2 − 30sin 2 = −35.6
a1 = −40sin 2 − 60 cos 2 = −11.40
v1 = −35.6 mm/s 
a1 = −11.40 mm/s 2 
(b) Maximum velocity occurs when a = 0.
−40sin 2t − 60cos 2t = 0
tan 2t = −
60
= −1.5
40
2t = tan −1 (−1.5) = −0.9828 and −0.9828 + π
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)
= −36.056
vmax = −36.1 mm/s 
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 defined by the
relation x = 60e−4.8t sin16t where x and t are expressed in
mm and seconds, respectively. Determine the position, the
velocity and the acceleration of the railroad car when
(a) t = 0, (b) t = 0.3 s.
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,
x0 = 0
v0 = 960 mm/s
a0 = −9216 mm/s 2
(b) At t = 0.3 s,
x0 = 0 mm 
v0 = 960 mm/s
a0 = 9220 mm/s 2


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
x0.3 = 14.16 mm

v0.3 = 87.9 mm/s

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 2

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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 − 24t + 3
dt
and
a=
dv
= 72t 2 − 12t − 24
dt
When a = 0:
72t 2 − 12t − 24 = 12(6t 2 − t − 2) = 0
(3t − 2)(2t + 1) = 0
or
t=
or
At t =
2
s:
3
2
1
s and t = − s (Reject)
3
2
4
3
2
2
2
2
2
x2/3 = 6   − 2   − 12   + 3   + 3
3
3
3
3
3
t = 0.667 s 
or
x2/3 = 0.259 m 
2
2
2
2
v2/3 = 24   − 6   − 24   + 3
3
3
3
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 inches
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)
When v = 0:
3 t 2 − 18t + 24 = 3(t 2 − 6t + 8) = 0
(t − 2)(t − 4) = 0
t = 2.00 s and t = 4.00 s 
(b)
When a = 0:
6t − 18 = 0 or t = 3 s
x3 = (3)3 − 9(3)2 + 24(3) − 8
At t = 3 s:
First observe that 0 ≤ t < 2 s:
v>0
2 s < t ≤ 3 s:
v<0
or
x3 = 10.00 in. 
Now
At t = 0:
x0 = −8 in.
At t = 2 s:
x2 = (2)3 − 9(2) 2 + 24(2) − 8 = 12 in.
Then
x2 − x0 = 12 − (−8) = 20 in.
| x3 − x2 | = |10 − 12| = 2 in.
Total distance traveled = (20 + 2) in.
Total distance = 22.0 in. 
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PROBLEM 11.7
The motion of a particle is defined by the relation x = 2t 3 − 15t 2 + 24t + 4, where x is expressed in meters
and t in seconds. 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
v=
dt
dv
a=
= 12t − 30
dt
(a)
v = 0 when
6t 2 − 30t + 24 = 0
6(t − 1)(t − 4) = 0
(b)
a = 0 when
12t − 30 = 0
t = 1.000 s and t = 4.00 s 
t = 2.5 s
x2.5 = 2(2.5)3 − 15(2.5) 2 + 24(2.5) + 4
For t = 2.5 s:
x2.5 = +1.500 m 
To find total distance traveled, we note that
x1 = 2(1)3 − 15(1)2 + 24(1) + 4
v = 0 when t = 1 s:
x1 = +15 m
For t = 0,
x0 = +4 m
Distance traveled
From t = 0 to t = 1 s:
From t = 1 s to t = 2.5 s:
x1 − x0 = 15 − 4 = 11 m
x2.5 − x1 = 1.5 − 15 = 13.5 m
Total distance traveled = 11 m + 13.5 m
Total distance = 24.5 m 
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PROBLEM 11.8
The motion of a particle is defined by the relation x = t 3 − 6t 2 − 36t − 40, where x and t are expressed in feet
and seconds, respectively. Determine (a) when the velocity is zero, (b) the velocity, the acceleration, and the
total distance traveled when x = 0.
SOLUTION
We have
x = t 3 − 6t 2 − 36t − 40
Then
v=
dx
= 3t 2 − 12t − 36
dt
and
a=
dv
= 6t − 12
dt
(a)
When v = 0:
3t 2 − 12t − 36 = 3(t 2 − 4t − 12) = 0
(t + 2)(t − 6) = 0
or
t = −2 s (Reject) and t = 6 s
or
(b)
When x = 0:
t = 6.00 s 
t 3 − 6t 2 − 36t − 40 = 0
Factoring
(t − 10)(t + 2)(t + 2) = 0 or t = 10 s
Now observe that
0 ≤ t < 6 s:
v<0
6 s < t ≤ 10 s:
v>0
and at t = 0:
t = 6 s:
x0 = −40 ft
x6 = (6)3 − 6(6)2 − 36(6) − 40
= −256 ft
t = 10 s:
Then
v10 = 3(10) 2 − 12(10) − 36
or v10 = 144.0 ft/s 
a10 = 6(10) − 12
or
a10 = 48.0 ft/s 2 
| x6 − x0 | = | − 256 − (−40)| = 216 ft
x10 − x6 = 0 − (−256) = 256 ft
Total distance traveled = (216 + 256) ft
Total distance = 472 ft 
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PROBLEM 11.9
The brakes of a car are applied, causing it to slow down at a
rate of 10 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 = −10 ft/s 2
(a)
Velocity at x = 0.
v

0
v0
0−
(b)
dv
= a = −10
dx
vdv = −

xf
0
(−10) dx
v02
= −10 x f = −(10)(300)
2
v02 = 6000
v0 = 77.5 ft/s 2 
Time to stop.
dv
= a = −10
dx

0
v0
dv = −

tf
0
−10dt
0 − v0 = −10t f
tf =
v0 77.5
=
10
10
t f = 7.75 s 
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PROBLEM 11.10
The acceleration of a particle is directly proportional to the time t. At t = 0, the velocity of the particle
is v = 16 in./s. Knowing that v = 15 in./s and that x = 20 in. when t = 1 s, determine the velocity, the position,
and the total distance traveled when t = 7 s.
SOLUTION
a = kt
We have
dv
= a = kt
dt
Now
v
At t = 0, v = 16 in./s:

or
v − 16 =
dv =
16
0
kt dt
1 2
kt
2
1 2
kt (in./s)
2
15 in./s = 16 in./s +
k = −2 in./s3
or
1
k (1 s) 2
2
and v = 16 − t 2
dx
= v = 16 − t 2
dt
Also
At t = 1 s, x = 20 in.:

t
v = 16 +
or
At t = 1 s, v = 15 in./s:
k = constant

x
20
dx =

t
1
(16 − t 2 ) dt
t
or
or
1 

x − 20 = 16t − t 3 
3 1

1
13
x = − t 3 + 16t + (in.)
3
3
Then
At t = 7 s:
When v = 0:
v7 = 16 − (7) 2
or v7 = −33.0 in./s 
1
13
x7 = − (7)3 + 16(7) +
3
3
or
x7 = 2.00 in. 
16 − t 2 = 0 or t = 4 s
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PROBLEM 11.10 (Continued)
At t = 0:
x0 =
13
3
1
13
x4 = − (4)3 + 16(4) + = 47 in.
3
3
t = 4 s:
Now observe that
Then
0 ≤ t < 4 s:
v>0
4 s < t ≤ 7 s:
v<0
13
= 42.67 in.
3
| x7 − x4 | = |2 − 47| = 45 in.
x4 − x0 = 47 −
Total distance traveled = (42.67 + 45) in.
Total distance = 87.7 in. 
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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
a = kt 2
We have
k = constant
dv
= a = kt 2
dt
Now
v

or
1
v − 18 = k (t 3 − 216)
3
18
dv =

t
At t = 6 s, v = 18 m/s:
6
kt 2 dt
1
v = 18 + k (t 3 − 216)(m/s)
3
or
dx
1
= v = 18 + k (t 3 − 216)
dt
3
Also
x
1

18 + k (t 3 − 216)  dt
3


or
1 1

x − 24 = 18t + k  t 4 − 216t 
3 4

24
dx =

t
At t = 0, x = 24 m:
0

Now
At t = 6 s, x = 96 m:
or
Then
or
and
or
1 1

96 − 24 = 18(6) + k  (6) 4 − 216(6) 
3 4

k=
1
m/s 4
9
1  1  1

x − 24 = 18t +   t 4 − 216t 
3  9  4

x(t ) =
1 4
t + 10t + 24
108

11
v = 18 +   (t 3 − 216)
3 9
v(t ) =
1 3
t + 10
27

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PROBLEM 11.12
The acceleration of a particle is defined by the relation a = kt 2 . (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
dv
= a = kt 2
dt
(1)
t = 0, v = −8 m/s and t = 2 s, v = +8 ft/s

(a)
8
−8
dv =

2
0
kt 2 dt
1
8 − ( −8) = k (2)3
3
(b)
k = 6.00 m/s 4 
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
(2t 3 − 8) dt ; x =
1 4
t − 8t
2
t
2
1
 1

x =  t 4 − 8t  −  (2) 4 − 8(2) 
2
 2

1
x = t 4 − 8t − 8 + 16
2
x=
1 4
t − 8t + 8 
2
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PROBLEM 11.13
The acceleration of Point A is defined by the relation a = −1.8sin kt , where a and t are
expressed in m/s 2 and seconds, respectively, and k = 3 rad/s. Knowing that x = 0 and
v = 0.6 m/s when t = 0, determine the velocity and position of Point A when t = 0.5 s.
SOLUTION
Given:
a = −1.8sin kt m/s 2 ,
v0 = 0.6 m/s, x0 = 0,
t
t
v − v0 =  0 a dt = −1.8  0 sin kt dt =
v − 0.6 =
0
v = 0.6cos kt m/s
t
t
x − x0 =  0 v dt = 0.6  0 cos kt dt =
x−0=
When t = 0.5 s,
t
1.8
cos kt
k
1.8
(cos kt − 1) = 0.6cos kt − 0.6
3
Velocity:
Position:
k = 3 rad/s
0.6
sin kt
k
t
0
0.6
(sin kt − 0) = 0.2sin kt
3
x = 0.2sin kt m
kt = (3)(0.5) = 1.5 rad
v = 0.6cos1.5 = 0.0424 m/s
v = 42.4 mm/s 
x = 0.2sin1.5 = 0.1995 m
x = 199.5 mm 
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PROBLEM 11.14
The acceleration of Point A is defined by the relation a = −1.08sin kt − 1.44cos 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
Given:
a = −1.08sin kt − 1.44 cos kt m/s 2 ,
x0 = 0.16 m,
k = 3 rad/s
v0 = 0.36 m/s
t
t
t
v − v0 =  0 a dt = −1.08  0 sin kt dt − 1.44  0 cos kt dt
v − 0.36 =
=
1.08
cos kt
k
t
0
−
1.44
sin kt
k
t
0
1.08
1.44
(cos kt − 1) −
(sin kt − 0)
3
3
= 0.36 cos kt − 0.36 − 0.48sin kt
Velocity:
v = 0.36 cos kt − 0.48sin kt m/s
t
t
t
x − x0 =  0 v dt = 0.36  0 cos kt dt − 0.48  0 sin kt dt
x − 0.16 =
0.36
sin kt
k
t
0
+
0.48
cos kt
k
t
0
0.36
0.48
(sin kt − 0) +
(cos kt − 1)
3
3
= 0.12sin kt + 0.16cos kt − 0.16
=
Position:
When t = 0.5 s,
x = 0.12sin kt + 0.16cos kt m
kt = (3)(0.5) = 1.5 rad
v = 0.36 cos1.5 − 0.48sin1.5 = −0.453 m/s
x = 0.12sin1.5 + 0.16 cos1.5 = 0.1310 m
v = −453 mm/s 
x = 131.0 mm 
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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
= −k x
dx
Separate and integrate.

vf
v0
vdv = −

xf
0
k x dx
1 2 1 2
1
v f − v0 = − kx 2
2
2
2
xf
0
1
= − k x 2f
2
Use v0 = 4 m/s, x f = 0.02 m, and v f = 0. Solve for k.
1
1
0 − (4) 2 = − 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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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 
0 = (900 ft/s) − k  ft 
 12 
4
ft, v = 0:
12
k = 2700
or
(a)
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 − kx) = −kv
dt dt
a0 = −2.43 × 106 ft/s 2 
or
(b)
dx
= v = v0 − kx
dt
We have
At t = 0, x = 0:
or

x
0
dx
=
v0 − kx
t
 dt
0
1
− [ln(v0 − kx)]0x = t
k
or
t=
1  v0  1  1 

ln 
 = ln 
k  v0 − kx  k  1 − vk x 
0


When x = 3.9 in.:
t=

1
ln 
1
2700 s 1 −
1
2700 1/s
900 ft/s
(
3.9
12

ft ) 
or
t = 1.366 × 10−3 s 
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PROBLEM 11.17
The acceleration of a particle is defined by the relation a = −k/x. It has been experimentally determined that
v = 15 ft/s when x = 0.6 ft and that v = 9 ft/s when x = 1.2 ft. Determine (a) the velocity of the particle
when x = 1.5 ft, (b) the position of the particle at which its velocity is zero.
SOLUTION
a=
vdv − k
=
dx
x
Separate and integrate using x = 0.6 ft, v = 15 ft/s.

v
15
vdv = − k

x
0.6
v
dx
x
1 2
v
= − k ln x
2 15
x
0.6
1 2 1
 x 
v − (15) 2 = − k ln 

2
2
 0.6 
(1)
When v = 9 ft/s, x = 1.2 ft
1 2 1
 1.2 
(9) − (15) 2 = −k ln 

2
2
 0.6 
Solve for k.
k = 103.874 ft 2 /s 2
(a)
Velocity when x = 65 ft.
Substitute
k = 103.874 ft 2 /s 2
and
x = 1.5 ft into (1).
1 2 1
 1.5 
v − (15) 2 = −103.874 ln 

2
2
 0.6 
v = 5.89 ft/s 
(b)
Position when for v = 0,
1
 x 
0 − (15) 2 = −103.874 ln 

2
 0.6 
 x 
ln 
 = 1.083
 0.6 


 x = 1.772 ft 
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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,
respectively, and k = 4 × 10−4 m3 /s 2 . Determine the maximum velocity and
acceleration of B.
SOLUTION
The maximum veolocity occurs when a = 0.
xm2 =
0 = −9.81 +
k
4 × 10−4
=
= 40.775 × 10−6 m 2
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
x0
dx
x2
1 1 
1 2
v = −9.81( x − x0 ) − k  − 
2
 x x0 
 1
1 2
1 
vm = −9.81( xm − x0 ) − k 
− 
2
 xm x0 
1
1 

= −9.81(0.0063855 − 0.004) − (4 × 10−4 ) 
−

0.0063855
0.004


= −0.023402 + 0.037358 = 0.013956 m 2 /s2
Maximum velocity:
vm = 0.1671 m/s
vm = 167.1 mm/s 
The maximum acceleration occurs when x is smallest, that is, x = 0.004 m.
am = −9.81 +
4 × 10−4
(0.004) 2
am = 15.19 m/s 2

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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 and x 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:
v

v
1
dv
x 

= a = −  0.1 + sin
dx
0.8 

vdv =

x 

−  0.1 + sin
dx
0 
0.8 
x
x
1 2
x 

(v − 1) = − 0.1x − 0.8 cos
2
0.8  0

or
x
1 2
v = −0.1x + 0.8 cos
− 0.3
2
0.8
or
(a)
When x = −1 m:
1 2
−1
v = −0.1(−1) + 0.8 cos
− 0.3
2
0.8
v = ± 0.323 m/s 
or
(b)
When v = vmax, a = 0:
or
(c)
x 

−  0.1 + sin
=0
0.8 

x = −0.080 134 m
x = −0.0801 m 
When x = −0.080 134 m:
1 2
−0.080 134
vmax = −0.1(−0.080 134) + 0.8 cos
− 0.3
2
0.8
= 0.504 m 2 /s 2
or
vmax = 1.004 m/s 
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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
lx
= −100  x −

dx
l 2 + x2





Separate variables and integrate:

vf
v0
vdv = −100


lx
x−
2
x0 
l + x2

0
 x2
1 2 1 2
− l l 2 + x2
v f − v0 = −100 
2
2
2





 dx


0
x0
 x2
1 2
v f − 0 = −100  − 0 − l 2 + l l 2 + x02
 2
2




1 2 100 2
(−l + x02 − l 2 − 2l l 2 + x02 )
vf =
2
2
100
( l 2 + x02 − l )2
=
2
v f = 10( l 2 + x02 − l ) 
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PROBLEM 11.21
The acceleration of a particle is defined by the relation a = −0.8v where a is expressed in m/s2 and
v in m/s. Knowing that at t = 0 the velocity is 1 m/s, determine (a) the distance the particle will travel
before coming to rest, (b) the time required for the particle’s velocity to be reduced by 50 percent of its
initial value.
SOLUTION
(a)
Determine relationship between x and v.
a=
vdv
= −0.8v
dx
dv = −0.8dx
Separate and integrate with v = 1 m/s when x = 0.

v
1
dv = −0.8

x
0
dx
v − 1 = −0.8 x
Distance traveled.
For v = 0,
(b)
x=
−1

−0.8
x = 1.25 m 
Determine realtionship between v and t.
a=

v
1
dv
= 0.8v
dt
dv
=−
v

x
0
0.8dt
v
ln   = −0.8t
1
1
t = 1.25ln  
v
For v = 0.5(1 m/s) = 0.5 m/s,
 1 
t = 1.25ln 

 0.5 
t = 0.866 s 
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PROBLEM 11.22
Starting from x = 0 with no initial velocity, a particle is given an acceleration a = 0.1 v 2 + 16,
where a and v are expressed in ft/s2 and ft/s, respectively. Determine (a) the position of the
particle when v = 3ft/s, (b) the speed and acceleration of the particle when x = 4 ft.
SOLUTION
a=
vdv
= 0.1(v 2 + 16)1/ 2
dx
(1)
Separate and integrate.

v
0
vdv
v 2 + 16
=

x
0
0.1 dx
v
(v 2 + 16)1/2 = 0.1 x
0
2
1/2
(v + 16)
(a)
(b)
− 4 = 0.1 x
x = 10[(v 2 + 16)1/2 − 4]
(2)
x = 10[(32 + 16)1/2 − 4]
x = 10.00 ft 
v = 3 ft/s.
x = 4 ft.
From (2),
(v 2 + 16)1/2 = 4 + 0.1x = 4 + (0.1)(4) = 4.4
v 2 + 16 = 19.36
v 2 = 3.36 ft 2 /s 2
From (1),
a = 0.1(1.8332 + 16)1/2
v = 1.833 ft/s 
a = 0.440 ft/s 2 
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PROBLEM 11.23
A ball is dropped from a boat so that it strikes the surface of a lake
with a speed of 16.5 ft/s. While in the water the ball experiences an
acceleration of a = 10 − 0.8v, where a and v are expressed in ft/s2
and ft/s, respectively. Knowing the ball takes 3 s to reach the
bottom of the lake, determine (a) the depth of the lake, (b) the speed
of the ball when it hits the bottom of the lake.
SOLUTION
a=
dv
= 10 − 0.8v
dt
Separate and integrate:

dv
=
v0 10 − 0.8v
v

t
0
dt
v
−
1
ln(10 − 0.8v) = t
0.8
v0
 10 − 0.8v 
ln 
 = −0.8t
 10 − 0.8v0 
10 − 0.8v = (10 − 0.8v0 )e −0.8t
or
0.8v = 10 − (10 − 0.8v0 )e−0.8t
v = 12.5 − (12.5 − v0 )e −0.8t
With v0 = 16.5ft/s
v = 12.5 + 4e−0.8t
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PROBLEM 11.23 (Continued)
Integrate to determine x as a function of t.
dx
= 12.5 + 4e−0.8t
v=
dt

x
0
dx =

t
0
(12.5 + 4e−0.8t )dt
x = 12.5t − 5e −0.8t
(a)
t
0
= 12.5t − 5e−0.8t + 5
At t = 35 s,
x = 12.5(3) − 5e −2.4 + 5 = 42.046 ft
(b)
v = 12.5 + 4e −2.4 = 12.863 ft/s
x = 42.0 ft 
v = 12.86 ft/s 
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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
v

or
2 3/2 v
[v ]81 = −kx
3
or
2 3/2
[v − 729] = −kx
3
When x = 18 m, v = 36 m/s:
v dv =
81

x
When x = 0, v = 81 m/s:
0
− k dx
2
(363/2 − 729) = − k (18)
3
k = 19 m/s 2
or
Finally
When x = 20 m:
2 3/2
(v − 729) = −19(20)
3
v3/2 = 159
or
(b)
We have
dv
= a = −19 v
dt
At t = 0, v = 81 m/s:

v
dv
81
v
=

t
0
−19dt
or
v
2[ v ]81
= −19t
or
2( v − 9) = −19t
When v = 0:
or
v = 29.3 m/s 
2(−9) = −19t
t = 0.947 s 
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PROBLEM 11.25
A particle is projected to the right from the position x = 0 with an initial velocity of 9 m/s. If the acceleration
of the particle is defined by the relation a = −0.6v3/ 2 , where a and v are expressed in m/s2 and m/s,
respectively, determine (a) the distance the particle will have traveled when its velocity is 4 m/s, (b) the time
when v = 1 m/s, (c) the time required for the particle to travel 6 m.
SOLUTION
(a)
We have
When x = 0, v = 9 m/s:
v

v
9
dv
= a = −0.6v3/2
dx
v − (1/2) dv =

x
0
−0.6dx
or
2[v1/2 ]9v = −0.6 x
or
x=
1
(3 − v1/ 2 )
0.3
When v = 4 m/s:
x=
1
(3 − 41/ 2 )
0.3
x = 3.33 m 
or
(b)
dv
= a = −0.6v3/2
dt
We have
v

or
−2[v − (1/2) ]9v = −0.6t
When v = 1 m/s:
9
v − (3/2) dv =
1
v
1
1

t
When t = 0, v = 9 m/s:
or
0
−0.6dt
−
1
= 0.3t
3
−
1
= 0.3t
3
t = 2.22 s 
or
(c)
We have
(1)
1
v
−
1
= 0.3t
3
2
or
Now
9
 3 
v=
 =
1
0.9
t
+
(1 + 0.9t ) 2


dx
9
=v=
dt
(1 + 0.9t ) 2
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PROBLEM 11.25 (Continued)

At t = 0, x = 0:
x
0

dx =
9
dt
0 (1 + 0.9t ) 2
t
t
1 
 1
x = 9 −

 0.9 1 + 0.9t  0
or
1 

= 10 1 −

 1 + 0.9t 
9t
=
1 + 0.9t
When x = 6 m:
6=
9t
1 + 0.9t
t = 1.667 s 
or
An alternative solution is to begin with Eq. (1).
x=
1
(3 − v1/ 2 )
0.3
dx
= v = (3 − 0.3x) 2
dt
Then
Now
At t = 0, x = 0:

x
0
dx
=
(3 − 0.3x) 2

t
0
dt
x
or
1  1 
x
=
t=


0.3  3 − 0.3x  0 9 − 0.9 x
which leads to the same equation as above.
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PROBLEM 11.26
The acceleration of a particle is defined by the relation a = 0.4(1 − kv), where k is a constant. Knowing that
at t = 0 the particle starts from rest at x = 4 m and that when t = 15 s, v = 4 m/s, determine (a) the constant k,
(b) the position of the particle when v = 6 m/s, (c) the maximum velocity of the particle.
SOLUTION
(a)
dv
= a = 0.4(1 − kv)
dt
We have

At t = 0, v = 0:
dv
=
0 1− kv
v

t
0
0.4dt
1
− [ln(1 − k v)]v0 = 0.4t
k
or
or
ln(1 − k v) = −0.4 kt
At t = 15 s, v = 4 m/s:
ln(1 − 4k ) = −0.4k (15)
(1)
= −6k
k = 0.145703 s/m
Solving yields
k = 0.1457 s/m 
or
(b)
We have
v

When x = 4 m, v = 0:
vdv
=
0 1− kv
v

x
4
0.4 dx
v
1
1/k
=− +
1− kv
k 1− kv
Now
Then
dv
= a = 0.4(1 − kv)
dx

v

1
1
− +

 dv =
0
 k k (1 − k v) 

x
4
0.4 dx
v
or
 v 1

x
 − k − k 2 ln(1 − k v)  = 0.4[ x]4

0
or
v 1

−  + 2 ln(1 − kv)  = 0.4( x − 4)
k
k


When v = 6 m/s:


6
1
ln(1 − 0.145 703 × 6)  = 0.4( x − 4)
−
+
2
 0.145 703 (0.145 703)

or
0.4( x − 4) = 56.4778
or
x = 145.2 m 
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PROBLEM 11.26 (Continued)
(c)
The maximum velocity occurs when a = 0.
a = 0: 0.4(1 − k vmax ) = 0
vmax =
or
1
0.145 703
vmax = 6.86 m/s 
or
An alternative solution is to begin with Eq. (1).
ln(1 − k v) = −0.4 kt
v=
Then
Thus, vmax is attained as t
1
(1 − k −0.4 kt )
k
∞
vmax =
1
k
as above.
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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
=−
a=−
When x = 2 m:
0.0324v02
x3
0.0324(3.6) 2
(2)3
a = −0.0525 m/s 2 
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 
 2 x  = 0.18v0 (t3 − t1 )

1
or
(t3 − t1 ) =
1
2
(9 − 1)
0.18(3.6)
or
t3 − t1 = 6.17 s 
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PROBLEM 11.28
Based on observations, the speed of a jogger can be approximated by the
relation v = 7.5(1 − 0.04 x)0.3 , where v and x are expressed in mi/h and miles,
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 ft/s2 at t = 0,
(c) the time required for the jogger to run 6 mi.
SOLUTION
(a)
dx
= v = 7.5(1 − 0.04 x)0.3
dt
We have

At t = 0, x = 0:
or
x
0
dx
=
(1 − 0.04 x)0.3
t
 7.5dt
0
1 
1 
[(1 − 0.04 x)0.7 ]0x = 7.5t
−

0.7  0.04 
1 − (1 − 0.04 x)0.7 = 0.21t
or
or
x=
1
[1 − (1 − 0.21t )1/0.7 ]
0.04
At t = 1 h:
x=
1
{1 − [1 − 0.21(1)]1/0.7 }
0.04
(1)
x = 7.15 mi 
or
(b)
We have
a=v
dv
dx
d
[7.5(1 − 0.04 x)0.3 ]
dx
= 7.52 (1 − 0.04 x)0.3 [(0.3)(−0.04)(1 − 0.04 x) −0.7 ]
= 7.5(1 − 0.04 x)0.3
= −0.675(1 − 0.04 x) −0.4
At t = 0, x = 0:
a0 = −0.675 mi/h 2 ×
5280 ft  1 h 
×

1 mi
 3600 s 
a0 = −275 × 10−6 ft/s 2 
or
(c)
From Eq. (1)
t=
When x = 6 mi:
t=
or
2
1
[1 − (1 − 0.04 x)0.7 ]
0.21
1
{1 − [1 − 0.04(6)]0.7 }
0.21
= 0.83229 h
t = 49.9 min 
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PROBLEM 11.29
The acceleration due to gravity at an altitude y above the surface of the earth can be
expressed as
a=
−32.2
[1 + ( y/20.9 × 106 )]2
where a and y are expressed in ft/s2 and feet, respectively. Using this expression,
compute the height reached by a projectile fired vertically upward from the surface
of the earth if its initial velocity is (a) 1800 ft/s, (b) 3000 ft/s, (c) 36,700 ft/s.
SOLUTION
We have
v
dv
=a=−
dy
(1 +
32.2
y
20.9 × 106
When
y = 0,
provided that v does reduce to zero,
y = ymax , v = 0

Then
v0
v dv =

v = v0
ymax
0
(1 +
−32.2
y
20.9 × 106
)
2
dy

1
v02 = 1345.96 × 106 1 −
 1 + ymax
20.9 × 106

or
ymax =
or
v0 = 1800 ft/s:
ymax =
v0 = 3000 ft/s:
or
y
 max


0




v02
64.4 −
v02
20.9 × 106
(1800)2
64.4 −
(1800)2
20.9 × 106
ymax = 50.4 × 103 ft 
or
(b)
2

1
1
− v02 = −32.2  −20.9 × 106
y

2
1+
× 106
20.9

or
(a)
0
)
ymax =
(3000) 2
64.4 −
(3000)2
20.9 × 106
ymax = 140.7 × 103 ft 
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PROBLEM 11.29 (Continued)
(c)
v0 = 36,700 ft/s:
ymax =
(36, 700) 2
64.4 −
(36,700)2
20.9 × 106
= −3.03 × 1010 ft
This solution is invalid since the velocity does not reduce to zero. The velocity 36,700 ft/s is above the
escape velocity vR from the earth. For vR and above.
ymax
∞ 
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PROBLEM 11.30
The acceleration due to gravity of a particle falling toward the earth is a = − gR 2 /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 = 3960 mi, calculate the
escape velocity, that is, the minimum velocity with which a particle must be projected
vertically upward from the surface of the earth if it is not to return to the earth. (Hint: v = 0
for r = ∞.)
SOLUTION
We have
v
dv
gR 2
=a=− 2
dr
r
r = R, v = ve
When
r = ∞, v = 0
0
gR 2
dr
r2

or
1
1 
− ve2 = gR 2  
2
 r R
or
ve = 2 gR
ve
vdv =

∞
then
R
−
∞
1/2
5280 ft 

=  2 × 32.2 ft/s 2 × 3960 mi ×

1 mi 

or
ve = 36.7 × 103 ft/s 
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PROBLEM 11.31
The velocity of a particle is v = v0 [1 − sin (π 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
 π t 
= v = v0 1 − sin   
dt
 T 

We have
At t = 0, x = 0:

x
0
dx =


 π t 
v0 1 − sin    dt
0
 T 

t
t
 T
 T
 π t 
 πt  T 
x = v0 t + cos    = v0 t + cos   − 
 T 0
T  π
 π
 π
At t = 3T :

T
2T 
 π × 3T  T 

−  = v0  3T −
x3T = v0 3T + cos 

π
π 
 T  π


a=
At t = 3T :
(b)
(1)
x3T = 2.36 v0T 
π
πt
dv d  
 π t   
= v0 1 − sin     = −v0 cos
dt dt  
T
T
 T   
a3T = −v0
π
T
cos
π × 3T
a3T =
T
π v0
T

Using Eq. (1)
At t = 0:
At t = T :
Now
T
T

x0 = v0 0 + cos(0) −  = 0
π
π



T
 πT
xT = v0 T + cos 
π
 T

vave =
2T
 T

 − π  = v0  T − π



xT − x0 0.363v0T − 0
=
Δt
T −0

 = 0.363v0T

vave = 0.363v0 
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PROBLEM 11.32
The velocity of a slider is defined by the relation v = v′ sin (ωn t + φ ). Denoting the velocity and the position
of the slider at t = 0 by v0 and x0 , respectively, and knowing that the maximum displacement of the slider
is 2 x0 , show that (a) v′ = (v02 + x02ωn2 )/2 x0ωn , (b) the maximum value of the velocity occurs when
x = x0 [3 − (v0 /x0ωn ) 2 ]/2.
SOLUTION
(a)
v0 = v′ sin (0 + φ ) = v′ sin φ
At t = 0, v = v0 :
Then
2
cos φ = v′ − v02 v′
dx
= v = v′ sin (ωn t + φ )
dt
Now
x

or
 1

x − x0 = v′  −
cos (ωn t + φ ) 
 ωn
0
x0
dx =

t
At t = 0, x = x0 :
0
v′ sin (ωn t + φ )dt
t
or
x = x0 +
v′
ωn
[cos φ − cos (ωn t + φ )]
Now observe that xmax occurs when cos (ωn t + φ ) = −1. Then
xmax = 2 x0 = x0 +
ωn
[cos φ − ( −1)]
2
2

v′  v′ − v0

x0 =
+
1

ωn 
v1


Substituting for cos φ
or
v′
x0ωn − v′ = v′2 − v02
Squaring both sides of this equation
x02ωn2 − 2 x0ωn + v′2 = v′2 − v02
or
v′ =
v02 + x02ωn2
2 x0ωn
Q. E. D.
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PROBLEM 11.32 (Continued)
(b)
First observe that vmax occurs when ωn t + φ = π2 . The corresponding value of x is
v′ 
 π 
cos φ − cos   
ωn 
 2 
v′
= x0 +
cos φ
xvmax = x0 +
ωn
Substituting first for cos φ and then for v′
xvmax = x0 +
2
v′ − v02
v′
v′
ωn
1/2
2


1  v02 + x02ωn2 
2
= x0 +

 − v0 

ωn  2 x0ωn 


1
v04 + 2v02 x02ωn2 + x04ωn4 − 4 x02ωn2 v02
= x0 +
2
2 x0ωn
(
= x0 +
= x0 +
x
= 0
2
1
(
 x 2ω 2 − v 2
0 n
0
2 x0ωn2 
)
1/2
1/ 2
2
) 
x02ωn2 − v02
2 x0ωn2
  v 2 
3 −  0  
  x0ωn  


Q. E. D.
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PROBLEM 11.33
A stone is thrown vertically upward from a point on a bridge located 40 m above the water. Knowing that it
strikes the water 4 s after release, determine (a) the speed with which the stone was thrown upward, (b) the
speed with which the stone strikes the water.
SOLUTION
Uniformly accelerated motion. Origin at water.
1
y = y0 + v0t + a t 2
2
v = v0 + a t
where y0 = 40 m and a = −9.81 m/s2 .
(a)
Initial speed.
y = 0 when t = 4 s.
1
0 = 40 + v0 (4) − (9.81)(4)2
2
v0 = 9.62 m/s
(b)
v0 = 9.62 m/s 
Speed when striking the water. (v at t = 4 s)
v = 9.62 − (9.81)(4) = −29.62 m/s
v = 29.6 m/s 
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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 v0 = 54 km/h = 15 m/s
(a)
x = x0 + v0t +
1 2
at
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/s 2 
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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PROBLEM 11.35
A motorist enters a freeway at 30 mi/h and accelerates uniformly
to 60 mi/h. From the odometer in the car, the motorist knows
that she traveled 550 ft while accelerating. Determine (a) the
acceleration of the car, (b) the time required to reach 60 mi/h.
SOLUTION
(a)
Acceleration of the car.
v12 = v02 + 2a( x1 − x0 )
a=
Data:
v12 − v02
2( x1 − x0 )
v0 = 30 mi/h = 44 ft/s
v1 = 60 mi/h = 88 ft/s
x0 = 0
x1 = 550 ft
a=
(b)
(88) 2 − (44) 2
(2)(55 − 0)
a = 5.28 ft/s 2 
Time to reach 60 mi/h.
v1 = v0 + a (t1 − t0 )
v1 − v0
a
88 − 44
=
5.28
= 8.333 s
t1 − t0 =
t1 − t0 = 8.33 s 
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PROBLEM 11.36
A group of students launches a model rocket in the vertical direction. Based on tracking
data, they determine that the altitude of the rocket was 89.6 ft at the end of the powered
portion of the flight and that the rocket landed 16 s later. Knowing that the descent
parachute failed to deploy so that the rocket fell freely to the ground after reaching its
maximum altitude and assuming that g = 32.2 ft/s 2 , determine (a) the speed v1 of the
rocket at the end of powered flight, (b) the maximum altitude reached by the rocket.
SOLUTION
(a)
1 2
at
2
We have
y = y1 + v1t +
At tland ,
y=0
Then
0 = 89.6 ft + v1 (16 s)
+
1
(−32.2 ft/s 2 )(16 s) 2
2
v1 = 252 ft/s 
or
(b)
We have
v 2 = v12 + 2a( y − y1 )
At
y = ymax , v = 0
Then
0 = (252 ft/s) 2 + 2( −32.2 ft/s 2 )( ymax − 89.6) ft
or
ymax = 1076 ft 
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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
D we have
v 2 = v02 + 2a( x − x0 )
2
vBC
= 0 + 2(4.8 m/s 2 )(3 − 0) m
Then, at B
= 28.8 m 2 /s 2
(vBC = 5.3666 m/s)
2
vD2 = vBC
+ 2aCD ( xD − xC )
and at D
d = xD − xC
(7.2 m/s)2 = (28.8 m 2 /s 2 ) + 2(4.8 m/s 2 )d
or
d = 2.40 m 
or
(b)
For A
B and C
D we have
v = v0 + at
Then A
B
5.3666 m/s = 0 + (4.8 m/s 2 )t AB
t AB = 1.11804 s
or
and C
7.2 m/s = 5.3666 m/s + (4.8 m/s 2 )tCD
D
tCD = 0.38196 s
or
Now, for B
C, we have
xC = xB + vBC 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
or
t D = 2.06 s 
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PROBLEM 11.38
A sprinter in a 100-m race accelerates uniformly for the first 35 m and then runs
with constant velocity. If the sprinter’s time for the first 35 m is 5.4 s, determine
(a) his acceleration, (b) his final velocity, (c) his time for the race.
SOLUTION
Given:
0 ≤ x ≤ 35 m, a = constant
35 m < x ≤ 100 m, v = constant
At t = 0, v = 0 when
x = 35 m, t = 5.4 s
Find:
(a)
a
(b)
v when x = 100 m
(c)
t when x = 100 m
(a)
We have
At t = 5.4 s:
or
x = 0 + 0t +
35 m =
1 2
at
2
for
0 ≤ x ≤ 35 m
1
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 ≤ x ≤ 100 m.
Now
(c)
v 2 = 0 + 2a( x − 0)
for
0 ≤ x ≤ 35 m
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:
or
x = x1 + v0 (t − t1 )
vmax = 12.96 m/s 
for
35 m < x ≤ 100 m
100 m = 35 m + (12.9628 m/s)(t2 − 5.4) s
t2 = 10.41 s 
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PROBLEM 11.39
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/s 2 
or
Also
At t = 1.82 s:
v A = (v A ) 0 + a A t
(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/s 2 
(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/s 2 )(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. 
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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 A t
(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
x A = ( x A ) 0 + (v A )0 t +
1
1
a At 2 = 50 m + (50 m/s)(8 s) + (1.5625 m/s 2 )(8 s) 2 = 500 m
2
2
and
xB = 0 + (vB )0 t +
At t = 8 s:
x A = xB
1
aB t 2
2
500 m = (50 m/s)(8 s) +
or
(vB )0 = 50 m/s
1
aB (8 s) 2
2
aB = 3.13 m/s 2 
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PROBLEM 11.41
A police officer in a patrol car parked in a 45 mi/h speed zone observes a passing automobile traveling at a
slow, constant speed. Believing that the driver of the automobile might be intoxicated, the officer starts his
car, accelerates uniformly to 60 mi/h in 8 s, and, maintaining a constant velocity of 60 mi/h, overtakes the
motorist 42 s after the automobile passed him. Knowing that 18 s elapsed before the officer began pursuing
the motorist, determine (a) the distance the officer traveled before overtaking the motorist, (b) the motorist’s
speed.
SOLUTION
(vP )18 = 0
(a)
(vP )26 = 60 mi/h = 88 ft/s
(vP ) 42 = 90 mi/h = 88 ft/s
Patrol car:
For 18 s < t ≤ 26 s:
At t = 26 s:
vP = 0 + aP (t − 18)
88 ft/s = aP (26 − 18) s
or
aP = 11 ft/s 2
Also,
xP = 0 + 0(t − 18) −
At t = 26 s:
For 26 s < t ≤ 42 s:
At t = 42 s:
(xP ) 26 =
1
aP (t − 18) 2
2
1
(11 ft/s 2 )(26 − 18) 2 = 352 ft
2
xP = ( xP )26 + (vP ) 26 (t − 26)
(xP ) 42 = 352 m + (88 ft/s)(42 − 26) s
= 1760 ft
( xP ) 42 = 1760 ft 
(b)
For the motorist’s car:
At t = 42 s, xM = xP :
or
xM = 0 + vM t
1760 ft = vM (42 s)
vM = 41.9048 ft/s
or
vM = 28.6 mi/h 
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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 1.8 ft/s 2 and
that B has a constant deceleration of 1.2 ft/s 2 , determine
(a) when and where A will overtake B, (b) the speed of each
automobile at that time.
SOLUTION
a A = +1.8 ft/s 2
aB = −1.2 ft/s 2
(v A )0 = 24 mi/h = 35.2 ft/s
A overtakes B
(vB )0 = 36 mi/h = 52.8 ft/s
Motion of auto A:
v A = (v A )0 + a At = 35.2 + 1.8t
x A = ( x A ) 0 + (v A ) 0 t +
(1)
1
1
a At 2 = 0 + 35.2t + (1.8)t 2
2
2
(2)
Motion of auto B:
vB = (vB )0 + aB t = 52.8 − 1.2t
x B = ( xB ) 0 + ( vB ) 0 t +
(a)
(3)
1
1
aB t 2 = 75 + 52.8t + (−1.2)t 2
2
2
(4)
A overtakes B at t = t1 .
x A = xB : 35.2t + 0.9t12 = 75 + 52.8t1 − 0.6t12
1.5t12 − 17.6t1 − 75 = 0
t1 = −3.22 s and t1 = 15.0546
Eq. (2):
x A = 35.2(15.05) + 0.9(15.05) 2
t1 = 15.05 s 
x A = 734 ft 
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PROBLEM 11.42 (Continued)
(b)
Velocities when t1 = 15.05 s
Eq. (1):
v A = 35.2 + 1.8(15.05)
v A = 62.29 ft/s
Eq. (3):
v A = 42.5 mi/h

vB = 23.7 mi/h

vB = 52.8 − 1.2(15.05)
vB = 34.74 ft/s
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PROBLEM 11.43
Two automobiles A and B are approaching each other in adjacent highway lanes. At t = 0, A and B are 3200 ft
apart, their speeds are v A = 65 mi/h and vB = 40 mi/h, and they are at Points P and Q, respectively. Knowing
that A passes Point Q 40 s after B was there and that B passes Point P 42 s after A was there, determine (a) the
uniform accelerations of A and B, (b) when the vehicles pass each other, (c) the speed of B at that time.
SOLUTION
(a)
x A = 0 + (v A ) 0 t +
We have
(x is positive
3200 m = (95.333 m/s)(40 s) +
Also, xB = 0 + (vB )0 t +
1
aB t 2
2
Then (95.333 ft/s)t AB +
(c)
1
aB (42 s) 2
2
aB = 0.83447 ft/s 2
When the cars pass each other
Solving
a A = −0.767 ft/s 2 
(vB )0 = 40 mi/h = 58.667 ft/s
3200 ft = (58.667 ft/s)(42 s) +
or
or
1
a A (40 s) 2
2
; origin at Q.)
At t = 42 s:
(b)
(v A )0 = 65 mi/h = 95.33 ft/s
; origin at P.)
At t = 40 s:
(xB is positive
1
a At 2
2
aB = 0.834 ft/s 2 
x A + xB = 3200 ft
1
1
2
2
+ (58.667 ft/s)t AB + (0.83447 ft/s 2 )t AB
= 3200 ft
(−0.76667 ft/s)t AB
2
2
2
0.03390t AB
+ 154t AB − 3200 = 0
t = 20.685 s and t = −4563 s
We have
vB = ( vB ) 0 + a B t
At t = t AB :
vB = 58.667 ft/s + (0.83447 ft/s 2 )(20.685 s)
= 75.927 ft/s
t >0
t AB = 20.7 s 
vB = 51.8 mi/h 
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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.
yB = ( yB )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.
y E = ( y E ) 0 + vE t
with ( yE )0 = −10 m and vE = 4 m/s.
(a)
Set yB = yE
Time of impact.
3t − 4.905t 2 = −10 + 4t
4.905t 2 + t − 10 = 0
t = 1.3295 and −1.5334
(b)
t = 1.330 s 
Location of impact.
yB = (3)(1.3295) − (4.905)(1.3295)2 = −4.68 m
yE = −10 + (4)(1.3295) = −4.68 m
(checks)
4.68 m below the man 
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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. Assuming a constant acceleration
g = 9.81 m/s2 , determine (a) the time t1, (b) the velocity of B relative to A at the
time of the explosion.
SOLUTION
Place origin at ground level. The motion of rockets A and B is
Rocket A:
v A = (v A )0 − gt = 100 − 9.81t
y A = ( y A ) 0 + (v A ) 0 t −
Rocket B:
(1)
1 2
gt = 100t − 4.905t 2
2
(2)
vB = (vB )0 − g (t − t1 ) = 100 − 9.81(t − t1 )
(3)
1
g (t − t1 ) 2
2
= 100(t − t1 ) − 4.905(t − t1 )2
(4)
yB = ( yB )0 + (vB )0 (t − t1 ) −
Time of explosion of rockets A and B. y A = yB = 300 ft
From (2),
300 = 100t − 4.905t 2
4.905t 2 − 100t + 300 = 0
t = 16.732 s and 3.655 s
From (4),
300 = 100(t − t1 ) − 4.905(t − t12 )
t − t1 = 16.732 s and 3.655 s
Since rocket A is falling,
Since rocket B is rising,
(a)
Time t1:
(b)
Relative velocity at explosion.
t = 16.732 s
t − t1 = 3.655 s
t1 = t − (t − t1 )
From (1),
v A = 100 − (9.81)(16.732) = −64.15 m/s
From (3),
vB = 100 − (9.81)(16.732 − 13.08) = 64.15 m/s
Relative velocity:
vB/A = vB − v A
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 60 mi/h. At t = 0, A starts and accelerates at a constant rate a A , while at t = 5 s, B begins to
slow down with a constant deceleration of magnitude a A /6. Knowing that when the cars pass each other
x = 294 ft and v A = vB , determine (a) the acceleration a A , (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
xA = 0 + 0 +
1
a At 2
2
0 ≤ t < 5 s:
xB = 0 + (vB )0 t (vB )0 = 60 mi/h = 88 ft/s
At t = 5 s:
xB = (88 ft/s)(5 s) = 440 ft
For t ≥ 5 s:
vB = (vB )0 + aB (t − 5)
1
aB = − a A
6
xB = ( xB ) S + (vB )0 (t − 5) +
1
aB (t − 5) 2
2
Assume t > 5 s when the cars pass each other.
At that time (t AB ),
v A = vB :
a At AB = (88 ft/s) −
x A = 294 ft:
294 ft =
Then
or
a A( 76 t AB − 65 )
1
2
2
a At AB
=
aA
(t AB − 5)
6
1
2
a At AB
2
88
294
2
44t AB
− 343t AB + 245 = 0
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PROBLEM 11.46 (Continued)
Solving
(a)
With t AB > 5 s,
t AB = 0.795 s and t AB = 7.00 s
294 ft =
1
a A (7.00 s) 2
2
a A = 12.00 ft/s 2 
or
(b)
t AB = 7.00 s 
From above
Note: An acceptable solution cannot be found if it is assumed that t AB ≤ 5 s.
(c)
We have
d = x + ( xB )t AB
= 294 ft + 440 ft + (88 ft/s)(2.00 s)
1 1

+  − × 12.00 ft/s 2  (2.00 s)2
2 6

or
d = 906 ft 
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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, (d ) the relative velocity of the counterweight W with respect to
the elevator.
SOLUTION
Choose the positive direction downward.
(a)
Velocity of cable C.
yC + 2 yE = constant
vC + 2vE = 0
vE = 4 m/s
But,
or
(b)
vC = −2vE = −8 m/s
vC = 8.00 m/s 
Velocity of counterweight W.
yW + yE = constant
vW + vE = 0 vW = −vE = −4 m/s
(c)
vW = 4.00 m/s 
Relative velocity of C with 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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PROBLEM 11.48
The elevator shown starts from rest and moves upward with a constant
acceleration. If the counterweight W moves through 30 ft in 5 s, determine
(a) the acceleration of the elevator and the cable C, (b) the velocity of the
elevator after 5 s.
SOLUTION
We choose positive direction downward for motion of counterweight.
yW =
At t = 5 s,
1
aW t 2
2
yW = 30 ft
30 ft =
1
aW (5 s) 2
2
aW = 2.4 ft/s 2
(a)
Accelerations of E and C.
Since
Thus:
Also,
Thus:
(b)
aW = 2.4 ft/s 2
yW + yE = constant vW + vE = 0, and aW + aE = 0
aE = − aW = −(2.4 ft/s 2 ),
a E = 2.40 ft/s 2 
yC + 2 yE = constant, vC + 2vE = 0, and aC + 2aE = 0
aC = −2aE = −2(−2.4 ft/s 2 ) = +4.8 ft/s 2 ,
aC = 4.80 ft/s 2 
Velocity of elevator after 5 s.
vE = (vE )0 + aE t = 0 + (−2.4 ft/s 2 )(5 s) = −12 ft/s
( v E )5 = 12.00 ft/s 
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PROBLEM 11.49
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 = −6 m/s
vC/D = vC − vD = (−6 m/s) − (2 m/s) = −8 m/s
vC/D = 8.00 m/s 
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PROBLEM 11.50
Block B starts from rest and moves downward with a constant
acceleration. Knowing that after slider block A has moved 9 in. its
velocity is 6 ft/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 yB = constant
Then
v A + 3vB = 0
(1)
and
a A + 3aB = 0
(2)
(a)
Eq. (2): a A + 3aB = 0 and a B is constant and
positive  a A is constant and negative
Also, Eq. (1) and (vB )0 = 0  (v A )0 = 0
v A2 = 0 + 2a A [ x A − ( x A )0 ]
Then
When |Δ x A | = 0.4 m:
(6 ft/s) 2 = 2a A (9/12 ft)
a A = 24.0 ft/s 2
or

Then, substituting into Eq. (2):
−24 ft/s 2 + 3aB = 0
or
aB =
24
ft/s 2
3
a B = 8.00 ft/s 2 
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PROBLEM 11.50 (Continued)
(b)
We have
vB = 0 + a B t
At t = 2 s:
 24

vB = 
ft/s 2  (2 s)
 3

v B = 16.00 ft/s 
or
yB = ( yB )0 + 0 +
Also
At t = 2 s:
or
y B − ( y B )0 =

1
aB t 2
2
1  24

ft/s 2  (2 s) 2 
2  3

y B − (y B )0 = 16.00 ft 
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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
From the diagram
xB + ( xB − x A ) − 2 x A = constant
Then
2vB − 3v A = 0
(1)
and
2aB − 3a A = 0
(2)
Also, we have
vD + v A = 0
Then
(a)
− xD − x A = constant
Substituting into Eq. (1)
2(300 mm/s) − 3v A = 0
or
(b)
From the diagram
Then
Substituting
or
(3)
v A = 200 mm/s

vC = 600 mm/s

xB + ( xB − xC ) = constant
2vB − vC = 0
2(300 mm/s) − vC = 0
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PROBLEM 11.51 (Continued)
(c)
From the diagram
Then
Substituting
( xC − x A ) + ( xD − x A ) = constant
vC − 2v A + vD = 0
600 mm/s − 2(200 mm/s) + vD = 0
v D = 200 mm/s
or
(d)
We have

vC/A = vC − v A
= 600 mm/s − 200 mm/s
or
vC/A = 400 mm/s

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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
xB + ( xB − x A ) − 2 x A = constant
From the diagram
Then
2vB − 3v A = 0
(1)
and
2aB − 3a A = 0
(2)
(a)
First observe that if block A moves to the right, v A → and Eq. (1)  v B → . Then, using
Eq. (1) at t = 0
2(150 mm/s) − 3(v A )0 = 0
(v A )0 = 100 mm/s
or
Also, Eq. (2) and aB = constant  a A = constant
v A2 = (v A )02 + 2a A [ x A − ( x A )0 ]
Then
When x A − ( x A )0 = 240 mm:
(60 mm/s) 2 = (100 mm/s) 2 + 2a A (240 mm)
or
or
aA = −
40
mm/s 2
3
a A = 13.33 mm/s 2

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PROBLEM 11.52 (Continued)
Then, substituting into Eq. (2)
 40

2aB − 3  −
mm/s 2  = 0
 3

aB = −20 mm/s 2
or
(b)
a B = 20.0 mm/s 2

From the diagram, − xD − x A = constant
vD + v A = 0
Then
Substituting
aD + a A = 0
 40

aD +  −
mm/s 2  = 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 = ( xB ) 0 + ( vB ) 0 t +

v B = 70.0 mm/s

1
aB t 2
2
xB − ( xB )0 = (150 mm/s)(4 s)
+
or
a D = 13.33 mm/s 2
1
(−20.0 mm/s 2 )(4 s) 2
2
x B − (x B )0 = 440 mm

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PROBLEM 11.53
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 24 in./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 + yB + ( yB − y A ) = constant
Then
v A + 2 vB = 0
(1)
and
a A + 2aB = 0
(2)
(a)
Eq. (1) and (v A )0 = 0  (vB )0 = 0
Also, Eq. (2) and a A is constant and negative  a B is constant and
positive.
Then
Now
From Eq. (2)
So that
v A = 0 + a At
vB = 0 + a B t
vB/A = vB − v A = (aB − a A )t
1
aB = − a A
2
3
vB/A = − a At
2
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PROBLEM 11.53 (Continued)
At t = 8 s:
3
24 in./s = − a A (8 s)
2
a A = 2.00 in./s 2 
or
and then
1
aB = − (−2 in./s 2 )
2
a B = 1.000 in./s 2 
or
(b)
At t = 6 s:
vB = (1 in./s 2 )(6 s)
v B = 6.00 in./s 
or

Now
At t = 6 s:
or
yB = ( yB )0 + 0 +
y B − ( y B )0 =
1
aB t 2
2
1
(1 in./s 2 )(6 s) 2
2
y B − (y B )0 = 18.00 in. 
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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
2v L + v B = 0
vM = 100 mm/s
with
vB = −
vL =
vM
= −33.333 m/s
3
vB
= −16.667 mm/s
2
v L = 16.67 mm/s 
(a)
Velocity of load L.
(b)
Velocity of pulley B with respect to load L. vB/L = vB − vL = −33.333 − (−16.667) = −16.667
v B/L = 16.67 mm/s 
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PROBLEM 11.55
Block C starts from rest at t = 0 and moves downward with a constant
acceleration of 4 in./s2. Knowing that block B has a constant velocity of 3 in./s
upward, determine (a) the time when the velocity of block A is zero, (b) the
time when the velocity of block A is equal to the velocity of block D, (c) the
change in position of block A after 5 s.
SOLUTION
From the diagram:
Cord 1:
2 y A + 2 y B + yC = constant
Then
2v A + 2vB + vC = 0
and
2a A + 2aB + aC = 0
Cord 2:
(1)
( y D − y A ) + ( y D − y B ) = constant
Then
2vD − v A − v B = 0
and
2aD − a A − aB = 0
(2)
Use units of inches and seconds.
Motion of block C:
vC = vC 0 + aC t
= 0 + 4t
where aC = −4 in./s 2
Motion of block B:
vB = −3 in./s;
Motion of block A:
From (1) and (2),
aB = 0
1
1
v A = −vB − vC = 3 − (4t ) = 3 − 2t in./s
2
2
1
1
a A = −aB − aC = 0 − (4) = −2 in./s 2
2
2
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PROBLEM 11.55 (Continued)
(a)
Time when vB is zero.
3 − 2t = 0
Motion of block D:
From (3),
vD =
(b)
1
1
1
1
v A + vB = (3 − 2t ) − (3) = −1t
2
2
2
2
Time when vA is equal to v0.
3 − 2t = −t
(c)
t = 1.500 s 
t = 3.00 s 
Change in position of block A (t = 5 s).
1
a At 2
2
1
= (3)(5) + (−2)(5)2 = −10 in.
2
Δy A = ( v A ) 0 t +
Change in position = 10.00 in. 
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PROBLEM 11.56
Block A starts from rest at t = 0 and moves downward with a constant
acceleration of 6 in./s2. Knowing that block B moves up with a constant
velocity of 3 in./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 inches and seconds.
Motion of block A:
v A = a At + 6t
Δy A =
Motion of block B:
1
1
a At 2 = (6)t 2 = 3t 2
2
2
v B = 3 in./s
vB = −3 in./s
ΔyB = vB t = −3t
Motion of block C:
From (1),
vC = −v A − 2vB = −6t − 2(−3) = 6 − 6t
ΔyC =
(a)
Time when vC is zero.
(b)
Corresponding position.

t
0
vC dt = 6t − 3t 2
6 − 6t = 0
ΔyC = (6)(1) − (3)(1)2 = 3 in.
t = 1.000 s 
ΔyC = 3.00 in. 
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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 yB + xC = constant
Then
3v A + 4vB + vC = 0
(1)
and
3a A + 4aB + aC = 0
(2)
(vB ) = 0,
Given:
a A = constent
(aC ) = 75 mm/s 2
At t = 2 s,
v B = 480 mm/s
vC = 280 mm/s
(a)
Eq. (2) and a A = constant and aC = constant  aB = constant
vB = 0 + a B t
Then
At t = 2 s:
480 mm/s = aB (2 s)
aB = 240 mm/s 2

or
a B = 240 mm/s 2 
or
a A = 345 mm/s 2 
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
( vC )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:
xC = ( xC )0 + (vC )0 t +
or
1
aC t 2
2
xC − ( xC )0 = (130 mm/s)(3 s) +
= 728 mm
(v A )0 = 43.3 mm/s 
1
(75 mm/s 2 )(3 s) 2
2
or
xC − (xC )0 = 728 mm

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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 of A and C, (c) the change in
position of block A after 5 s.
SOLUTION
From the diagram
3 y A + 4 yB + xC = constant
Then
3v A + 4vB + vC = 0
(1)
and
3a A + 4aB + aC = 0
(2)
v B = 20 mm/s ;
Given:
( v A )0 = 30 mm/s
(a)
Substituting into Eq. (1) at t = 0
3(−30 mm/s) + 4(20 mm/s) + (vC )0 = 0
(vC )0 = 10 mm/s
(b)
We have
At t = 3 s:
( vC )0 = 10.00 mm/s
or
xC = ( xC )0 + (vC )0 t +
1
aC t 2
2
57 mm = (10 mm/s)(3 s) +
1
aC (3 s)2
2
aC = 6 mm/s 2

Now

or
aC = 6.00 mm/s 2

v B = constant → aB = 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:
y A = ( y A )0 + (v A )0 t +
a A = 2.00 mm/s 2 
or
1
a At 2
2
y A − ( y A )0 = (−30 mm/s)(5 s) +
1
(−2 mm/s 2 )(5 s)2
2
= −175 mm
or
y A − (y A )0 = 175.0 mm 
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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
2 y A + 2 yB + yC = constant
Cable 1:
Then
2v A + 2vB + vC = 0
(1)
and
2a A + 2aB + aC = 0
(2)
Cable 2:
( yD − y A ) + ( yD − yB ) = constant
Then
and
− v A − v B + 2 vD = 0
(3)
− a A − aB + 2aD = 0
(4)
Given: At t = 0, v = 0; all accelerations constant;
aC/B = 60 mm/s 2 , aD /A = 110 mm/s 2
(a)
We have
aC/B = aC − aB = −60 or aB = aC + 60
and
aD/A = aD − a A = 110 or a A = aD − 110
Substituting into Eqs. (2) and (4)
Eq. (2):
or
Eq. (4):
or
2(aD − 110) + 2( aC + 60) + aC = 0
3aC + 2aD = 100
(5)
−(aD − 110) − ( aC + 60) + 2aD = 0
− aC + aD = −50
(6)
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PROBLEM 11.59 (Continued)
Solving Eqs. (5) and (6) for aC and aD
aC = 40 mm/s 2
aD = −10 mm/s 2
Now
vC = 0 + aC t
At t = 3 s:
vC = (40 mm/s 2 )(3 s)
vC = 120.0 mm/s 
or
(b)
We have
At t = 5 s:
or
yD = ( yD )0 + (0)t +
yD − ( yD )0 =
1
aD t 2
2
1
( −10 mm/s 2 )(5 s) 2
2
y D − (y D )0 = 125.0 mm 
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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 yB + yC = constant
Then
2v A + 2vB + vC = 0
(1)
and
2a A + 2aB + aC = 0
(2)
Cable 2: ( yD − y A ) + ( yD − yB ) = constant
Then
− v A − vB − 2 vD = 0
(3)
and
− a A − aB + 2aD = 0
(4)
Given: At
t =0
v=0
( y A )0 = ( yB )0 = ( yC )0
All accelerations constant.
At t = 2 s
yC /A = 280 mm
When
vB /A = 80 mm/s
y A − ( y A )0 = 160 mm
yB − ( yB )0 = 320 mm
aB > 10 mm/s 2
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PROBLEM 11.60* (Continued)
(a)
We have
y A = ( y A )0 + (0)t +
1
a At 2
2
and
yC = ( yC )0 + (0)t +
1
aC t 2
2
Then
yC/A = yC − y A =
1
(aC − a A )t 2
2
At t = 2 s, yC/A = −280 mm:
−280 mm =
or
1
(aC − a A )(2 s) 2
2
aC = a A − 140
(5)
Substituting into Eq. (2)
2a A + 2aB + (a A − 140) = 0
or
1
a A = (140 − 2aB )
3
Now
vB = 0 + a B t
(6)
v A = 0 + a At
vB/A = vB − v A = (aB − a A )t
Also
When

yB = ( yB )0 + (0)t +
v B /A = 80 mm/s :
1
aB t 2
2
80 = (aB − a A )t
Δy A = 160 mm :
160 =
1
a At 2
2
ΔyB = 320 mm :
320 =
1
aB t 2
2
(7)
1
(aB − a A )t 2
2
Then
160 =
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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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 + aC t
When vC = −600 mm/s:
or
Also
At t = 5 s:
or
−600 mm/s = ( −120 mm/s 2 )t
t =5s
yD = ( yD )0 + (0)t +
yD − ( yD )0 =
1
aD t 2
2
1
(30 mm/s 2 )(5 s)2
2
y D − (y D )0 = 375 mm 
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PROBLEM 11.61
A particle moves in a straight line with the acceleration shown in the figure.
Knowing that it starts from the origin with v0 = −14 ft/s, plot the v−t and x−t
curves for 0 < t < 15 s and determine (a) the maximum value of the velocity
of the particle, (b) the maximum value of its position coordinate.
SOLUTION
Change in v = area under a−t curve.
v0 = −14 ft/s
t = 0 to t = 2 s:
v2 − v0 = (3 ft/s 2 )(2 s) = +6 ft/s
v2 = −8 ft/s
t = 2 s to t = 5 s:
v5 − v2 = (8 ft/s2 )(3 s) = +24 ft/s
v5 = +16 ft/s
t = 5 s to t = 8 s:
v8 − v5 = (3 ft/s 2 )(3 s) = +9 ft/s
v8 = +25 ft/s
t = 8 s to t = 10 s:
v10 − v8 = (−5 ft/s2 )(2 s) = −10 ft/s
v10 = +15 ft/s
t = 10 s to t = 15 s:
v15 − v10 = (−5 ft/s2 )(5 s) = −25 ft/s
v15 = −10 ft/s
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PROBLEM 11.61 (Continued)
Plot v−t curve. Then by similar triangles Δ’s find t for v = 0.
Change in x = area under v−t curve
x0 = 0
t = 0 to t = 2 s:
1
x2 − x0 = (−14 − 8)(2) = −22 ft
2
x2 = −22 ft
t = 2 s to t = 3 s:
1
x3 − x2 = (−8)(1) = −4 ft
2
x8 = −26 ft
t = 3 s to t = 5 s:
1
x5 − x3 = (+16)(2) = +16 ft
2
x5 = −10 ft
t = 5 s to t = 8 s:
1
x8 − x5 = (+16 + 25)(3) = +61.5 ft
2
x8 = +51.6 ft
t = 8 s to t = 10 s:
1
x10 − x8 = (+25 + 15)(2) = + 40 ft
2
x10 = +91.6 ft
t = 10 s to t = 13 s:
1
x13 − x10 = (+15)(3) = +22.5 ft
2
x13 = +114 ft
t = 13 s to t = 15 s:
1
x15 − x13 = (−10)(2) = −10 ft
2
x15 = +94 ft
(a)
Maximum velocity: When t = 8 s,
(b)
Maximum x: When t = 13 s,
vm = 25.0 ft/s 
xm = 114.0 ft 
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PROBLEM 11.62
For the particle and motion of Problem 11.61, plot the v−t and x−t curves for
0 < t < 15 s and determine the velocity of the particle, its position, and the
total distance traveled after 10 s.
PROBLEM 11.61 A particle moves in a straight line with the acceleration
shown in the figure. Knowing that it starts from the origin with
v0 = −14 ft/s, plot the v−t and x−t curves for 0 < t < 15 s and determine
(a) the maximum value of the velocity of the particle, (b) the maximum
value of its position coordinate.
SOLUTION
Change in v = area under a−t curve.
v0 = −14 ft/s
t = 0 to t = 2 s:
v2 − v0 = (3 ft/s 2 )(2 s) = +6 ft/s
v2 = −8 ft/s
t = 2 s to t = 5 s:
v5 − v2 = (8 ft/s2 )(3 s) = +24 ft/s
v5 = +16 ft/s
t = 5 s to t = 8 s:
v8 − v5 = (3 ft/s 2 )(3 s) = +9 ft/s
v8 = +25 ft/s
t = 8 s to t = 10 s:
v10 − v8 = (−5 ft/s2 )(2 s) = −10 ft/s
v10 = +15 ft/s
t = 10 s to t = 15 s:
v15 − v10 = (−5 ft/s2 )(5 s) = −25 ft/s
v15 = −10 ft/s
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PROBLEM 11.62 (Continued)
Plot v−t curve. Then by similar triangles Δ’s find t for v = 0.
x0 = 0
Change in x = area under v−t curve
t = 0 to t = 2 s:
1
x2 − x0 = (−14 − 8)(2) = −22 ft
2
x2 = −22 ft
t = 2 s to t = 3 s:
1
x3 − x2 = (−8)(1) = −4 ft
2
x8 = −26 ft
t = 3 s to t = 5 s:
1
x5 − x3 = (+16)(2) = +16 ft
2
x5 = −10 ft
t = 5 s to t = 8 s:
1
x8 − x5 = (+16 + 25)(3) = +61.5 ft
2
x8 = +51.6 ft
t = 8 s to t = 10 s:
1
x10 − x8 = (+25 + 15)(2) = + 40 ft
2
x10 = +91.6 ft
t = 10 s to t = 13 s:
1
x13 − x10 = (+15)(3) = +22.5 ft
2
x13 = +114 ft
t = 13 s to t = 15 s:
1
x15 − x13 = (−10)(2) = −10 ft
2
x15 = +94 ft
when t = 10 s:
v10 = +15 ft/s 
x10 = +91.5 ft/s 
Distance traveled: t = 0 to t = 105
t = 0 to t = 3 s:
Distance traveled = 26 ft
t = 3 s to t = 10 s
Distance traveled = 26 ft + 91.5 ft = 117.5 ft
Total distance traveled = 26 + 117.5 = 143.5 ft 
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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  a = 0
−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  a = 0
t = 41 s to t = 46 s:
a=
t = 46 s:
−5 − (−20)
= 3 m/s 2
46 − 41
v = constant  a = 0
x2 = x1 + (area under v −t curve from t1 to t2 )
At t = 10 s:
x10 = −540 + 10(60) + 60 m
Next, find time at which v = 0. Using similar triangles
tv = 0 − 10
60
At
t = 22 s:
t = 26 s:
t = 41 s:
t = 46 s:
t = 50 s:
=
26 − 10
80
or
tv = 0 = 22 s
1
x22 = 60 + (12)(60) = 420 m
2
1
x26 = 420 − (4)(20) = 380 m
2
x41 = 380 − 15(20) = 80 m
 20 + 5 
x46 = 80 − 5 
 = 17.5 m
 2 
x50 = 17.5 − 4(5) = −2.5 m
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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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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
v = constant  a = 0
From t = 0 to t = 10 s:
−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  a = 0
t = 41 s to t = 46 s:
a=
−5 − (−20)
= 3 m/s 2
46 − 41
v = constant  a = 0
t = 46 s:
x2 = x1 + (area under v −t curve from t1 to t2 )
At t = 10 s:
x10 = −540 + 10(60) = 60 m
Next, find time at which v = 0. Using similar triangles
tv = 0 − 10
60
At
t = 22 s:
t = 26 s:
t = 41 s:
t = 46 s:
t = 50 s:
=
26 − 10
80
or
tv = 0 = 22 s
1
x22 = 60 + (12)(60) = 420 m
2
1
x26 = 420 − (4)(20) = 380 m
2
x41 = 380 − 15(20) = 80 m
 20 + 5 
x46 = 80 − 5 
 = 17.5 m
 2 
x50 = 17.5 − 4(5) = −2.5 m
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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
100 = 420 −
1
(22 − t1 )(v1 )
2
(22 − t1 )(v1 ) = 640
Using similar triangles
v1
60
=
22 − t1 12
Then
or
v1 = 5(22 − t1 )
(22 − t1 )[5(22 − t1 )] = 640
t1 = 10.69 s and t1 = 33.3 s
10 s < t1 < 22 s 
We have
t1 = 10.69 s 
Between 26 s and 41 s:
Using similar triangles
41 − t2
15
=
20
300
t2 = 40.0 s 
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PROBLEM 11.65
During a finishing operation the bed of an industrial planer moves alternately 30 in. to the right and 30 in. to
the left. The velocity of the bed is limited to a maximum value of 6 in./s to the right and 12 in./s to the left; the
acceleration is successively equal to 6 in./s2 to the right, zero 6 in./s2 to the left, zero, etc. Determine the time
required for the bed to complete a full cycle, and draw the v−t and x−t curves.
SOLUTION
We choose positive to the right, thus the range of permissible velocities is −12 in./s < v < 6 in./s since
acceleration is −6 in./s 2 , 0, or + 6 in./s 2 . The slope the v−t curve must also be −6 in./s2, 0, or +6 in./s2.
Planer moves = 30 in. to right: +30 in. = 3 + 6t1 + 3
t1 = 4.00 s
Planer moves = 30 in. to left: −30 in. = −12 − 12t2 + 12
t2 = 0.50 s
Total time = 1 s + 4 s + 1 s + 2 s + 0.5 s + 2 s = 10.5 s
ttotal = 10.50 s 
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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 (Continued)
(a)
Now Δ x = area under v −t curve for given time interval
Then
 55.555 + 13.888 
(586 − 600) m = −t1 
 m/s
2


t1 = 0.4032 s
(30 − 586) m = −t2 (13.888 m/s)
t2 = 40.0346 s
1
(0 − 30) m = − (t3 )(13.888 m/s)
2
t3 = 4.3203 s
ttotal = 44.8 s 
ttotal = (0.4032 + 40.0346 + 4.3203) s
(b)
We have
ainitial =
=
Δ 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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PROBLEM 11.67
A commuter train traveling at 40 mi/h is 3 mi
from a station. The train then decelerates so
that its speed is 20 mi/h when it is 0.5 mi from
the station. Knowing that the train arrives at
the station 7.5 min after beginning to decelerate
and assuming constant decelerations, determine
(a) the time required for the train to travel the
first 2.5 mi, (b) the speed of the train as it
arrives at the station, (c) the final constant
deceleration of the train.
SOLUTION
Given: At t = 0, v = 40 mi/h, x = 0; when x = 2.5 mi, v = 20 mi/h;
at t = 7.5 min, x = 3 mi; constant decelerations.
The v −t curve is first drawn as shown.
(a)
A1 = 2.5 mi
We have
1h
 40 + 20 
(t1 min) 
mi/h ×
= 2.5 mi

60 min
 2 
t1 = 5.00 min 
(b)
A2 = 0.5 mi
We have
 20 + v2
(7.5 − 5) min × 
 2
1h

 mi/h × 60 min = 0.5 mi

v2 = 4.00 mi/h 
(c)
We have
afinal = a12
=
(4 − 20) mi/h 5280 ft 1 min
1h
×
×
×
(7.5 − 5) min
mi
60 s 3600 s
afinal = −0.1564 ft/s 2 
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PROBLEM 11.68
A temperature sensor is attached to slider AB which moves back
and forth through 60 in. The maximum velocities of the slider
are 12 in./s to the right and 30 in./s to the left. When the slider is
moving to the right, it accelerates and decelerates at a constant
rate of 6 in./s2; when moving to the left, the slider accelerates
and decelerates at a constant rate of 20 in./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 =
vright
aright
=
12 in./s
=2s
6 in./s 2
vleft 30 in./s
=
aleft 20 in./s
= 1.5 s
td =
A1 = 60 in.
Now
[(t1 − 2) s](12 in./s) = 60 in.
or
t1 = 7 s
or
A2 = 60 in.
and
or
{[(t2 − 7) − 1.5] s}(30 in./s) = 60 in.
or
t2 = 10.5 s
tcycle = t2
Now
tcycle = 10.5 s 
We have xii = xi + (area under v −t curve from ti to tii )
At
t = 5 s:
1
(2) (12) = 12 in.
2
x5 = 12 + (5 − 2)(12)
t = 7 s:
= 48 in.
x7 = 60 in.
t = 2 s:
t = 8.5 s:
t = 9 s:
t = 10.5 s:
x2 =
1
x8.5 = 60 − (1.5)(30)
2
= 37.5 in.
x9 = 37.5 − (0.5)(30)
= 22.5 in.
x10.5 = 0
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PROBLEM 11.69
In a water-tank test involving the launching of a small model boat,
the model’s initial horizontal velocity is 6 m/s, and its horizontal
acceleration varies linearly from −12 m/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 and the position of the model at t = 1.4 s.
SOLUTION
Given:
v0 = 6 m/s; for 0 < t < t1 ,
for
t1 < t < 1.4 s a = −2 m/s 2 ;
at
t = 0 a = −12 m/s 2 ;
at t = t1
a = −2 m/s 2 , v = 1.8 m/s 2
The a −t and v −t curves are first drawn as shown. The time axis is not
drawn to scale.
(a)
We have
vt1 = v0 + A1
 12 + 2 
1.8 m/s = 6 m/s − (t1 s) 
m/s2

 2 
t1 = 0.6 s 
(b)
We have
v1.4 = vt1 + A2
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,
for 0 < t < t1:
Now
a=
−2 − (−12)
50
t − 12 = t − 12
0.6
3
dv
50
= a = t − 12
dt
3
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PROBLEM 11.69 (Continued)
At t = 0, v = 6 m/s:
or

v
6
dv =

t
50

t − 12  dt

0 3

v=6+
25 2
t − 12t
3
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
25 

= 6t − 6t 2 + t 3  = 2.04 m
9 0

Also
 1.8 + 0.2 
A4 = (1.4 − 0.6) 
 = 0.8 m
2


Then
x1.4 = (2.04 + 0.8) m
or
x1.4 = 2.84 m 
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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” 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:
= Δx = 6 m
(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 
Δt = 8 s
Δ x = (60 m/s)(14 s − 6 s) − (shaded area)
= (60 m/s)(8 s) − 6 m = 480 m − 6 m = 474 m
vaverage =
Δ x 474 m
=
Δt
8s
vaverage = 59.25 m/s 
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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 vB in 5 s with constant acceleration and maintains
that velocity until she reaches the half-way point with a split time
of 25.2 s. Both runners then run the second half of the race with the
same constant deceleration of 0.1 m/s 2. Determine (a) the race times
for both runners, (b) the position of the winner relative to the loser
when the winner reaches the finish line.
SOLUTION
Sketch v −t curves for first 200 m.
Runner A:
t1 = 4 s, t2 = 25 − 4 = 21 s
A1 =
1
(4)(v A )max = 2(vA ) max
2
A2 = 21(v A )max
A1 + A2 = Δ 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(vB ) max
A1 + A2 = Δ x = 200 m
22.7(vB ) max = 200
Sketch v −t curve for second 200 m.
A3 = vmaxt3 −
t3 =
Runner A:
or
(vB ) max = 8.8106 m/s
Δv = | a |t3 = 0.1t3
1
Δvt3 = 200
2
or
0.05t32 − vmaxt3 + 200 = 0
(
vmax ± (vmax ) 2 − (4)(0.05)(200)
= 10 vmax ± (vmax )2 − 40
(2)(0.05)
(vmax ) A = 8.6957,
Reject the larger root. Then total time
(t3 ) A = 146.64 s
and
)
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
Reject the larger root. Then total time
and
26.765 s
t B = 25.2 + 26.765 = 51.965 s
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 Δ x
Δ x = vave (Δt ) =
1
(5.999 + 5.968)(52.279 − 51.965)
2
Δ x = 1.879 m 
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PROBLEM 11.72
A car and a truck are both traveling at the constant
speed of 35 mi/h; the car is 40 ft behind the truck.
The driver of the car wants to pass the truck, i.e.,
he wishes to place his car at B, 40 ft in front of the
truck, and then resume the speed of 35 mi/h. The
maximum acceleration of the car is 5 ft/s2 and
the maximum deceleration obtained by applying
the brakes is 20 ft/s 2 . What is the shortest time in
which the driver of the car can complete the
passing operation if he does not at any time exceed
a speed of 50 mi/h? Draw the v −t curve.
SOLUTION
Relative to truck, car must move a distance:
Allowable increase in speed:
Δ x = 16 + 40 + 50 + 40 = 146 ft
Δvm = 50 − 35 = 15 mi/h = 22 ft/s
Acceleration Phase:
t1 = 22/5 = 4.4 s
A1 =
1
(22)(4.4) = 48.4 ft
2
Deceleration Phase:
t3 = 22/20 = 1.1 s
A3 =
1
(22)(1.1) = 12.1 ft
2
But: Δ x = A1 + A2 + A3 :
146 ft = 48.4 + (22)t2 + 12.1
ttotal = t1 + t2 + t3 = 4.4 s + 3.89 s + 1.1 s = 9.39 s
t2 = 3.89 s
t B = 9.39 s 
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PROBLEM 11.73
Solve Problem 11.72, assuming that the driver of the car does not pay any attention to the speed limit while
passing and concentrates on reaching position B and resuming a speed of 35 mi/h in the shortest possible time.
What is the maximum speed reached? Draw the v −t curve.
PROBLEM 11.72 A car and a truck are both traveling at the constant speed of 35 mi/h; the car is 40 ft
behind the truck. The driver of the car wants to pass the truck, i.e., he wishes to place his car at B, 40 ft in
front of the truck, and then resume the speed of 35 mi/h. The maximum acceleration of the car is 5 ft/s2 and
the maximum deceleration obtained by applying the brakes is 20 ft/s 2 . What is the shortest time in which the
driver of the car can complete the passing operation if he does not at any time exceed a speed of 50 mi/h?
Draw the v −t curve.
SOLUTION
Relative to truck, car must move a distance:
Δ x = 16 + 40 + 50 + 40 = 146 ft
Δvm = 5t1 = 20t2 ;
Δ x = A1 + A2 :
t2 =
146 ft =
1
(Δvm )(t1 + t2 )
2
146 ft =
1
1 

(5t1 )  t1 + t1 
2
4 

t12 = 46.72
1
t1
4
t1 = 6.835 s
t2 =
1
t1 = 1.709
4
ttotal = t1 + t2 = 6.835 + 1.709
t B = 8.54 s 
Δvm = 5t1 = 5(6.835) = 34.18 ft/s = 23.3 mi/h
Speed vtotal = 35 mi/h, vm = 35 mi/h + 23.3 mi/h
vm = 58.3 mi/h 
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PROBLEM 11.74
Car A is traveling on a highway at a constant
speed (v A )0 = 60 mi/h, and is 380 ft from
the entrance of an access ramp when car B
enters the acceleration lane at that point at
a speed (vB )0 = 15 mi/h. Car B accelerates
uniformly and enters the main traffic lane
after traveling 200 ft in 5 s. It then continues
to accelerate at the same rate until it reaches
a speed of 60 mi/h, which it then maintains.
Determine the final distance between the
two cars.
SOLUTION
Given:
(v A )0 = 60 mi/h, (vB )0 = 1.5 mi/h; at t = 0,
( x A )0 = −380 ft, ( xB ) 0 = 0; at t = 5 s,
xB = 200 ft; for 15 mi/h < vB ≤ 60 mi/h,
aB = constant; for vB = 60 mi/h,
aB = 0
First note
60 mi/h = 88 ft/s
15 mi/h = 22 ft/s
The v −t curves of the two cars are then drawn as shown.
Using the coordinate system shown, we have
at t = 5 s, xB = 200 ft:
 22 +(vB )5 
(5 s) 
 ft/s = 200 ft
2


(vB )5 = 58 ft/s
or
Then, using similar triangles, we have
(88 − 22) ft/s (58 − 22) ft/s
=
( = aB )
t1
5s
or
t1 = 9.16 67 s
Finally, at t = t1


 22 + 88 
ft/s 
xB/A = xB − x A =  (9.1667 s) 

 2 


− [−380 ft + (9.1667 s) (88 ft/s)]
or
xB/A = 77.5 ft 
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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 vE = 0; For 0 < vE ≤ 7.8 m/s, aE = 1.2 m/s 2 ↑;
For vE = 7.8 m/s, aE = 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/s2 , while the slope of the curve for the ball is − g (−9.81 m/s 2 ).
The time t1 is the time when vE reaches 7.8 m/s.
Thus,
vE = (0) + aE 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,
or
or
vB = (vB )0 − g (t − 2)
0 = 20 m/s − (9.81 m/s 2 ) (t top − 2) s
ttop = 4.0387 s
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PROBLEM 11.75 (Continued)
Using the coordinate system shown, we have
0 < t < t1 :
1

yE = −12 m +  aE t 2  m
2

At t = ttop :
yB =
and
y E = −12 m +
At
and at t = t1 ,
1
(4.0387 − 2) s × (20 m/s)
2
= 20.387 m
1
(1.2 m/s 2 )(4.0387 s) 2
2
= −2.213 m
t = [2 + 2(4.0387 − 2)] s = 6.0774 s, yB = 0
yE = −12 m +
1
(6.5 s) (7.8 m/s) = 13.35 m
2
The ball hits the elevator ( yB = yE ) when ttop ≤ t ≤ t1.
For t ≥ ttop :
1

yB = 20.387 m −  g (t − ttop )2  m
2


Then,
when
yB = yE
1
20.387 m − (9.81 m/s 2 ) (t − 4.0387)2
2
1
= −12 m + (1.2 m/s 2 ) (t s)2
2
or
Solving
5.505t 2 − 39.6196t + 47.619 = 0
t = 1.525 s and t = 5.67 s
Since 1.525 s is less than 2 s,
t = 5.67 s 
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PROBLEM 11.76
Car A is traveling at 40 mi/h when it enters
a 30 mi/h speed zone. The driver of car A
decelerates at a rate of 16 ft/s2 until
reaching a speed of 30 mi/h, which she
then maintains. When car B, which was
initially 60 ft behind car A and traveling at
a constant speed of 45 mi/h, enters the
speed zone, its driver decelerates at a rate
of 20 ft/s2 until reaching a speed of
28 mi/h. Knowing that the driver of car B
maintains a speed of 28 mi/h, determine
(a) the closest that car B comes to car A,
(b) the time at which car A is 70 ft in front
of car B.
SOLUTION
(v A )0 = 40 mi/h; For 30 mi/h < v A ≤ 40 mi/h, a A = −16 ft/s 2 ; For v A = 30 mi/h, a A = 0;
Given:
( x A /B )0 = 60 ft; (vB )0 = 45 mi/h;
When xB = 0, aB = −20 ft/s 2 ;
For vB = 28 mi/h, aB = 0
First note
40 mi/h = 58.667 ft/s 30 mi/h = 44 ft/s
45 mi/h = 66 ft/s
28 mi/h = 41.067 ft/s
At t = 0
The v −t curves of the two cars are as shown.
At
t = 0:
Car A enters the speed zone.
t = (tB )1:
Car B enters the speed zone.
t = tA:
Car A reaches its final speed.
t = tmin :
v A = vB
t = (t B )2 :
Car B reaches its final speed.
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PROBLEM 11.76 (Continued)
(a)
(v A )final − (v A )0
tA
aA =
We have
(44 − 58.667) ft/s
tA
−16 ft/s 2 =
or
t A = 0.91669 s
or
Also
60 ft = (tB )1 (vB )0
or
60 ft = (tB )1 (66 ft/s)
aB =
and
−20 ft/s 2 =
or
or
(t B )1 = 0.90909 s
(vB )final − (vB )0
(t B ) 2 − (t B )1
(41.067 − 66) ft/s
[(t B )2 − 0.90909] s
Car B will continue to overtake car A while vB > v A . Therefore, ( x A/B ) min will occur when v A = vB ,
which occurs for
(tB )1 < tmin < (tB ) 2
For this time interval
v A = 44 ft/s
vB = (vB )0 + aB [t − (tB )1 ]
Then
at t = tmin :
44 ft/s = 66 ft/s + (−20 ft/s2 )(tmin − 0.90909) s
tmin = 2.00909 s
or
Finally ( x A/B )min = ( x A )tmin − ( xB )tmin
  (v ) + (v A )final 

= t A  A 0
+ (tmin − t A )(v A )final 

2

 


 (v ) + (v A )final  
− ( xB )0 + (t B )1 (vB )0 + [tmin − (t B )1 ]  B 0

2





 58.667 + 44 
= (0.91669 s) 
ft/s + (2.00909 − 0.91669) s × (44 ft/s) 

2





 66 + 44  
−  −60 ft + (0.90909 s)(66 ft/s) + (2.00909 − 0.90909) s × 
 ft/s 
 2  

= (47.057 + 48.066) ft − (−60 + 60.000 + 60.500) ft
= 34.623 ft
or ( x A/B ) min = 34.6 ft 
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PROBLEM 11.76 (Continued)
(b)
Since ( x A/B ) ≤ 60 ft for t ≤ tmin , it follows that x A/B = 70 ft for t > (tB ) 2
[Note (tB ) 2  tmin ]. Then, for t > (tB ) 2
x A/B = ( x A/B )min + [(t − tmin )(vA )final ]


+ (vB )final 
 (v )
− [(t B )2 − (tmin )]  A final
+ [t − (t B )2 ](vB )final 

2




or
70 ft = 34.623 ft + [(t − 2.00909) s × (44 ft/s)]


 44 + 41.06 
− (2.15574 − 2.00909) s × 
 ft/s + (t − 2.15574) s × (41.067) ft/s 
2




or
t = 14.14 s 
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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 = 0.8 ft when t = 0.4 s, (a) construct
the v −t curve for 0 ≤ t ≤ 0.4 s, (b) determine the position of the
part at t = 0.3 s and t = 0.2 s.
SOLUTION
Divide the area of the a −t curve into the four areas A1, A2 , A3 and A4.
2
(8)(0.2) = 1.0667 ft/s
3
A2 = (16)(0.2) = 3.2 ft/s
A1 =
1
(16 + 8)(0.1) = 1.2 ft/s
2
1
A4 = (8)(0.1) = 0.4 ft/s
2
A3 =
Velocities: v0 = 0
v0.2 = v0 + A1 + A2
v0.2 = 4.27 ft/s 
v0.3 = v0.2 + A3
v0.3 = 5.47 ft/s 
v0.4 = v0.3 + A4
v0.4 = 5.87 ft/s 
Sketch the v −t curve and divide its area into A5 , A6 , and A7 as shown.
0.8
0.4
 x dx = 0.8 − x =  t vdt
At t = 0.3 s,
With A5 =
With A5 + A6 =
0.4
x = 0.8 −  t vdt
x0.3 = 0.8 − A5 − (5.47)(0.1)
2
(0.4)(0.1) = 0.0267 ft,
3
At t = 0.2 s,
and
or
x0.3 = 0.227 ft 
x0.2 = 0.8 − ( A5 + A6 ) − A7
2
(1.6)(0.2) = 0.2133 ft,
3
A7 = (4.27)(0.2) = 0.8533 ft
x0.2 = 0.8 − 0.2133 − 0.8533
x0.2 = −0.267 ft 
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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, (c) the average velocity of
the car during the interval 1 s ≤ t ≤ t1.
SOLUTION
Given:
At
t = 0, x = 0, v = 54 km/h;
For t = t1 ,
v = 54 km/h
First note
(a)
54 km/h = 15 m/s
vb = 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 :
15 = 1.5 +
1
(t1 − 4.5)(2)
2
or
(b)
t1 = 18.00 s 
Using the above values of the velocities, the v −t curve is drawn as shown.
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PROBLEM 11.78 (Continued)
Now
x at t = 18 s
x18 = 0 + Σ (area under the v −t curve from t = 0 to t = 18 s)
 15 + 9 
= (1 s)(15 m/s) + (1 s) 
 m/s
 2 
1


+ (2.5 s)(1.5 m/s)+ (2.5 s)(7.5 m/s) 
3


2


+ (13.5 s)(1.5 m/s) + (13.5 s)(13.5 m/s) 
3


= [15 + 12 + (3.75 + 6.25) + (20.25 + 121.50)] m
= 178.75 m
(c)
First note
or
x18 = 178.8 m 
x1 = 15 m
x18 = 178.75 m
Now
vave =
Δ x (178.75 − 15) m
=
= 9.6324 m/s
Δt
(18 − 1) s
or
vave = 34.7 km/h 
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PROBLEM 11.79
An airport shuttle train travels between two terminals that are 1.6 mi apart. To maintain passenger comfort,
the acceleration of the train is limited to ±4 ft/s2, and the jerk, or rate of change of acceleration, is limited to
±0.8 ft/s2 per second. If the shuttle has a maximum speed of 20 mi/h, determine (a) the shortest time for the
shuttle to travel between the two terminals, (b) the corresponding average velocity of the shuttle.
SOLUTION
xmax = 1.6 mi; | amax | = 4 ft/s 2
Given:
 da 
= 0.8 ft/s2 /s; vmax = 20 mi/h
 
 dt max
First note
(a)
20 mi/h = 29.333 ft/s
1.6 mi = 8448 ft
To obtain tmin , the train must accelerate and decelerate at the maximum rate to maximize the
time for which v = vmax . The time Δ t required for the train to have an acceleration of 4 ft/s2 is found
from
a
 da 
= max
 dt 
Δt
 max
or
or
4 ft/s2
0.8 ft/s2 /s
Δt = 5 s
Δt =
Now,
da


 since dt = constant 


after 5 s, the speed of the train is
v5 =
1
(Δt )(amax )
2
or
v5 =
1
(5 s)(4 ft/s 2 ) = 10 ft/s
2
Then, since v5 < vmax , the train will continue to accelerate at 4 ft/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 0.8 ft/s2/s.
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PROBLEM 11.79 (Continued)
Now
at t = (10 + Δt1 ) s, v = vmax :
1

2  (5 s)(4 ft/s2 )  + (Δt1 )(4 ft/s2 ) = 29.333 ft/s
2

or
Δt1 = 2.3333 s
Then
at t = 5 s:
t = 7.3333 s:
1
(5)(4) = 10 ft/s
2
v = 10 + (2.3333)(4) = 19.3332 ft/s
v =0+
1
t = 12.3333 s: v = 19.3332 + (5)(4) = 29.3332 ft/s
2
Using symmetry, the v −t curve is then drawn as shown.
Noting that A1 = A2 = A3 = A4 and that the area under the v −t curve is equal to xmax , we have

 10 + 19.3332  
2 (2.3333 s) 
 ft/s 
2

 

+ (10 + Δt2 ) s × (29.3332 ft/s) = 8448 ft
or
Δt2 = 275.67 s
Then
tmin = 4(5 s) + 2(2.3333 s) + 275.67 s
= 300.34 s
tmin = 5.01 min 
or
(b)
We have
vave =
Δx
1.6 mi 3600 s
=
×
Δt 300.34 s
1h
or
vave = 19.18 mi/h 
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PROBLEM 11.80
During a manufacturing process, a conveyor belt starts from rest and travels a total of 1.2 ft before
temporarily coming to rest. Knowing that the jerk, or rate of change of acceleration, is limited to ±4.8 ft/s2 per
second, determine (a) the shortest time required for the belt to move 1.2 ft, (b) the maximum and average
values of the velocity of the belt during that time.
SOLUTION
Given:
(a)
At
t = 0, x = 0, v = 0; xmax = 1.2 ft;
when
 da 
x = xmax , v = 0;  
= 4.8 ft/s 2
 dt max
Observing that vmax must occur at t = 12 tmin , the a −t curve must have the shape shown. Note that the
magnitude of the slope of each portion of the curve is 4.8 ft/s2/s.
We have
at t = Δt :
t = 2Δt :
1
1
v = 0 + (Δt )(amax ) = amax Δt
2
2
vmax =
1
1
amax Δt + ( Δt )( amax ) = amax Δt
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
(2Δt )(vmax ) = xmax
vmax = amax Δt  2amax Δt 2 = xmax
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PROBLEM 11.80 (Continued)
Now
amax
= 4.8 ft/s2 /s so that
Δt
2(4.8Δt ft/s3 )Δt 2 = 1.2 ft
or
Then
Δt = 0.5 s
tmin = 4Δt
tmin = 2.00 s 
or
(b)
We have
vmax = amax Δt
= (4.8 ft/s 2 /s × Δt)Δt
= 4.8 ft/s 2 /s × (0.5 s) 2
vmax = 1.2 ft/s 
or
Also
vave =
Δx
1.2 ft
=
Δttotal 2.00 s
or
vave = 0.6 ft/s 
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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 Δt = 0.25 s. The area
(Δt)(aave) of each rectangle is approximately equal to the change in velocity Δv for the specified
interval of time. Thus,
Δv ≅ aave Δt
where the values of aave and Δv are given in columns 1 and 2, respectively, of the following table.
2.
v(2) = v0 +
Now
and approximating the area

2
0

2
0
a dt = 0
a dt under the a −t curve by Σaave Δt ≈ ΣΔv, the initial velocity is then
equal to
v0 = −ΣΔv
Finally, using
v2 = v1 + Δv12
where Δv12 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
(Δt )(vave ) of each rectangle is approximately equal to the change in position Δ x for the specified
interval of time. Thus
Δ x ≈ vave Δt
where vave and Δ 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 + Δ x12
where Δ 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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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 Δt = 2 s. The
area (Δt )(aave ) of each rectangle is approximately equal to the change in velocity Δv for the
specified interval of time. Thus,
Δv ≅ aave Δt
where the values of aave and Δv are given in columns 1 and 2, respectively, of the following table.
2.
Noting that v0 = 0 and that
v2 = v1 + Δv12
where Δv12 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 Δt = 2 s. The
area (Δt )(vave ) of each rectangle is approximately equal to the change in position Δx for the
specified interval of time.
Thus,
Δx ≅ vave Δt
where vave and Δx are given in columns 4 and 5, respectively, of the table.
4.
With x0 = 0 and noting that
x2 = x1 + Δ x12
where Δ 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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PROBLEM 11.82 (Continued)
(a)
At t = 8 s, v = 32.58 m/s
(b)
At t = 20 s
or
v = 117.3 km/h 
x = 660 m 
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PROBLEM 11.83
A training airplane has a velocity of 126 ft/s when it lands on
an aircraft carrier. As the arresting gear of the carrier
brings the airplane to rest, the velocity and the acceleration
of the airplane are recorded; the results are shown (solid
curve) in the figure. Determine by approximate means (a)
the time required for the airplane to come to rest, (b) the
distance traveled in that time.
SOLUTION
Given: a −v curve:
v0 = 126 ft/s
The given curve is approximated by a series of uniformly accelerated motions (the horizontal dashed lines on
the figure).
For uniformly accelerated motion
v22 = v12 + 2a( x2 − x1 )
v2 = v1 + a(t2 − t1 )
v22 − v12
2a
v −v
Δt = 2 1
a
Δx =
or
For the five regions shown above, we have
Region
v1 , ft/s
v2 , ft/s
a, ft/s 2
Δx, ft
Δt , s
1
126
120
−12.5
59.0
0.480
2
120
100
−33
66.7
0.606
3
100
80
−45.5
39.6
0.440
4
80
40
−54
44.4
0.741
5
40
0
−58
13.8
0.690
223.5
2.957
Σ
(a)
From the table, when v = 0
t = 2.96 s 
(b)
From the table and assuming x0 = 0, when v = 0
x = 224 ft 
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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 = 10 in., (b) when v = 80 in./s.
SOLUTION
Given: v − x curve
First note that the slope of the above curve is
dv
.
dx
a=v
(a)
Now
dv
dx
When
x = 10 in., v = 55 in./s
Then
 40 in./s 
a = 55 in./s 

 13.5 in. 
a = 163.0 in./s 2 
or
(b)
When v = 80 in./s, we have
 40 in./s 
a = 80 in./s 

 28 in. 
or
a = 114.3 in./s 2 
Note: To use the method of measuring the subnormal outlined at the end of Section 11.8, it is necessary
that the same scale be used for the x and v axes (e.g., 1 in. = 50 in., 1 in. = 50 in./s). In the above
solution, Δv and Δ x were measured directly, so different scales could be used.
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PROBLEM 11.85
Using the method of Section 11.8, derive the formula x = x0 + v0t + 12 at 2 for the position coordinate of a
particle in uniformly accelerated rectilinear motion.
SOLUTION
The a −t curve for uniformly accelerated motion is as shown.
Using Eq. (11.13), we have
x = x0 + v0t + (area under a −t curve) (t − t )
 1 
= x0 + v0t + (t × a)  t − t 
 2 
1
= x0 + v0 t + at 2
Q.E.D.
2

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PROBLEM 11.86
Using the method of Section 11.8 determine the position of the
particle of Problem 11.61 when t = 8 s.
PROBLEM 11.61 A particle moves in a straight line with the
acceleration shown in the figure. Knowing that it starts from the
origin with v0 = −14 ft/s, plot the v−t and x−t curves for
0 < t < 15 s and determine (a) the maximum value of the velocity
of the particle, (b) the maximum value of its position coordinate.
SOLUTION
x0 = 0
v0 = −14 ft/s
when t = 8s:
x = x0 + v0t + Σ A(t1 − t )
= 0 − (14 ft/s)(8 s) + [(3 ft/s2 )(2 s)](7 s) + [(8 ft/s2 )(3 s)](4.5 s) + [(3 ft/s)(3 s)](1.5 s)
x8 = −112 ft + 42 ft + 108 ft + 13.5 ft
x8 = 51.5 ft 
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PROBLEM 11.87
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 
Position when t = t1 = 2.4 s
2
x = x0 + v0t1 + 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 
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PROBLEM 11.88
For the particle of Problem 11.63, draw the
a −t curve and determine, using the method
of Section 11.8, (a) the position of the
particle when t = 52 s, (b) the maximum
value of its position coordinate.
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=
We have
where
dv
dt
dv
dt
is the slope of the v −t curve. Then
t=0
from
to t = 10 s:
v = constant  a = 0
t = 10 s to t = 26 s: a =
t = 26 s to t = 41 s:
v = constant  a = 0
t = 41 s to t = 46 s: a =
t > 46 s:
−20 − 60
= −5 m/s 2
26 − 10
−5 − (−20)
= 3 m/s 2
46 − 41
v = constant  a = 0
The a −t curve is then drawn as shown.
(a)
From the discussion following Eq. (11.13),
we have
x = x0 + v0t1 + ΣA(t1 − t )
where A is the area of a region and t is the distance to its centroid. Then, for t1 = 52 s
x = −540 m + (60 m/s)(52 s) + {−[(16 s)(5 m/s 2 )](52 − 18) s
+ [(5 s)(3 m/s 2 )](52 − 43.5)s}
= [−540 + (3120) + (−2720 + 127.5)] m
or
x = −12.50 m 
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PROBLEM 11.88 (Continued)
(b)
Noting that xmax occurs when v = 0 ( dx
= 0 ), it is seen from the v–t curve that xmax occurs for
dt
10 s < t < 26 s. Although similar triangles could be used to determine the time at which x = xmax
(see the solution to Problem 11.63), the following method will be used.
For
10 s < t1 < 26 s, we have
x = −540 + 60t1
1

− [(t1 − 10)(5)]  (t1 − 10)  m
2

5
= −540 + 60t1 − (t1 − 10) 2
2
When x = xmax :
or
Then
or
dx
= 60 − 5(t1 − 10) = 0
dt
(t1 ) xmax = 22 s
5
xmax = −540 + 60(22) − (22 − 10)2
2
xmax = 420 m 
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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 α of the acceleration of block B?
(a)
(b)
(c)
(d)
(e)
SOLUTION



Since aB = a A + aB/A we get
Answer: (d ) 
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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)
Velocity at t = 1 s.
vx = 5
v y = 6 − 9.8 = −3.8
v = vx2 + v y2 = 52 + 3.82 = 6.28 m/s
tan θ =
(b)
vy
vx
=
−3.8
5
Horizontal distance:
θ = −37.2°
v = 6.28 m/s
37.2° 
( y = 0)
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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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 −1.5 i + [60e−1 cos 7.5 − 8e−1 sin 7.5]
= 6.694i + 4.8906 j mm/s
a = 90e−1.5 i + [ −240e −1 cos 7.5 − 884e−1 sin 7.5 j]
= −20.08i − 335.65 j mm/s 2
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PROBLEM 11.91
The motion of a vibrating particle is defined by the position vector
r = (4sin π t )i − (cos 2π t ) j, where r is expressed in inches 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 π t )i − (cos 2π t ) j
v = (4π cos π t )i + (2π sin 2π t ) j
a = −(4π 2 sin π t )i + (4π 2 cos 2π t ) j
(a)
When t = 1 s:
v = (4π cos π )i + (2π sin 2π ) j
v = −(4π in/s)i 
a = −(4π 2 sin π )i − (4π 2 cos π ) j
(b)
a = −(4π 2 in/s 2 ) j 
Path of particle:
Since r = xi + y j ;
x = 4sin π t ,
y = − cos 2π t
Recall that cos 2θ = 1 − 2sin 2 θ and write
y = − cos 2π t = −(1 − 2sin 2 π t )
But since x = 4sin π t or sin π t =
(1)
1
x, Eq.(1) yields
4
2

1  
y = − 1 − 2  x  
 4  

y=
1 2
x − 1 (Parabola) 
8
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PROBLEM 11.92
The motion of a particle is defined by the equations x = 10t − 5sin t and y = 10 − 5cos t , where x and y are
expresed in feet and t is expressed in seconds. Sketch the path of the particle for the time interval 0 ≤ t ≤ 2π ,
and determine (a) the magnitudes of the smallest and largest velocities reached by the particle, (b) the
corresponding times, positions, and directions of the velocities.
SOLUTION
Sketch the path of the particle, i.e., plot of y versus x.
Using x = 10t − 5sin t , and y = 10 − 5cos t obtain the values in the table below. Plot as shown.
t(s)
x(ft)
y(ft)
0
0.00
5
10.71
10
31.41
15
52.12
10
62.83
5
π
2
π
3
π
2
2π
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PROBLEM 11.92 (Continued)
(a)
Differentiate with respect to t to obtain velocity components.
vx =
dx
= 10 − 5cos t and v y = 5sin t
dt
v 2 = vx2 + v y2 = (10 − 5cos t )2 + 25sin 2 t = 125 − 100cos t
d (v ) 2
= 100sin t = 0 t = 0. ± π . ± 2π  ± N π
dt
(b)
When t = 2 N π .
cos t = 1.
and
v2 is minimum.
When t = (2 N + 1)π .
cos t = −1.
and
v2 is maximum.
(v 2 )min = 125 − 100 = 25(ft/s) 2
vmin = 5 ft/s 
(v 2 )max = 125 + 100 = 225(ft/s) 2
vmax = 15 ft/s 
When v = vmin .
When N = 0,1, 2,
x = 10(2π N ) − 5sin(2π N )
x = 20π N ft 
y = 10 − 5cos(2π N )
y = 5 ft 
vx = 10 − 5cos(2π N )
vx = 5 ft/s 
v y = 5sin(2π N )
tan θ =
vy
vx
= 0,
vy = 0 
θ =0 
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PROBLEM 11.92 (Continued)
t = (2 N + 1)π s 
When v = vmax .
x = 10[2π ( N − 1)] − 5sin[2π ( N + 1)]
x = 20π ( N + 1) ft 
y = 10 − 5cos[2π ( N + 1)]
y = 15 ft 
vx = 10 − 5cos[2π ( N + 1)]
vx = 15 ft/s 
v y = 5sin[2π ( N + 1)]
tan θ =
vy
vx
= 0,
vy = 0 
θ =0 
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PROBLEM 11.93
The damped motion of a vibrating particle is defined by
the position vector r = x1[1 − 1/(t + 1)]i + ( y1e−π t/2 cos 2π 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
1 

−π t/2
r = 30 1 −
cos 2π t ) j
 i + 20(e
 t +1
Then
v=
= 30
1
 π

i + 20  − e −π t/2 cos 2π t − 2π e−π t/2 sin 2π t  j
2
(t + 1)
 2

= 30

1
1

i − 20π e −π t/ 2  cos 2π t + 2 sin 2π t   j
2
2
(t + 1)



a=
and
dr
dt
dv
dt
 π

2
1

i − 20π  − e−π t/2  cos 2π t + 2 sin 2π t  + e−π t/2 (−π sin 2π t + 4 cos 2π t )  j
3
(t + 1)
2

 2

60
=−
i + 10π 2 e−π t/2 (4 sin 2π t − 7.5 cos 2π t ) j
(t + 1)3
= −30
(a)
At t = 0:
 1
r = 30 1 −  i + 20(1) j
 1
r = 20 mm 
or
 1
1

v = 30   i − 20π (1)  + 0   j
1

 2
v = 43.4 mm/s
or
a=−
or
46.3° 
60
i + 10π 2 (1)(0 − 7.5) j
(1)
a = 743 mm/s 2
85.4° 
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PROBLEM 11.93 (Continued)
(b)
At t = 1.5 s:
1 

−0.75π
r = 30 1 −
(cos 3π ) j
 i + 20e
2.5


= (18 mm)i + ( −1.8956 mm) j
or
r = 18.10 mm
6.01° 
v = 5.65 mm/s
31.8° 
a = 70.3 mm/s 2
86.9° 
30
1

i − 20π e−0.75π  cos 3π + 0  j
2
(2.5) 2


= (4.80 mm/s)i + (2.9778 mm/s) j
v=
or
a=−
60
i + 10π 2 e−0.75π (0 − 7.5 cos 3π ) j
(2.5)3
= (−3.84 mm/s 2 )i + (70.1582 mm/s 2 ) j
or
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PROBLEM 11.94
The motion of a particle is defined by the position vector
r = A(cos t + t sin t )i + A(sin t − t cos t ) j, where t is expressed in
seconds. Determine the values of t for which the position vector
and the acceleration are (a) perpendicular, (b) parallel.
SOLUTION
We have
r = A(cos t + t sin t )i + A(sin t − t cos t ) j
Then
v=
and
a=
(a)
dr
= A( − sin t + sin t + t cos t )i
dt
+ A(cos t − cos t + t sin t ) j
= A(t cos t )i + A(t sin t ) j
dv
= A(cos t − t sin t )i + A(sin t + t cos t ) j
dt
When r and a are perpendicular, r ⋅ a = 0
A[(cos t + t sin t )i + (sin t − t cos t ) j] ⋅ A[(cos t − t sin t )i + (sin t + t cos t ) j] = 0
(b)
or
(cos t + t sin t )(cos t − t sin t ) + (sin t − t cos t )(sin t + t cos t ) = 0
or
(cos 2 t − t 2 sin 2 t ) + (sin 2 t − t 2 cos 2 t ) = 0
or
1− t2 = 0
or
t =1 s 
or
t=0 
When r and a are parallel, r × a = 0
A[(cos t + t sin t )i + (sin t − t cos t ) j] × A[(cos t − t sin t )i + (sin t + t cos t ) j] = 0
or
Expanding
or
[(cos t + t sin t )(sin t + t cos t ) − (sin t − t cos t )(cos t − t sin t )]k = 0
(sin t cos t + t + t 2 sin t cos t ) − (sin t cos t − t + t 2 sin t cos t ) = 0
2t = 0
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PROBLEM 11.95
The three-dimensional motion of a particle is defined by the position vector r = (Rt cos ωnt)i + ctj + (Rt sin
ωnt)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 ωn t )i + ctj + ( Rt sin ωn t )k
Then
v=
and
a=
dr
= R(cos ωn t − ωn t sin ωn t )i + cj + R(sin ωn t + ωn t cos ωn t )k
dt
dv
dt
= R(−ωn sin ωn t − ωn sin ωn t − ωn2t cos ωn t )i
+ R(ωn cos ωn t + ωn cos ωn t − ωn2 t sin ωn t )k
= R(−2ωn sin ωn t − ωn2 t cos ωn t )i + R(2ωn cos ωn t − ωn2 t sin ωn t )k
Now
v 2 = vx2 + v 2y + vz2
= [ R(cos ωn t − ωn t sin ωn t )]2 + (c)2 + [ R(sin ωn t + ωn t cos ωn t )]2
(
)
= R 2  cos 2 ωn t − 2ωn t sin ωn t cos ωn t + ωn2 t 2 sin 2 ωn t

+ sin 2 ωn t + 2ωn t sin ωn t cos ωn t + ωn2 t 2 cos 2 ωn t  + c 2

(
(
)
)
= R 2 1 + ωn2 t 2 + c 2
(
Also,
)
v = R 2 1 + ωn2 t 2 + c 2 
or
a 2 = ax2 + a 2y + az2
(
)
cos ω t − ω t sin ω t ) 

2
=  R −2ωn sin ωn t − ωn2 t cos ωn t  + (0) 2


(
(
2
2
+  R 2ωn
n
n
n

= R 2  4ωn2 sin 2 ωn t + 4ωn3t sin ωn t cos ωn t + ωn4t 2 cos 2 ωn t

+ 4ωn2 cos 2 ωn t − 4ωn3t sin ωn t cos ωn t + ωn4 t 2 sin 2 ωn t 

(
(
= R 2 4ωn2 + ωn4 t 2
or
)
)
)
a = Rωn 4 + ωn2 t 2 
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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
cos t =
Then
x
At
y = A t 2 + 1 z = Bt sin t
sin t =
2
 y
t2 =   −1
 A
z
Bt
2
2
2
2
2
x z
t2 =   +  
 A  B
or
2
Then
 y
x z
 A  −1 =  A  +  B 
 
   
or
 y  x  z
 A  −  A  −  B  =1
     
2
(a)
2
 x   z 
cos 2 t + sin 2 t = 1    +   = 1
 At   Bt 
Now
2
2

Q.E.D.
With A = 3 and B = 1, we have
v=
dr
t
j + (sin t + t cos t )k
= 3(cos t − t sin t )i + 3
2
dt
t +1
( )
t
and
t 2 + 1 − t t 2 +1
dv
a=
j
= 3( − sin t − sin t − t cos t )i + 3
dt
(t 2 + 1)
+ (cos t + cos t − t sin t )k
1
j + (2 cos t − t sin t )k
= −3(2 sin t + t cos t )i + 3 2
(t + 1)3/2
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PROBLEM 11.96 (Continued)
At t = 0:
v = 3(1 − 0)i + (0) j + (0)k
v = vx2 + v y2 + vz2
v = 3 ft/s 
or
and
Then
a = −3(0)i + 3(1) j + (2 − 0)k
a 2 = (0) 2 + (3) 2 + (2) 2 = 13
a = 3.61 ft/s 2 
or
(b)
If r and v are perpendicular, r ⋅ v = 0
t


[(3t cos t )i + (3 t 2 + 1) j + (t sin t )k ] ⋅ [3(cos t − t sin t )i +  3
 j + (sin t + t cos t )k ] = 0
2
 t +1 
or
Expanding
t


(3t cos t )[3(cos t − t sin t )] + (3 t 2 + 1)  3
 + (t sin t )(sin t + t cos t ) = 0
2
t
1
+


(9t cos 2 t − 9t 2 sin t cos t ) + (9t ) + (t sin 2 t + t 2 sin t cos t ) = 0
or (with t ≠ 0)
10 + 8 cos 2 t − 8t sin t cos t = 0
or
7 + 2 cos 2t − 2t sin 2t = 0
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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PROBLEM 11.97
An airplane used to drop water on brushfires is flying
horizontally in a straight line at 180 mi/h at an altitude of
300 ft. Determine the distance d at which the pilot should
release the water so that it will hit the fire at B.
SOLUTION
First note
v0 = 180 km/h = 264 ft/s
Place origin of coordinates at Point A.
Vertical motion. (Uniformly accelerated motion)
y = 0 + (0)t −
At B:
or
1 2
gt
2
1
−300 ft = − (32.2 ft/s 2 )t 2
2
t B = 4.31666 s
Horizontal motion. (Uniform)
x = 0 + (v x ) 0 t
At B:
or
d = (264 ft/s)(4.31666 s)
d = 1140 ft 
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PROBLEM 11.98
A helicopter is flying with a constant horizontal velocity of
180 km/h and is directly above Point A when a loose part
begins to fall. The part lands 6.5 s later at Point B on an
inclined surface. Determine (a) the distance d between
Points A and B, (b) the initial height h.
SOLUTION
Place origin of coordinates at Point A.
Horizontal motion:
(vx )0 = 180 km/h = 50 m/s
x = x0 + (vx )0 t = 0 + 50t m
At Point B where t B = 6.5 s,
(a)
Distance AB.
325
cos10°
From geometry
d=
Vertical motion:
y = y0 + ( v y ) 0 t −
At Point B
(b)
xB = (50)(6.5) = 325 m
d = 330 m 
1 2
gt
2
1
− xB tan10° = h + 0 − (9.81)(6.5) 2
2
Initial height.
h = 149.9 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 α corresponding
to h = 788 mm and h = 1068 mm.
SOLUTION
(a)
y0 = 1.5 m, (v y )0 = 0
Vertical motion:
t =
2( y0 − y )
g
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
or
y =h
x0 = 0, (vx )0 = v0 ,
x = v0t
or
v0 =
x
x
= B
t
tB
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PROBLEM 11.99 (Continued)
v0 =
12.2
= 32.02 m/s
0.3810
and
v0 =
12.2
= 41.11 m/s
0.2968
32.02 m/s ≤ v0 ≤ 41.11 m/s
or
Vertical motion:
v y = (v y )0 − gt = − gt
Horizontal motion:
vx = v0
With xB = 12.2 m,
(b)
we get
tan α = −
115.3 km/h ≤ v0 ≤ 148.0 km/h 
(v y ) B
dy
gt
=−
= B
dx
(vx ) B
v0
For h = 0.788 m,
tan α =
(9.81)(0.3810)
= 0.11673,
32.02
α = 6.66° 
For h = 1.068 m,
tan α =
(9.81)(0.2968)
= 0.07082,
41.11
α = 4.05° 
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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 + (vx )0 t = v0 t
At B:
y:
t B = 0.455016 s
or
Then
x:
y:
or
1
−2 ft = − (32.2 ft/s 2 )t 2
2
tC = 0.352454 s
or
Then
7 ft = (v0 ) B (0.455016 s)
(v0 ) B = 15.38 ft/s
or
At C:
1
1
−3 ft = − (32.2 ft/s 2 )t 2
3
2
x:
1
12 ft = (v0 )C (0.352454 s)
3
(v0 )C = 35.0 ft/s
15.38 ft/s < v0 < 35.0 ft/s 
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PROBLEM 11.101
Water flows from a drain spout with an initial velocity of
2.5 ft/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
First note
(vx )0 = (2.5 ft/s) cos 15° = 2.4148 ft/s
(v y )0 = −(2.5 ft/s) sin 15° = −0.64705 ft/s
Vertical motion. (Uniformly accelerated motion)
y = 0 + (v y ) 0 t −
1 2
gt
2
At the top of the trough
1
−8.8 ft = (−0.64705 ft/s) t − (32.2 ft/s 2 ) t 2
2
or
t BC = 0.719491 s
(the other root is negative)
Horizontal motion. (Uniform)
x = 0 + (v x ) 0 t
In time t BC
xBC = (2.4148 ft/s)(0.719491 s) = 1.737 ft
Thus, the trough must be placed so that
xB < 1.737 ft or xC ≥ 1.737 ft
Since the trough is 2 ft wide, it then follows that
0 < d < 1.737 ft 
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PROBLEM 11.102
Milk is poured into a glass of height 140 mm and inside
diameter 66 mm. If the initial velocity of the milk is 1.2 m/s at
an angle of 40° with the horizontal, determine the range of
values of the height h for which the milk will enter the glass.
SOLUTION
First note
(vx )0 = (1.2 m/s) cos 40° = 0.91925 m/s
(v y )0 = −(1.2 m/s) sin 40° = −0.77135 m/s
Horizontal motion. (Uniform)
x = 0 + (v x ) 0 t
Vertical motion. (Uniformly accelerated motion)
y = y0 + ( v y ) 0 t −
1 2
gt
2
Milk enters glass at B.
x:
0.08 m = (0.91925 m/s) t or tB = 0.087028 s
y : 0.140 m = hB + (−0.77135 m/s)(0.087028 s)
1
− (9.81 m/s 2 )(0.087028 s) 2
2
or
hB = 0.244 m
Milk enters glass at C.
x : 0.146 m = (0.91925 m/s) t or tC = 0.158825 s
y : 0.140 m = hC + (−0.77135 m/s)(0.158825 s)
1
− (9.81 m/s 2 )(0.158825 s) 2
2
or

hC = 0.386 m
0.244 m < h < 0.386 m 
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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 + (v x ) 0 t
At C
9 m = (12.5919 m/s) t or tC = 0.71475 s
Vertical motion. (Uniformly accelerated motion)
y = y0 + ( v y ) 0 t −
At C:
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 > 2.43 m (height of net)  ball clears net 
(b)
At B, y = 0:
1
0 = 2.1 m + (4.5831 m/s)t − (9.81 m/s 2 )t 2
2
Solving
t B = 1.271175 s (the other root is negative)
Then
d = (vx )0 t B = (12.5919 m/s)(1.271175 s)
= 16.01 m
The ball lands
b = (16.01 − 9.00) m = 7.01 m from the net 
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PROBLEM 11.104
A golfer hits a golf ball with an initial velocity of 160 ft/s at an angle of 25° with the horizontal. Knowing that
the fairway slopes downward at an average angle of 5°, determine the distance d between the golfer and Point B
where the ball first lands.
SOLUTION
(vx )0 = (160 ft/s) cos 25°
First note
(v y )0 = (160 ft/s) sin 25°
xB = d cos 5°
and at B
Now
yB = − d sin 5°
Horizontal motion. (Uniform)
x = 0 + (v x ) 0 t
d cos 5° = (160 cos 25°)t or t B =
At B
cos 5°
d
160 cos 25°
Vertical motion. (Uniformly accelerated motion)
y = 0 + (v y )0 t −
1 2
gt
2
( g = 32.2 ft/s 2 )
1 2
gt B
2
At B:
− d sin 5° = (160 sin 25°)t B −
Substituting for t B
 cos 5° 
1  cos 5°  2
− d sin 5° = (160 sin 25°) 
d − g 
 d
2  160 cos 25° 
 160 cos 25° 
2
or
or
2
(160 cos 25°) 2 (tan 5° + tan 25°)
32.2 cos 5°
= 726.06 ft
d=
d = 242 yd 
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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
(vx )0 = v0 cos 40°
(v y )0 = v0 sin 40°
Horizontal motion. (Uniform)
x = 0 + (v x ) 0 t
At B:
14 = (v0 cos 40°) t or t B =
14
v0 cos 40°
Vertical motion. (Uniformly accelerated motion)
y = 0 + (v y )0 t −
At B:
1 2
gt
2
1.5 = (v0 sin 40°) t B −
( g = 32.2 ft/s 2 )
1 2
gt B
2
Substituting for t B

 1 

14
14
1.5 = (v0 sin 40°) 
− g

 v0 cos 40°  2  v0 cos 40° 
or
v02
=
1
2
2
(32.2)(196)/ cos 2 40°
−1.5 + 14 tan 40°
or
v0 = 22.9 ft/s 
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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:
x = x0 + (vx )0 t = (v0 cos 40°)t
Vertical motion:
y = y0 + ( v y ) 0 t =
From (1),
Then
1 2
gt
2
1
= 2 + (v0 sin 40°) = − (9.81)t 2
2
v0 t =
(2)
x
cos 40°
(3)
y = 2 + x tan 40° − 4.905t 2
t2 =
Point B:
(1)
2 + x tan 40° − y
4.905
(4)
x = 8 + 10 cos 35° = 16.1915 m
y = 1.5 + 10sin 35° = 7.2358 m
16.1915
v0 t =
= 21.1365 m
cos 40°
2 + 16.1915 tan 40° − 7.2358
t2 =
4.905
21.1365
v0 =
1.3048
t = 1.3048 s
v0 = 16.199 m/s
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PROBLEM 11.106 (Continued)
Point C:
Range of values of v0.
x = 8 + (10 + 7) cos 35° = 21.9256 m
y = 1.5 + (10 + 7)sin 35° = 11.2508 m
v0 t =
21.9256
= 28.622 m
cos 40°
t2 =
2 + 21.9256 tan 40° − 11.2508
4.905
v0 =
28.622
1.3656
t = 1.3656 s
v0 = 20.96 m/s
16.20 m/s < v0 < 21.0 m/s 
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PROBLEM 11.107
A basketball player shoots when she is 16 ft from the
backboard. Knowing that the ball has an initial velocity v0
at an angle of 30° with the horizontal, determine the value
of v0 when d is equal to (a) 9 in., (b) 17 in.
SOLUTION
First note
(vx )0 = v0 cos 30°
Horizontal motion. (Uniform)
(v y )0 = v0 sin 30°
x = 0 + (v x ) 0 t
(16 − d ) = (v0 cos 30°) t or t B =
At B:
16 − d
v0 cos 30°
y = 0 + (v y )0 t −
Vertical motion. (Uniformly accelerated motion)
1 2
gt
2
1 2
gt B
2
At B:
3.2 = (v0 sin 30°) t B −
Substituting for tB
 16 − d  1  16 − d 
3.2 = (v0 sin 30°) 
− g

 v0 cos 30°  2  v0 cos 30° 
or
v02 =
(a)
d = 9 in.:
v02
d = 17 in.:
v02
(b)
=
=
( g = 32.2 ft/s 2 )
2
2 g (16 − d ) 2
3

1
3
(16 − d ) − 3.2 

2(32.2) (16 − 129 )
3

1
3
(16 − 129 ) − 3.2
2(32.2) (16 − 17
12 )
3

1
3
2
v0 = 29.8 ft/s 
2
(16 − 1217 ) − 3.2
v0 = 29.6 ft/s 
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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 cos 5° and (v y )0 = v0 sin 5°.
Horizontal motion:
x = x0 + (vx )0 t = (v0 cos 5°)t
Vertical motion:
y = y0 + ( v y ) 0 t =
(1)
1 2
gt
2
1
= 2.5 − (v0 sin 5°)t = − (9.81)t 2
2
From (1),
Then
v0 t =
x
cos 5°
(2)
(3)
y = 2.5 − x tan 5° − 4.905t 2
t2 =
2.5 − x tan 5° − y
4.905
(4)
At the minimum speed the ball just clears the net.
x = 12.2 m,
y = 0.914 m
12.2
= 12.2466 m
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
v0 t =
= 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
Range for v0.
37.7 m/s < 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.
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/s 2 )
x = (0.175 m) sin 10°
y = (0.175 m) cos 10°
At Point B:
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°)t B −
1 2
gtB
2
Substituting for t B
 0.050388  1
 0.050388 
−0.032659 = (−v0 sin 6°) 
 − (9.81) 

 v0 cos 6°  2
 v0 cos 6° 
2
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PROBLEM 11.109 (Continued)
v02 =
or
1
2
(9.81)(0.050388)2
cos 2 6°(0.032659 − 0.050388 tan 6°)
(v0 ) B = 0.678 m/s
or
x = (0.175 m) cos 30°
y = (0.175 m) sin 30°
At Point C:
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  1
 0.171554 
−0.117500 = (−v0 sin 6°) 
 − (9.81) 

 v0 cos 6°  2
 v0 cos 6° 
or
or

v02 =
1
2
2
(9.81)(0.171554) 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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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 + (v x ) 0 t
At either B or C, x = 5 m
s = (v0 cos 65°)t B,C
or
t B ,C =
5
(v0 cos 65°)
Vertical motion. (Uniformly accelerated motion)
y = 0 + (v y )0 t −
1 2
gt
2
( g = 9.81 m/s 2 )
At the tree limbs, t = tB ,C

 1 

5
5
yB ,C = (v0 sin 65°) 
− g

 v0 cos 65°  2  v0 cos 65° 
2
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PROBLEM 11.110 (Continued)
or
v02 =
=
1
2
(9.81)(25)
2
cos 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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PROBLEM 11.111
The pitcher in a softball game throws a ball with an initial velocity v0 of 72 km/h at an angle α with the
horizontal. If the height of the ball at Point B is 0.68 m, determine (a) the angle α, (b) the angle θ that the
velocity of the ball at Point B forms with the horizontal.
SOLUTION
First note
v0 = 72 km/h = 20 m/s
(vx )0 = v0 cos α = (20 m/s) cos α
and
(v y )0 = v0 sin α = (20 m/s) sin α
(a)
Horizontal motion. (Uniform)
x = 0 + (vx )0 t = (20 cos α ) t
At Point B:
14 = (20 cos α )t or t B =
7
10 cos α
Vertical motion. (Uniformly accelerated motion)
y = 0 + (v y )0 t −
At Point B:
1 2
1
gt = (20 sin α )t − gt 2
2
2
0.08 = (20 sin α )t B −
( g = 9.81 m/s 2 )
1 2
gt B
2
Substituting for t B

 1 

7
7
0.08 = (20 sin α ) 
− g

 10 cos α  2  10 cos α 
or
Now
8 = 1400 tan α −
2
1
49
g
2 cos 2 α
1
= sec2 α = 1 + tan 2 α
2
cos α
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PROBLEM 11.111 (Continued)
Then
or
Solving
8 = 1400 tan α − 24.5 g (1 + tan 2 α )
240.345 tan 2 α − 1400 tan α + 248.345 = 0
α = 10.3786° and α = 79.949°
Rejecting the second root because it is not physically reasonable, we have
α = 10.38° 
(b)
We have
vx = (vx )0 = 20 cos α
and
v y = (v y )0 − gt = 20 sin α − gt
At Point B:
(v y ) B = 20 sin α − gt B
= 20 sin α −
7g
10 cos α
Noting that at Point B, v y < 0, we have
tan θ =
=
=
or
|(v y ) B |
vx
7g
10 cos α
− 20 sin α
20 cos α
7
9.81
200 cos 10.3786°
− sin 10.3786°
cos 10.3786°
θ = 9.74° 
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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 α 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 = v0 t sin α
Vertical motion:
y = v0t cos α −
cos α =
sin 2 α + cos 2 α =
sin α =
x
v0t
1 2
gt
2
1 
1

y + gt 2 

v0 t 
2

(2)
2
1  2 
1 2 
x
y
gt
+
+


  =1
2
(v0 t )2 

 
x 2 + y 2 + gyt 2 +
1 24
g t = v02t 2
4
1 24
g t − v02 − gy t 2 + ( x 2 + y 2 ) = 0
4
(
At Point B,
(1)
)
(3)
x 2 + 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
t = 15.8916 s
and 1.6458 s 2
and 1.2829 s
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PROBLEM 11.112 (Continued)
Restrictions on α :
0 < α < 120°
tan α =
x
2
y + gt
1
2
=
100 cos 30°
= 0.0729
−50 + (4.905)(15.8916) 2
α = 4.1669°
100 cos 30°
= −2.0655
−50 + (4.905)(1.2829)2
α = 115.8331°
and
Use α = 4.1669° corresponding to the steeper possible trajectory.
(a)
Angle α .
(b)
Maximum height.
α = 4.17° 
v y = 0 at
y = ymax
v y = v0 cos α − gt = 0
t=
v0 cos α
g
ymax = v0 t cos α −
=
(c)

Duration of the flight.
v 2 cos 2 α
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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PROBLEM 11.113
The initial velocity v0 of a hockey puck is 105 mi/h. Determine (a) the largest value (less than 45°) of
the angle α for which the puck will enter the net, (b) the corresponding time required for the puck to reach
the net.
SOLUTION
First note
v0 = 105 mi/h = 154 ft/s
(vx )0 = v0 cos α = (154 ft/s) cos α
and
(v y )0 = v0 sin α = (154 ft/s) sin α
(a)
Horizontal motion. (Uniform)
x = 0 + (vx )0 t = (154 cos α )t
At the front of the net,
Then
or
x = 16 ft
16 = (154 cos α )t
tenter =
8
77 cos α
Vertical motion. (Uniformly accelerated motion)
1 2
gt
2
1
= (154 sin α ) t − gt 2
2
y = 0 + (v y )0 t −
( g = 32.2 ft/s 2 )
At the front of the net,
yfront = (154 sin α ) tenter −
1 2
gtenter
2

 1 

8
8
= (154 sin α ) 
− g

77
cos
2
77
cos
α
α




32 g
= 16 tan α −
5929 cos 2 α
2
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PROBLEM 11.113 (Continued)
Now
Then
or
1
= sec2 α = 1 + tan 2 α
cos 2 α
yfront = 16 tan α −
tan 2 α −
32 g
(1 + tan 2 α )
5929
 5929

5929
tan α + 1 +
yfront  = 0
2g
32 g


5929
2g
(

±  − 5929
2g

)
2
(
)
1/2

− 4 1 + 5929
y
32 g front 

2
Then
tan α =
or
 5929 2 
5929
5929

±  −
tan α =
yfront  
 − 1 +
4 × 32.2  4 × 32.2   32 × 32.2
 
or
tan α = 46.0326 ± [(46.0326) 2 − (1 + 5.7541 yfront )]1/2
1/2
Now 0 < yfront < 4 ft so that the positive root will yield values of α > 45° for all values of yfront.
When the negative root is selected, α increases as yfront is increased. Therefore, for α max , set
yfront = yC = 4 ft
(b)
Then
tan α = 46.0326 − [(46.0326) 2 − (1 + 5.7541 + 4)]1/ 2
or
α max = 14.6604°
We had found
or
α max = 14.66° 
8
77 cos α
8
=
77 cos 14.6604°
tenter =
tenter = 0.1074 s 
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PROBLEM 11.114
A worker uses high-pressure water to clean the inside of a long drainpipe. If the water is discharged with an
initial velocity v0 of 11.5 m/s, determine (a) the distance d to the farthest Point B on the top of the pipe that the
worker can wash from his position at A, (b) the corresponding angle α.
SOLUTION
First note
(vx )0 = v0 cos α = (11.5 m/s) cos α
(v y )0 = v0 sin α = (11.5 m/s) sin α
By observation, d max occurs when
ymax = 1.1 m.
Vertical motion. (Uniformly accelerated motion)
v y = (v y )0 − gt
= (11.5 sin α ) − gt
y = ymax
When
Then
at
1 2
gt
2
1
= (11.5 sin α ) t − gt 2
2
y = 0 + (v y ) 0 t −
B , (v y ) B = 0
(v y ) B = 0 = (11.5 sin α ) − gt
11.5 sin α
g
( g = 9.81 m/s 2 )
or
tB =
and
yB = (11.5 sin α ) t B −
1 2
gt B
2
Substituting for t B and noting yB = 1.1 m
 11.5 sin α
1.1 = (11.5 sin α ) 
g

1
=
(11.5) 2 sin 2 α
2g
or
sin 2 α =
2.2 × 9.81
11.52
 1  11.5 sin α 
− g

g
 2 

2
α = 23.8265°
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PROBLEM 11.114 (Continued)
(a)
Horizontal motion. (Uniform)
x = 0 + (vx )0 t = (11.5 cos α ) t
At Point B:
where
Then
or
(b)
From above
x = d max
tB =
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 
α = 23.8° 
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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 α 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 α = (8 m/s) cos α
(v y )0 = v0 sin α = (8 m/s) sin α
Horizontal motion. (Uniform)
x = 0 + (vx )0 t = (8 cos α ) t
At Point B:
or
x = d : d = (8 cos α ) t
d
tB =
8 cos α
Vertical motion. (Uniformly accelerated motion)
1 2
gt
2
1
= (8 sin α ) t − gt 2 ( g = 9.81 m/s 2 )
2
1
0 = (8 sin α ) t B − gt B2
2
y = 0 + (v y )0 t −
At Point B:
Simplifying and substituting for t B
0 = 8 sin α −
d=
or
(a)
1  d 
g

2  8 cos α 
64
sin 2α
g
(1)
When h = 0, the water can follow any physically possible trajectory. It then follows from
Eq. (1) that d is maximum when 2α = 90°
α = 45° 
or
Then
d=
64
sin (2 × 45°)
9.81
or
d max = 6.52 m 
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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 α increases
in value from 0 to 45° and then d decreases as α is further increased. Thus, d max occurs for the value
of α 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 α
Also, with ycorn = h, we have
h = (8 sin α ) tcorn −
1 2
gtcorn
2
Substituting for tcorn and noting h = 1.8 m,
 1.5  1  1.5 
1.8 = (8 sin α ) 
− g

 8 cos α  2  8 cos α 
or
Now
Then
or
1.8 = 1.5 tan α −
2
2.25 g
128 cos 2 α
1
= sec2 α = 1 + tan 2 α
cos 2 α
1.8 = 1.5 tan α −
2.25(9.81)
(1 + tan 2 α )
128
0.172441 tan 2 α − 1.5 tan α + 1.972441 = 0
α = 58.229° and α = 81.965°
Solving
From the above discussion, it follows that d = d max when
α = 58.2° 
Finally, using Eq. (1)
d=
or
64
sin (2 × 58.229°)
9.81
d max = 5.84 m 
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PROBLEM 11.116*
A mountain climber plans to jump from A to B over a
crevasse. Determine the smallest value of the climber’s initial
velocity v0 and the corresponding value of angle α so that he
lands at B.
SOLUTION
First note
(vx )0 = v0 cos α
(v y )0 = v0 sin α
Horizontal motion. (Uniform)
x = 0 + (vx )0 t = (v0 cos α ) t
At Point B:
or
1.8 = (v0 cos α )t
tB =
1.8
v0 cos α
Vertical motion. (Uniformly accelerated motion)
1 2
gt
2
1
= (v0 sin α ) t − gt 2
2
y = 0 + (v y )0 t −
At Point B:
−1.4 = (v0 sin α ) t B −
( g = 9.81 m/s 2 )
1 2
gt B
2
Substituting for t B
 1.8
−1.4 = (v0 sin α ) 
 v0 cos α
or
 1  1.8 
− g

 2  v0 cos α 
v02 =
1.62 g
cos α (1.8 tan α + 1.4)
=
1.62 g
0.9 sin 2α + 1.4 cos 2 α
2
2
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PROBLEM 11.116* (Continued)
Now minimize v02 with respect to α.
We have
dv02
−(1.8 cos 2α − 2.8 cos α sin α )
= 1.62 g
=0
dα
(0.9 sin 2α + 1.4 cos 2 α ) 2
1.8 cos 2α − 1.4 sin 2α = 0
or
tan 2α =
or
18
14
α = 26.0625° and α = 206.06°
or
Rejecting the second value because it is not physically possible, we have
α = 26.1° 
Finally,
or
v02 =
1.62 × 9.81
cos 26.0625°(1.8 tan 26.0625° + 1.4)
2
(v0 )min = 2.94 m/s 
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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 = 302 + 452 − 2(30)(45) cos 15°
or
vA /B = 17.80450 ft/s
and
or
30
17.80450
=
sin α
sin 15°
α = 25.8554°
α + 25° = 50.8554°
vA /B = 17.8 ft/s
50.9° 
Alternative solution.
vA /B = v A − vB
= 30 cos 10° i − 30 sin 10° j − (45 cos 25° i − 45° sin 25° j)
= 11.2396i + 13.8084 j
= 5.05 m/s = 17.8 ft/s
50.9°
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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
Cable 1:
y A + yD = constant
Then
v A + vD = 0
Cable 2:
(1)
( yB − yD ) + ( yC − yD ) = constant
vB + vC − 2vD = 0
Then
(2)
Combining Eqs. (1) and (2) to eliminate vD ,
2v A + vB + vC = 0
(3)
Now
v A/C = v A − vC = −300 mm/s
(4)
and
vB/A = vB − v A = 200 mm/s
(5)
(3) + (4) − (5) 
Then
(2v A + vB + vC ) + (v A − vC ) − (vB − v A ) = (−300) − (200)
v A = 125 mm/s 
or
and using Eq. (5)
vB − (−125) = 200
v B = 75 mm/s 
or
Eq. (4)
or
−125 − vC = −300
vC = 175 mm/s 
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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
v A = 45 mi/h = 66 ft/s
vB = 30 mi/h = 44 ft/s
Law of cosines
vB2/A = 662 + 442 − 2(66)(44) cos110°
vB/A = 90.99 ft/s
vB = v A + vB/A
Law of sines
sin β sin110°
=
66
90.99
β = 42.97°
α = 90° − β = 90° − 42.97° = 47.03°
v B/A = 91.0 ft/s
(a)
Relative velocity:
(b)
Change in position for Δt = 4 s.
ΔrB/A = vB/A Δt = (91.0 ft/s)(4 s)
(c)
rB/A = 364 ft
47.0° 
47.0° 
Distance between autos 2 seconds after auto A has passed intersection.
Auto A travels for 2 s.
v A = 66 ft/s
20°
rA = v At = (66 ft/s)(2 s) = 132 ft
rA = 132 ft
20°
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PROBLEM 11.119 (Continued)
Auto B
v B = 44 ft/s
rB = v B t = (44 ft/s)(5 s) = 220 ft
rB = rA + rB/A
Law of cosines
rB2/A = (132) 2 + (220) 2 − 2(132)(220) cos110°
rB/A = 292.7 ft
Distance between autos = 293 ft 
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PROBLEM 11.120
Shore-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
v F = v R + v F/R
or
v F = v F/R + v R
The graphical representation of the second equation is then as shown.
We have
vR2 = 182 + 18.42 − 2(18)(18.4) cos 10°
or
vR = 3.1974 km/h
and
or
18
3.1974
=
sin α sin 10°
α = 77.84°
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 vC/A = 350 km/h
75°, and the relative velocity of C with respect
to B is vC/B = 400 km/h
40°. Determine (a) the relative velocity of
B with respect to A, (b) the velocity of A if ground-based radar indicates
that the hurricane is moving at a speed of 30 km/h due north, (c) the
change in position of C with respect to B during a 15-min interval.
SOLUTION
(a)
We have
v C = v A + v C/ A
and
v C = v B + v C/ B
Then
v A + vC/A = v B + v C/B
or
v B − v A = v C/A − vC/B
Now
v B − 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 + 4002 − 2(350)(400) cos 65°
or
vB/A = 405.175 km/h
and
or
400 405.175
=
sin α
sin 65°
α = 63.474°
75° − α = 11.526°
v B/A = 405 km/h
11.53° 
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PROBLEM 11.121 (Continued)
(b)
We have
v C = v A + v C/ A
or
v A = (30 km/h) j − (350 km/h)(− cos 75°i − sin 75° j )
v A = (90.587 km/h)i + (368.07 km/h) j
v A = 379 km/h
or
(c)
76.17° 
Noting that the velocities of B and C are constant, we have
rB = (rB )0 + v B t
Now
rC = (rC )0 + v C t
rC/B = rC − rB = [(rC )0 − (rB )0 ] + ( vC − v B )t
= [(rC )0 − (rB )0 ] + v C/B t
Then
ΔrC/B = (rC/B )t2 − (rC/B )t1 = v C/B (t2 − t1 ) = vC/B Δt
For Δt = 15 min:
1 
ΔrC/B = (400 km/h)  h  = 100 km
4 
ΔrC/B = 100 km
40° 
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PROBLEM 11.122
Pin P moves at a constant speed of 150 mm/s in a counterclockwise sense along a
circular slot which has been milled in the slider block A shown. Knowing that the
block moves downward at a constant speed 100 mm/s determine the velocity of
pin P when (a) θ = 30°, (b) θ = 120°.
SOLUTION
v P = v A + v P/A
v P = 100 ms (− j) + 150(cos θ i + sin θ j)mm/s
(a)
For θ = 30°
v P = −100 mm/s ( j) + 150( − cos(30°)i + sin(30°) j)mm/s
v P = (−75i + 29.9038 j)mm/s
v P = 80.7 mm/s
(b)
For θ = 120°
21.7° 
v P = −100 mm/s ( j) + 150(− cos(120°)i + sin(120°) j)mm/s
v P = (−129.9038i + −175 j) mm/s
v P = 218 mm/s
53.4° 
Alternative Solution
(a)
For θ = 30°, vP /A = 7.5 in./s
30°
vP = v A + vP/A
Law of cosines
vP2 = (150) 2 + (100) 2 − 2(100)(150) cos 30°
vP = 80.7418 mm/s
Law of sines
sin β
sin 30°
=
150 80.7418
β = 111.7°
vP = 80.7 mm/s
21.7° 
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PROBLEM 11.122 (Continued)
(b)
For θ = 120°, vP/A = 150 mm/s
30°
Law of cosines
vP2 = (150) 2 + (100) 2 − 2(100)(150) cos120°
vP = 217.9449 mm/s
Law of sines
sin β
sin120°
=
β = 36.6°
150
217.9449
α = 90 − β = 90° − 36.6 = 53.4°
vP = 218 mm/s
53.4° 
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PROBLEM 11.123
Knowing that at the instant shown assembly A has a velocity of 9 in./s and an
acceleration of 15 in./s2 both directed downward, determine (a) the velocity
of block B, (b) the acceleration of block B.
SOLUTION
Length of cable = constant
L = x A + 2 xB/A = constant
v A + 2vB/A = 0
(1)
a A + 2aB/A = 0
(2)
a A = 15 in./s 2
Data:
v A = 9 in./s
Eqs. (1) and (2)
a A = −2aB/A
v A = −2vB/A
15 = −2aB/A
9 = −2vB/A
aB/A = −7.5 in./s 2
a B/A = 7.5 in./s 2
(a)
vB/A = −4.5 in./s
40°
v B/A = −4.5 in./s
Velocity of B.
v B = v A + v B/A
Law of cosines:
vB2 = (9) 2 + (4.5) 2 − 2(9)(4.5) cos 50°
40°
vB = 7.013 in./s
Law of sines:
sin β sin 50°
=
4.5
7.013
β = 29.44°
α = 90° − β = 90° − 29.44° = 60.56°
v B = 7.01 in./s
60.6° 
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PROBLEM 11.123 (Continued)
(b)
Acceleration of B. a B may be found by using analysis similar to that used above for vB . An alternate
method is
a B = a A + a B/A
a B = 15 in./s 2 ↓ +7.5 in./s 2
40°
= −15 j − (7.5 cos 40°)i + (7.5 sin 40°) j
= −15 j − 5.745i + 4.821j
a B = −5.745i − 10.179 j
a B = 11.69 in./s 2
60.6° 
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PROBLEM 11.124
Knowing that at the instant shown block A has a velocity of 8 in./s and
an acceleration of 6 in./s2 both directed down the incline, determine
(a) the velocity of block B, (b) the acceleration of block B.
SOLUTION
From the diagram
2 x A + xB/A = constant
Then
2v A + vB/A = 0
| vB/A | = 16 in./s
or
2a A + aB/A = 0
and
| aB/A | = 12 in./s 2
or
Note that v B/A and a B/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
vB2 = 82 + 162 − 2(8)(16) cos 15°
or
vB = 8.5278 in./s
and
or
8
8.5278
=
sin α sin 15°
α = 14.05°
v B = 8.53 in./s
(b)
54.1° 
The same technique that was used to determine v B can be used to determine a B . An alternative method
is as follows.
We have
a B = a A + a B/A
= (6i ) + 12(− cos 15°i + sin 15° j)*
= −(5.5911 in./s 2 )i + (3.1058 in./s 2 ) j
or
a B = 6.40 in./s 2
54.1° 
* Note the orientation of the coordinate axes on the sketch of the system.
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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 ,
Vertical motion of the ball:
v y = (v y )0 − gt
yB = (v y )0 t −
At maximum height,
xball = (vx )0 t
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 m 2 /s 2
(v y )0 = 12.528 m/s
At time of catch,
tcatch = 2.554 s
or
Motion of the deck:
1
(9.81)t 2
2
and
v y = 12.528 m/s
y = 0 = 12.528 −
vx = (vx )0 + aDt ,
xdeck = (vx )0 t +
1
a Dt 2
2
Motion of the ball relative to the deck:
(vB/D ) x = (vx )0 − [(vx )0 + aDt ] = −aDt
1
1


xB/D = (vx )0 t − (vx )0 t + aDt 2  = − aDt 2
2
2


(vB/D ) y = (v y )0 − gt , yB/D = yB
(a)
(b)
At time of catch,
1
d = xD/B = − (− 0.3)(2.554) 2
2
(vB/D ) x = −(− 0.3)(2.554) = + 0.766 m/s
(vB/D ) y = 12.528 m/s
d = 0.979 m 
or 0.766 m/s
v B/D = 12.55 m/s
86.5° 
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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
a C = a B + a C/ B
The graphical representation of this equation is then as shown.
First note
α = 180° − (20° + 105°)
= 55°
Then
aC
2
=
sin 20° sin 55°
aC = 0.83506 mm/s 2
(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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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 = −5 ft/s
0
vs y = 2 gh = (2)(32.2 ft/s 2 )(3 ft) = 13.90 ft/s ↓
y-direction.
v S/B = v S − v B
Relative velocity.
(a)
(1)
If vS/B is vertical,
−vS /B j = −5i − 13.9 j − ( −vB cos 15°i + vB sin 15° j)
= −5i − 13.9 j + vB cos 15°i − vB sin 15° j
Equate components.
i : 0 = −5 + vB cos 15°
vB =
5
= 5.176 ft/s
cos 15°
v B = 5.18 ft/s
(b)
15° 
vS/C is as small as possible, so make vS/B ⊥ to vB into (1).
−vS/B sin 15°i − vS/B cos 15° j = − 5i − 13.9 j + vB cos 15°i − vB sin 15° j
Equate components and transpose terms.
(sin 15°) vS/B + (cos 15°) vB = 5
(cos 15°) vS/B − (sin 15°) vB = 13.90
Solving,
vS/B = 14.72 ft/s
vB = 1.232 ft/s
v B = 1.232 ft/s
15° 
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PROBLEM 11.128
Conveyor belt A, which forms a 20°
angle with the horizontal, moves at
a constant speed of 4 ft/s and is
used to load an airplane. Knowing
that a worker tosses duffel bag B
with an initial velocity of 2.5 ft/s at
an angle of 30° with the horizontal,
determine the velocity of the bag
relative to the belt as it lands on the
belt.
SOLUTION
First determine the velocity of the bag as it lands on the belt. Now
[(vB ) x ]0 = (vB )0 cos 30°
= (2.5 ft/s) cos 30°
[(vB ) y ]0 = (vB )0 sin 30°
= (2.5 ft/s) sin 30°
Horizontal motion. (Uniform)
x = 0 + [(vB ) x ]0 t
(vB ) x = [(vB ) x ]0
= (2.5 cos 30°) t
= 2.5 cos 30°
Vertical motion. (Uniformly accelerated motion)
y = y0 + [(vB ) y ]0 t −
1 2
gt
2
= 1.5 + (2.5 sin 30°) t −
(vB ) y = [(vB ) y ]0 − gt
1 2
gt
2
= 2.5 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
1.5 + (2.5 sin 30°) t −
1 2
gt = [(2.5 cos 30°) t ] tan 20°
2
or
1
(32.2) t 2 + 2.5(cos 30° tan 20° − sin 30°) t − 1.5 = 0
2
or
16.1t 2 − 0.46198t − 1.5 = 0
Solving
t = 0.31992 s and t = −0.29122 s (Reject)
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PROBLEM 11.128 (Continued)
The velocity v B of the bag as it lands on the belt is then
v B = (2.5 cos 30°)i + [2.5 sin 30° − 32.2(0.319 92)] j
= (2.1651 ft/s)i − (9.0514 ft/s) j
Finally
or
v B = v A + v B/A
v B/A = (2.1651i − 9.0514 j) − 4(cos 20°i + sin 20° j)
= −(1.59367 ft/s)i − (10.4195 ft/s) j
or
v B/A = 10.54 ft/s
81.3° 
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PROBLEM 11.129
During a rainstorm the paths of the raindrops appear to form an angle of 30° with the vertical and to be directed
to the left when observed from a side window of a train moving at a speed of 15 km/h. A short time later, after
the speed of the train has increased to 24 km/h, the angle between the vertical and the paths of the drops
appears to be 45°. If the train were stopped, at what angle and with what velocity would the drops be observed
to fall?
SOLUTION
vrain = vtrain + vrain/train
Case :
vT = 15 km/h
;
vR / T
30°
Case :
vT = 24 km/h
;
vR / T
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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PROBLEM 11.129 (Continued)
3.706
21.294
β = 7.24°
21.294
= 21.47 km/h = 5.96 m/s
vR =
cos 7.24°
tan β =
vR = 5.96 m/s
82.8° 
Alternate solution
Alternate, vector equation
v R = vT + v R / T
For first case,
v R = 15i + vR / T −1 (− sin 30°i − cos 30° j)
For second case,
v R = 24i + vR /T − 2 (− sin 45°i − cos 45° j)
Set equal
15i + vR /T −1 (− sin 30°i − cos 30° 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° + vR /T − 2 sin 45° = 9
(3)
−vR /T −1 cos 30° = −vR / T − 2 cos 45°
j:
vR / T −1 cos 30° + 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
 −21.29 
 = −82.7585°
 2.71 
θ = tan −1 
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
As observed from a ship moving due east at 9 km/h, the wind appears to blow from the south. After the ship
has changed course and speed, and as it is moving north at 6 km/h, the wind appears to blow from the
southwest. Assuming that the wind velocity is constant during the period of observation, determine the
magnitude and direction of the true wind velocity.
SOLUTION
v wind = v ship + v wind/ship
v w = v s + v w/s
Case 
v s = 9 km/h →; v w /s
Case 
v s = 6 km/h ↑ ; v w /s
15
= 1.6667
9
α = 59.0°
tan α =
vw = 92 + 152 = 17.49 km/h
v w = 17.49 km/h
59.0° 
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PROBLEM 11.131
When a small boat travels north at 5 km/h, a flag mounted on its
stern forms an angle θ = 50° with the centerline of the boat as
shown. A short time later, when the boat travels east at 20 km/h,
angle θ is again 50°. Determine the speed and the direction of the
wind.
SOLUTION
We have
vW = v B + vW/B
Using this equation, 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
θ = 180° − (50° + 90° + α )
= 40° − α
We have
vW
5
=
sin 50° sin (40° − α )
φ = 180° − (50° + α )
= 130° − α
vW
20
=
sin 50° sin (130° − α )
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PROBLEM 11.131 (Continued)
Therefore
or
or
5
20
=
sin (40° − α ) sin (130° − α )
sin 130° cos α − cos 130° sin α = 4(sin 40° cos α − cos 40° sin α )
tan α =
sin 130° − 4 sin 40°
cos 130° − 4 cos 40°
or
α = 25.964°
Then
vW =
5 sin 50°
= 15.79 km/h
sin (40° − 25.964°)
vW = 15.79 km/h
26.0° 
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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 3 ft/s, determine the speed
and the direction of the train.
SOLUTION
v D = v B + v D/B
We have
v D = vC + v D/C
The graphical representations of these equations are then as shown.
Then
vD
3
=
sin 8° sin (22° + α )
Equating the expressions for
vD
3
=
sin 7° sin (23° − α )
vD
3
sin 8°
sin 7°
=
sin (22° + α ) sin (23° − α )
or
or
or
Then
sin 8° (sin 23° cos α − cos 23° sin α )
= sin 7° (sin 22° cos α + cos 22° sin α )
tan α =
sin 8° sin 23° − sin 7° sin 22°
sin 8° cos 23° + sin 7° cos 22°
α = 2.0728°
vD =
3 sin 8°
= 1.024 ft/s
sin (22° + 2.0728°)
v D = 1.024 ft/s
2.07° 
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PROBLEM 11.132 (Continued)
Alternate solution using components.
v B = (3 ft/s)
30° = (2.5981 ft/s)i + (1.5 ft/s) j
vC = (3 ft/s)
30° = (2.5981 ft/s)i − (1.5 ft/s) j
v D/B = u1
22° = −(u1 cos 22°)i − (u1 sin 22°) j
v D/C = u2
23° = −(u2 cos 23°)i + (u2 sin 23°) j
v D = vD
α = −(vD cos α )i + (vD sin α ) j
v D = v B + v D/B = vC + v D/C
2.5981i + 1.5 j − (u1 cos 22°)i − (u1 sin 22°) j = 2.5981i − 1.5 j − (u2 cos 23°)i + (u2 sin 23°) j
Separate into components, transpose, and change signs.
u1 cos 22° − u2 cos 23° = 0
u1 sin 22° + u1 sin 23° = 3
Solving for u1 and u2 ,
u1 = 3.9054 ft/s
u2 = 3.9337 ft/s
v D = 2.5981i + 1.5 j − (3.9054 cos 22°)i − (3.9054 sin 22°) j
= −(1.0229 ft/s)i + (0.0370 ft/s) j
or
v D = 2.5981i − 1.5 j − (3.9337 cos 23°)i + (3.9337 sin 23°) j
= −(1.0229 ft/s)i + (0.0370 ft/s) j
v D = 1.024 ft/s
2.07° 
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PROBLEM 11.CQ8
The Ferris wheel is rotating with a constant angular velocity ω. 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-go-round A with a constant speed u relative to A. The
merry-go-round undergoes fixed axis rotation about its center with a constant
angular velocity ω 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 − rθ 2 )er + (rθ + 2rθ)eθ
(1)
r = 0, θ = 0, r = 0, θ = ω, r = -u. When we substitute
From the information given, we know 
these values into (1), we will only have a term in the −θ direction.
Answer: (d ) 
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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/s 2 .
SOLUTION
an =
v2
an = 0.8 m/s 2
ρ
v = 72 km/h = 20 m/s
0.8 m/s 2 =
(20 m/s) 2
ρ
ρ = 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 ρ is 25 m
and the normal component of their acceleration cannot exceed 3 g.
SOLUTION
We have
an =
v2
ρ
Then
(vmax ) 2AB = (3 × 9.81 m/s 2 )(25 m)
or
(vmax ) AB = 27.124 m/s
or
(vmax ) AB = 97.6 km/h 
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PROBLEM 11.135
A bull-roarer is a piece of wood that produces a roaring sound when attached to the
end of a string and whirled around in a circle. Determine the magnitude of the
normal acceleration of a bull-roarer when it is spun in a circle of radius 0.9 m at
a speed of 20 m/s.
SOLUTION
an =

v2
ρ
=
(20 m/s) 2
= 444.4 m/s 2
0.9 m
an = 444 m/s 2 
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PROBLEM 11.136
To test its performance, an automobile is driven around a circular test track of diameter d. Determine (a) the
value of d if when the speed of the automobile is 45 mi/h, the normal component of the acceleration
is 11 ft/s 2 , (b) the speed of the automobile if d = 600 ft and the normal component of the acceleration is
measured to be 0.6 g.
SOLUTION
(a)
First note
Now
v = 45 mi/h = 66 ft/s
an =
ρ=
v2
ρ
(66 ft/s) 2
= 396 ft
11 ft/s 2
d = 2ρ
(b)
d = 792 ft 
v2
We have
an =
Then
1

v 2 = (0.6 × 32.2 ft/s 2 )  × 600 ft 
2

ρ
v = 76.131 ft/s
v = 51.9 mi/h 
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PROBLEM 11.137
An outdoor track is 420 ft in diameter. A runner increases her speed at a constant
rate from 14 to 24 ft/s over a distance of 95 ft. Determine the magnitude of the total
acceleration of the runner 2 s after she begins to increase her speed.
SOLUTION
We have uniformly accelerated motion
v22 = v12 + 2at Δs12
Substituting
or
(24 ft/s)2 = (14 ft/s)2 + 2at (95 ft)
at = 2 ft/s 2
Also
v = v1 + at t
At t = 2 s:
v = 14 ft/s + (2 ft/s 2 )(2 s) = 18 ft/s
v2
Now
an =
At t = 2 s:
an =
Finally
a 2 = at2 + an2
At t = 2 s:
a 2 = 22 + 1.542862
or
ρ
(18 ft/s) 2
= 1.54286 ft/s 2
210 ft
a = 2.53 ft/s 2 
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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:
Speed:
at = 10 mm/s 2
v = at t
v2
an =
where
ρ = 0.8 m = 800 mm
(a)
When t = 4 s
ρ
v = (10)(4) = 40 mm/s
an =
Acceleration:
ρ
=
at2 t 2
Normal 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/s 2
a 2 = an2 + at2
2
 (10) 2 t 2 
2
(80) = 
 + 10
800


2
t 4 = 403200 s 4
t = 25.2 s 
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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
maximum total acceleration of the train must not exceed 1.5 m/s 2 , 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 ρ = 400 m,
an =
v2
ρ
=
(20)2
= 1.000 m/s 2
400
a = an2 + at2
But
at = a 2 − an2 = (1.5) 2 − (1.000) 2 = ± 1.11803 m/s 2
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/s 2 
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PROBLEM 11.140
A motorist starts from rest at Point A on a circular entrance ramp when
t = 0, increases the speed of her automobile at a constant rate and
enters the highway at Point B. Knowing that her speed continues to
increase at the same rate until it reaches 100 km/h at Point C,
determine (a) the speed at Point B, (b) the magnitude of the total
acceleration when t = 20 s.
SOLUTION
v0 = 0
Speeds:
v1 = 100 km/h = 27.78 m/s
s =
Distance:
π
2
(150) + 100 = 335.6 m
v12 = v02 + 2at s
Tangential component of acceleration:
at =
At Point B,
v12 − v02
(27.78) 2 − 0
=
= 1.1495 m/s 2
2s
(2)(335.6)
vB2 = v02 + 2at sB
where
sB =
π
2
(150) = 235.6 m
vB2 = 0 + (2)(1.1495)(235.6) = 541.69 m 2 /s 2
vB = 23.27 m/s
(a)
At t = 20 s,
vB = 83.8 km/h 
v = v0 + at t = 0 + (1.1495)(20) = 22.99 m/s
ρ = 150 m
Since v < vB , the car is still on the curve.
(b)
Normal component of acceleration:
an =
Magnitude of total acceleration:
|a| =
v2
ρ
=
(22.99)2
= 3.524 m/s 2
150
at2 + an2 =
(1.1495) 2 + (3.524) 2
| a | = 3.71 m/s 2 
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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.
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)
50°
vB/A = vB − vA
Relative velocity:
vB/A = (55.56 cos 50°) ← + 55.56sin 50° ↓ + 66.67
= 30.96 → + 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:
ρ
( ρ = ∞)
Car B:
( ρ = 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 ) n = 10.288 m/s 2
40°
aB/A = aB − aA
a B/A = (a B )t + (a B ) n − (a A )t − (a A ) n
=6
50° + 10.288
40° + 10 → + 0
= (6cos 50° + 10.288cos 40° + 10)
+ (6sin 50° − 10.288sin 40°)
= 21.738 → + 2.017
a B/A = 21.8 m/s 2
5.3° 
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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 radius. Knowing that at the given instant the
speed of B is being decreased at the rate of 3 m/s 2 , 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)
v B = v A + v B/A
We have
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 α sin 60°
α = 68.9°
v B/A = 501 km/h
(b)
First note
Now
a A = 8 m/s 2
( aB ) n =
v 2B
ρB
=
Then
60°
(150 m/s) 2
300 m
(a B ) n = 75 m/s 2

(a B )t = 3 m/s 2
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
a B = 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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PROBLEM 11.143
From a photograph of a homeowner using a snowblower, it
is determined that the radius of curvature of the trajectory of
the snow was 30 ft as the snow left the discharge chute at A.
Determine (a) the discharge velocity v A of the snow, (b) the
radius of curvature of the trajectory at its maximum height.
SOLUTION
(a) The acceleration vector is 32.2 ft/s .
At Point A, tangential and normal components of a are as shown
in the sketch.
an = a cos 40° = 32.2 cos 40° = 24.67 ft/s 2
v A2 = ρ A (a A ) n = (30)(24.67) = 740.0 ft 2 /s 2
v A = 27.2 ft/s
40° 
vx = 27.20 cos 40° = 20.84 ft/s
(b)
At maximum height,
v = vx = 20.84 ft/s
an = g = 32.2 ft/s 2 ,
ρ =
v2
(20.84) 2
=
32.2
an
ρ = 13.48 ft 
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PROBLEM 11.144
A basketball is bounced on the ground at Point A and rebounds with a velocity v A of
magnitude 2.5 m/s as shown. 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
or
(a A ) n =
ρA =
v 2A
ρA
(2.5 m/s) 2
(9.81 m/s 2 ) sin 15°
ρ A = 2.46 m 
or
(b)
We have
( aB ) n =
vB2
ρB
where Point B is the highest point of the trajectory, so that
vB = (v A ) x = v A sin 15°

Then
ρB =
[(2.5 m/s) sin 15°]2
= 0.0427 m
9.81 m/s 2
or
ρ B = 42.7 mm 
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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
or
(a A ) n =
ρA =
v 2A
ρA
(50 m/s)2
(9.81 m/s 2 ) cos 25°
ρ A = 281 m 
or
(b)
We have
( aB ) n =
vB2
ρB
where Point B is the highest point of the trajectory, so that
vB = (v A ) x = v A cos 25°
Then
or
ρB =
[(50 m/s) cos 25°]2
9.81 m/s 2
ρ B = 209 m 
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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 = v0 t
Vertical motion:
y0 = 0,
(v y ) = 0
v y = − gt
1
y = − gt 2
2
2h
,
g
t=
where h is the vertical distance fallen.
| v y| = 2 gh
Speed:
v 2 = vx2 + v 2y = v02 + 2 gh
Direction of velocity.
cos θ =
v0
v
Direction of normal acceleration.
an = g cos θ =
Radius of curvature:
At Point B,
ρ=
gv0 v 2
=
v
ρ
v3
gv0
hB = 1 m; xB = 7 m
tB =
(2)(1 m)
= 0.45152 s
9.81 m/s 2
xB = v0t B
v0 =
xB
7m
=
= 15.504 m/s
t B 0.45152 s
vB2 = (15.504) 2 + (2)(9.81)(1) = 259.97 m 2 /s 2
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PROBLEM 11.146 (Continued)
(a)
Radius of curvature at Point B.
ρB =
At Point C
(259.97 m 2 /s 2 )3/ 2
(9.81 m/s 2 )(15.504 m/s)
ρ B = 27.6 m 
hC = 1 m + 2 m = 3 m
vC2 = (15.504) 2 + (2)(9.81)(3) = 299.23 m 2 /s 2
(b)
Radius of curvature at Point C.
ρC =
(299.23 m 2 /s 2 )3/2
(9.81 m/s 2 )(15.504 m/s)
ρC = 34.0 m 
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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 A = g = 9.81 m/s 2
(a) At Point A.
Sketch tangential and normal components of acceleration at A.
(a A ) n = g cos 50°
ρA =
vA2
(2) 2
=
(a A ) n
9.81cos 50°
ρ A = 0.634 m 
(b) At Point B, 1 meter below Point A.
Horizontal motion: (vB ) x = (v A ) x = 2 cos 50° = 1.286 m/s
Vertical motion:
(vB ) 2y = (v A ) 2y + 2a y ( yB − y A )
= (2 cos 40°)2 + (2)(−9.81)(−1)
= 21.97 m 2 /s 2
(vB ) y = 4.687 m/s
tan θ =
(vB ) y
(vB ) x
=
4.687
,
1.286
or
θ = 74.6°
aB = g cos 74.6°
(vB ) 2x + (vB ) 2y
vB 2
ρB =
=
(aB )n
g cos 74.6°
=
(1.286) 2 + 21.97
9.81cos 74.6°
ρ B = 9.07 m 
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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 2A
ρA
or
4

v A2 =  (9.81 m/s 2 )  (25 m)
5


or
v A = 14.0071 m/s
vA = 14.01 m/s
(b)
We have
Where
Then
or
( aB ) n =
vB2
ρB
vB = ( v A ) x =
ρB
36.9° 
4
vA
5
( 4 × 14.0071 m/s )
= 5
2
9.81 m/s 2
ρ B = 12.80 m 
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PROBLEM 11.149
A child throws a ball from Point A with an initial velocity v A of
20 m/s at an angle of 25° with the horizontal. Determine the
velocity of the ball at the points of the trajectory described by the
ball where the radius of curvature is equal to three-quarters of its
value at A.
SOLUTION
Assume that Points B and C are the points of interest, where yB = yC and vB = vC .
Now
(a A ) n =
v 2A
ρA
or
ρA =
v 2A
g cos 25°
Then
ρB =
v A2
3
3
ρA =
4
4 g cos 25°
vB2
We have
( aB ) n =
where
(aB ) n = g cos θ
so that
v A2
vB2
3
=
4 g cos 25° g cos θ
or
vB2 =
ρB
3 cos θ 2
vA
4 cos 25°
(1)
Noting that the horizontal motion is uniform, we have
( v A ) x = ( vB ) x
where
Then
or
(v A ) x = v A cos 25°
(vB ) x = vB cos θ
v A cos 25° = vB cos θ
cos θ =
vA
cos 25°
vB
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PROBLEM 11.149 (Continued)
Substituting for cos θ in Eq. (1), we have
or
vB2 =
 v A2
3  vA
 cos 25° 
4  vB
 cos 25°
vB3 =
3 3
vA
4
3
vB = 3 v A = 18.17 m/s
4
4
cos 25°
3
θ = ± 4.04°
cos θ = 3
and
v B = 18.17 m/s
4.04° 
v B = 18.17 m/s
4.04° 
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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 θ the angle formed by
the trajectory and the horizontal at a given
Point C, show that the radius of curvature
of the trajectory at C is ρ = ρ min /cos3 θ .
SOLUTION
For the arbitrary Point C, we have
(aC ) n =
ρC =
or
vC2
ρC
vC2
g cos θ
Noting that the horizontal motion is uniform, we have
(v A ) x = (vC ) x
(v A ) x = v0 cos α
where
v0 cos α = vC cos θ
Then
cos α
v0
cos θ
or
vC =
so that
1
ρC =
g cos θ
(a)
2
 cos α 
v 2 cos 2 α
v0  = 0

g cos3 θ
 cos θ 
In the expression for ρC , v0 , α , and g are constants, so that ρC is minimum where cos θ is
maximum. By observation, this occurs at Point B where θ = 0.
ρ min = ρ B =
(b)
(vC ) x = vC cos θ
ρC =
1
cos3 θ
ρC =
ρ min
cos3 θ
v02 cos 2 α
g
 v02 cos 2 α

g

Q.E.D.

Q.E.D.




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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 ωnt)i + ctj + (Rt sin ωnt)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 ωn t − ωn t sin ωn t )i + cj + R (sin ωn t + ωn t cos ωn t )k
dt
and
a=
dv
= R − ωn sin ωn t − ωn sin ωn t − ωn2t cos ωn t i
dt
(
)
(
)
+ R ωn cos ωn t + ωn cos ωn t − ωn2 t sin ωn t k
or
Now
a = ωn R [−(2 sin ωn t + ωn t cos ωn t )i + (2 cos ωn t − ωn t sin ωn t ) k ]
v 2 = R 2 (cos ωn t − ωn t sin ωn t )2 + c 2 + R 2 (sin ωn t + ωn t cos ωn t )2
(
)
= R 2 1 + ωn2 t 2 + c 2
Then
and
Now
At t = 0:
(
)
1/2
v =  R 2 1 + ωn2 t 2 + c 2 


R 2ωn2 t
dv
=
1/ 2
dt  2
R 1 + ωn2 t 2 + c 2 


(
2
a =
at2
+
an2
)
2
2
 dv   v 
=   +  
 dt   ρ 
2
dv
=0
dt
a = ωn R(2 k ) or a = 2ωn R
v2 = R2 + c2
Then, with
we have
or
dv
= 0,
dt
a=
2ωn R =
v2
ρ
R 2 + c2
ρ
ρ=
R 2 + c2

2ωn R
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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


2
dv


t
t
+
1
−
a=
= 3(− sin t − sin t − t cos t )i + 3


dt
2
t
+
1

(
)
 3t 
dr
= 3(cos t − t sin t )i +  2
j + (sin t + t cos t )k
 t + 1 
dt


t
t2 + 1

 j


+ (cos t + cos t − t sin t )k
= − 3(2sin t + t cos t )i + 3
1
j
2
t
+
(
1)1/2
+ (2 cos t − t sin t )k
v 2 = 9 (cos t − t sin t )2 + 9
Then
t2
+ (sin t + t cos t )2
t2 + 1
Expanding and simplifying yields
v 2 = t 4 + 19t 2 + 1 + 8(cos 2 t + t 4 sin 2 t ) − 8(t 3 + t )sin 2t
Then
v = [t 4 + 19t 2 + 1 + 8(cos 2 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)sin 2t + 2(t 3 + t ) cos 2t ]
=
dt
2[t 4 + 19t 2 + 1 + 8(cos 2 t + t 4 sin 2 t ) − 8(t 3 + t ) sin 2t ]1/ 2
Now
2
2
 dv   v 
a 2 = at2 + an2 =   +  
 dt   ρ 
2
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PROBLEM 11.152* (Continued)
At t = 0:
a = 3j + 2k
or
a = 13 ft/s 2
dv
=0
dt
v 2 = 9 (ft/s) 2
Then, with
dv
= 0,
dt
we have
a=
or
ρ=
v2
ρ
9 ft 2 /s 2
13 ft/s 2
ρ = 2.50 ft 
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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: (umean )orbit = 107 Mm/h.
SOLUTION
g = 274 m/s 2 ,
For the sun,
R=
and
Given that an =
gR 2
v2
and
that
for
a
circular
orbit
=
a
n
r
r2
Eliminating an and solving for r,
r =
gR 2
v2
v = 107 × 106 m/h = 29.72 × 103 m/s
For the planet Earth,
Then
1
1
D =   (1.39 × 109 ) = 0.695 × 109 m
2
2
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: (umean )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:
=
a
n
r
r2
gR 2
v2
Eliminating an and solving for r,
r =
For the planet Saturn,
v = 34.7 × 106 m/h = 9.639 × 103 m/s
Then,
r =
(274)(0.695 × 109 ) 2
= 1.425 × 1012 m
(9.639 × 103 )2
r = 1425 Gm 
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PROBLEM 11.155
Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a
circular orbit 100 mi above the surface of the planet. (See information given in Problems 11.153–11.154).
Venus: g = 29.20 ft/s 2 , R = 3761 mi.
SOLUTION
From Problems 11.153 and 11.154,
an =
gR 2
r2
For a circular orbit,
an =
v2
r
v= R
Eliminating an and solving for v,
For Venus,
g
r
g = 29.20 ft/s 2
R = 3761 mi = 19.858 × 106 ft.
r = 3761 + 100 = 3861 mi = 20.386 × 106 ft
Then,
v = 19.858 × 106
29.20
= 23.766 × 103 ft/s
20.386 × 106
v = 16200 mi/h 
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PROBLEM 11.156
Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a
circular orbit 100 mi above the surface of the planet. (See information given in Problems 11.153–11.154).
Mars: g = 12.17 ft/s 2 , R = 2102 mi.
SOLUTION
From Problems 11.153 and 11.154,
an =
gR 2
r2
For a circular orbit,
an =
v2
r
v= R
Eliminating an and solving for v,
For Mars,
g
r
g = 12.17 ft/s 2
R = 2102 mi = 11.0986 × 106 ft
r = 2102 + 100 = 2202 mi = 11.6266 × 106 ft
Then,
v = 11.0986 × 106
12.17
= 11.35 × 103 ft/s
11.6266 × 106
v = 7740 mi/h 
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PROBLEM 11.157
Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a
circular orbit 100 mi above the surface of the planet. (See information given in Problems 11.153–11.154).
Jupiter: g = 75.35 ft/s 2 , R = 44, 432 mi.
SOLUTION
From Problems 11.153 and 11.154,
an =
gR 2
r2
For a circular orbit,
an =
v2
r
v= R
Eliminating an and solving for v,
For Jupiter,
g
r
g = 75.35 ft/s 2
R = 44432 mi = 234.60 × 106 ft
r = 44432 + 100 = 44532 mi = 235.13 × 106 ft
Then,
v = (234.60 × 106 )
75.35
= 132.8 × 103 ft/s
6
235.13 × 10
v = 90600 mi/h 
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PROBLEM 11.158
A satellite is traveling in a circular orbit around Mars at an altitude of 300 km. After the altitude of the satellite
is adjusted, it is found that the time of one orbit has increased by 10 percent. Knowing that the radius of Mars
is 3382 km, determine the new altitude of the satellite. (See information given in Problems 11.153–11.155.)
SOLUTION
an = g
We have
Then
g
R2
r2
and an =
v2
r
R2 v2
=
r
r2
v=R
g
r
where
r =R+h
The circumference s of a circular orbit is equal to
s = 2π r
Assuming that the speed of the satellite in each orbit is constant, we have
s = vtorbit
Substituting for s and v
2π r = R
torbit =
=
Now
g
torbit
r
2π r 3/2
R g
2π ( R + h)3/2
R
g
(torbit ) 2 = 1.1(torbit )1
2π ( R + h2 )3/2
2π ( R + h1 )3/2
= 1.1
R
R
g
g
h2 = (1.1)2/3 ( R + h1 ) − R
= (1.1)2/3 (3382 + 300) km − (3382 km)
h2 = 542 km 
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PROBLEM 11.159
Knowing that the radius of the earth is 6370 km, determine the time of one orbit of the Hubble Space
Telescope, knowing that the telescope travels in a circular orbit 590 km above the surface of the earth.
(See information given in Problems 11.153–11.155.)
SOLUTION
an = g
We have
Then
or
g
R2
r2
and an =
v2
r
R2 v2
=
r
r2
v=R
g
r
where
r =R+h
The circumference s of the circular orbit is equal to
s = 2π r
Assuming that the speed of the telescope is constant, we have
s = vtorbit
Substituting for s and v
2π r = R
or
torbit =
=
g
torbit
r
2π r 3/2
R g
2π
[(6370 + 590) km]3/2
1h
×
−3
2 1/2
6370 km [9.81 × 10 km/s ]
3600 s
or
torbit = 1.606 h 
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PROBLEM 11.160
Satellites A and B are traveling in the same plane in circular orbits around
the earth at altitudes of 120 and 200 mi, respectively. If at t = 0 the
satellites are aligned as shown and knowing that the radius of the earth is
R = 3960 mi, determine when the satellites will next be radially aligned.
(See information given in Problems 11.153–11.155.)
SOLUTION
an = g
We have
Then
g
R2
r2
R2 v2
=
r
r2
and an =
or
v2
r
v=R
g
r
r =R+h
where
The circumference s of a circular orbit is
equal to
s = 2π r
Assuming that the speeds of the satellites are constant, we have
s = vT
Substituting for s and v
2π r = R
or
Now
T=
g
T
r
2π r 3/ 2 2π ( R + h)3/2
=
R g
R
g
hB > hA  (T ) B > (T ) A
Next let time TC be the time at which the satellites are next radially aligned. Then, if in time TC satellite B
completes N orbits, satellite A must complete ( N + 1) orbits.
Thus,
TC = N (T ) B = ( N + 1)(T ) A
or
 2π ( R + hB )3/2 
 2π ( R + hA )3/2 
(
1)
N
=
N
+



g
g
 R

 R

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PROBLEM 11.160 (Continued)
N=
or
=
Then
( R + hA )3/2
( R + hB )
3/ 2
− ( R + hA )
1
(
3960 + 200
3960 +120
)
3/2
TC = N (T ) B = N
−1
3/2
=
1
(
R + hB
R + hA
)
3/2
−1
= 33.835 orbits
2π ( R + hB )3/2
R
g
[(3960 + 200) mi] × 1 h
2π
3960 mi 32.2 ft/s 2 × 1 mi 1/2 3600s
5280 ft
3/2
= 33.835
(
)
TC = 51.2 h 
or
Alternative solution
From above, we have (T ) B > (T ) A . Thus, when the satellites are next radially aligned, the angles θ A and θ B
swept out by radial lines drawn to the satellites must differ by 2π . That is,
θ A = θ B + 2π
For a circular orbit
s = rθ
From above
s = vt and v = R
Then
θ=
At time TC :
or
R g
( R + hA )
3/2
TC =
TC =
=
g
r
R g
R g
s vt 1 
g
= = R
t
 t = 3/2 t =

r r r
r 
r
( R + h)3/2
R g
( R + hB )3/ 2
TC + 2π
2π

1
1
R g
−
 ( R + hA )3/ 2 ( R + hB )3/ 2 
2π
(
1 mi
(3960 mi) 32.2 ft/s 2 × 5280
ft
×
×
1
[(3960 + 120) mi ]3/ 2
1
−
)
1/2
1
[(3960 + 200) mi ]3/ 2
1h
3600 s
or
TC = 51.2 h 
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PROBLEM 11.161
The oscillation of rod OA about O is defined by the relation θ = (3/π )(sin π t ),
where θ and t are expressed in radians and seconds, respectively. Collar B slides
along the rod so that its distance from O is r = 6(1 − e−2t ) where r and t
are expressed in inches and seconds, respectively. When t = 1 s, determine (a) the
velocity of the collar, (b) the acceleration of the collar, (c) the acceleration of the
collar relative to the rod.
SOLUTION
Calculate the derivatives with respect to time.
3
r = 6 − 6e −2t in.
θ=
r = 12e −2t in/s
θ = 3cosπ t rad/s
r = −24e−2t in/s 2
θ = −3π sin π t rad/s2
π
sin π t rad
At t = 1 s,
(a)
3
r = 6 − 6e −2 = 5.1880 in.
θ=
r = 12e −2 = 1.6240 in/s
θ = 3cos π = −3 rad/s

r = −24e−2 = −3.2480 in/s 2
θ = −3π sin π = 0
π
sin π = 0
Velocity of the collar.
v = rer + rθeθ = 1.6240 e r + (5.1880)(−3)eθ
v = (1.624 in/s)er + (15.56 in/s)eθ 
(b)
Acceleration of the collar.
a = (r − rθ 2 )e r + (rθ + 2rθ)eθ
= [ −3.2480 − (5.1880)( −3) 2 ]er + (5.1880)(0) + (2)(1.6240)(−3)]eθ
(−49.9 in/s 2 )er + (−9.74 in/s 2 )eθ 
(c)
Acceleration of the collar relative to the rod.
a B /OA = 
re r
a B /OA = (−3.25 in/s 2 )er 
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PROBLEM 11.162
The rotation of rod OA about O is defined by the relation θ = t 3 − 4t , where
θ and t are expressed in radians and seconds, respectively. Collar B slides along
the rod so that its distance from O is r = 2.5t 3 − 5t 2 , where r and t are expressed
in inches and seconds, respectively. When t = 1 s, determine (a) the velocity of the
collar, (b) the acceleration of the collar, (c) the radius of curvature of the path of
the collar.
SOLUTION
Calculate the derivatives with respect to time.
r = 2.5t 3 − 5t 2
θ = t 3 − 4t
r = 7.5t 2 − 10t
θ = 3t 2 − 4

r = 15t − 10
θ = 6t
At t = 1 s,
(a)
r = 2.5 − 5 = −2.5 in.
θ = 1 − 4 = −3 rad
r = 7.5 − 10 = −2.5 in./s
θ = 3 − 4 = −1 rad/s
r = 15 − 10 = 5 in./s 2
θ = 6 rad/s 2
Velocity of the collar.
v = rer + rθeθ = −2.5er + (−2.5)( −1)eθ
v = ( −2.50 in./s)er + (2.50 in./s)eθ 
v = (2.50) 2 + (2.50) 2 = 3.5355 in./s
Unit vector tangent to the path.
et =
(b)
v
= −0.70711e r + 0.70711eθ
v
Acceleration of the collar.
a = (r − rθ 2 )e r + (rθ + 2rθ)eθ
= [5 − (−2.5)(−1)2 ]er + [(−2.5)(6) + (2)(−2.5)(−1)]eθ
a = (7.50 in/s 2 )er + (−10.00 in/s 2 )eθ 
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PROBLEM 11.162 (Continued)
Magnitude:
a = (7.50) 2 + (10.00) 2 = 12.50 in./s 2
at = aet
Tangential component:
at = (7.50)(−0.70711) + (−10.00)(0.70711) = −12.374 in./s 2
Normal component:
(c)
an = a 2 − at2 = 1.7674 in./s 2
Radius of curvature of path.
an =
ρ=
v2
ρ
v 2 (3.5355 in./s) 2
=
an
1.7674 in./s 2
ρ = 7.07 in. 
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PROBLEM 11.163
The path of particle P is the ellipse defined by the relations
r = 2/(2 − cos π t ) and θ = π t , where r is expressed in meters, t is
in seconds, and θ is in radians. Determine the velocity and the
acceleration of the particle when (a) t = 0, (b) t = 0.5 s.
SOLUTION
We have
r=
2
2 − cos π t
θ = πt
Then
r =
−2π sin π t
(2 − cos π t )2
θ = π
and
r = −2π
π cos π t (2 − cos π t ) − sin π t (2π sin π t )
(2 − cos π t )3
= −2π 2
(a)
At t = 0:
Now
2cos π t − 1 − sin 2 π t
(2 − cos π t )3
r=2m
θ =0
r = 0
θ = π rad/s
r = −2π 2 m/s 2
θ = 0
v = rer + rθeθ = (2)(π )eθ
v = (2π m/s)eθ 
or
and
θ = 0
a = (r − rθ 2 )e r + (rθ + 2rθ)eθ
= [ −2π 2 − (2)(π )2 ]er
a = −(4π 2 m/s 2 )er 
or
(b)
At t = 0.5 s:
θ=
r =1 m
r =
−2π
π
= − m/s
2
2
(2)
r = −2π 2
−1 − 1 π 2
=
m/s 2
3
2
(2)
π
rad
2
θ = π rad/s
θ = 0
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PROBLEM 11.163 (Continued)
Now
 π
v = rer + rθeθ =  −  e r + (1)(π )eθ
 2
π

v = −  m/s  er + (π m/s)eθ 
2


or
and
a = (r − rθ 2 )e r + (rθ + 2rθ)eθ
π 2

  π 
=
− (1)(π ) 2  e r +  2  −  (π )  eθ
  2 
 2

or
π2

a = − 
m/s 2  er − (π 2 m/s 2 )eθ 
 2

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PROBLEM 11.164
The two-dimensional motion of a particle is defined by the relations r = 2a cos θ and θ = bt 2 /2, where a and b
are constants. Determine (a) the magnitudes of the velocity and acceleration at any instant, (b) the radius of
curvature of the path. What conclusion can you draw regarding the path of the particle?
SOLUTION
(a)
We have
r = 2a cos θ
1
θ = bt 2
2
Then
r = −2aθ sin θ
θ = bt
and

r = −2a (θ sin θ + θ 2 cos θ )
θ = b
Substituting for θ and θ
r = −2abt sin θ

r = −2ab(sin θ + bt 2 cos θ )
Now
Then
vr = r = −2abt sin θ
vθ = rθ = 2abt cos θ
v = vr2 + vθ2 = 2abt[( − sin θ ) 2 + (cos θ ) 2 ]1/2
v = 2abt 
or
Also
ar = r − rθ 2 = −2ab(sin θ + bt 2 cos θ ) − 2ab 2 t 2 cos θ
= −2ab(sin θ + 2bt 2 cos θ )
and
aθ = rθ + 2rθ = 2ab cos θ − 4ab 2 t 2 sin θ
= −2ab(cos θ − 2bt 2 sin θ )
Then
a = ar2 + aB2 = 2ab[(sin θ + 2bt 2 cos θ ) 2
+ (cos θ − 2bt 2 sin θ ) 2 ]1/2
a = 2ab 1 + 4b 2 t 4 
or
(b)
2
at2
+
an2
2
2
 dv   v 
=   +  
 dt   ρ 
Now
a =
Then
dv d
= (2abt ) = 2ab
dt dt
2
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PROBLEM 11.164 (Continued)
so that
or
(
2ab 1 + 4b 2 t 4
)
2
= (2ab) 2 + an2
4a 2b 2 (1 + 4b 2t 4 ) = 4a 2 b 2 + an2
or
an = 4ab 2 t 2
Finally
an =
v2
(2abt ) 2
ρ=
ρ
4ab 2 t 2
or
ρ =a 
Since the radius of curvature is a constant, the path is a circle of radius a.

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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 θ ) and that θ = kt , determine the
velocity and acceleration of P when (a) θ = 0, (b) θ = 90°.
SOLUTION
2b
θ = kt
1 + cos kt
2bk sin kt
θ = k θ = 0
r =
(1 + cos kt ) 2
2bk
[(1 + cos kt )2 k cos kt + (sin kt )2(1 + cos kt )( k sin kt )]
r =
(1 + cos kt ) 4
r=
(a)
When θ = kt = 0:
2bk
1
[(2)2 k (1) + 0] = bk 2
4
2
(2)
r =b
r = 0

r=
θ =0
θ = k
θ = 0
vθ = rθ = bk
vr = r = 0
v = bk eθ 
1
1

ar = 
r − rθ 2 = bk 2 − bk 2 = − bk 2 
2
2


aθ = rθ + 2rθ = b(0) + 2(0) = 0

(b)
1
a = − bk 2 er 
2
When θ = kt = 90°:
r = 2b
r = 2bk
θ = 90°
θ = k
vr = r = 2bk
2bk
[0 + 2k ] = 4bk 2
19
θ = 0
r =
vθ = rθ = 2bk
ar = r − rθ 2 = 4bk 2 − 2bk 2 = 2bk 2
a = rθ + 2rθ = 2b(0) + 2(2bk )k = 4bk 2
θ
v = 2bk er + 2bk eθ 
a = 2bk 2 e r + 4bk 2 eθ 
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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 θ,
(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 θ
r = −2b sin θ θ
Since θ = constant, θ = 0
r = −2b cos θ θ 2
ar = r − rθ 2 = −2b cos θ θ 2 − (2b cos θ )θ 2
a = −4b cos θ θ 2
r
aθ = rθ + 2rθ = (2b cos θ )(0) + 2(−2b sin θ )θ 2
a = −4b sin θ θ 2
θ
a = ar2 + aθ2 = 4bθ 2 (− cos θ ) 2 + (− sin θ )2
a = 4bθ 2
Since both b and θ are constant, we find that
a = constant 
γ = tan −1
 −4b sin θ θ 2
aθ
= tan −1 
2
ar
 −4b cos θ θ



γ = tan −1 (tan θ )
γ =θ
Thus, a is directed toward A 
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PROBLEM 11.167
To study the performance of a racecar, a high-speed camera is
positioned at Point A. The camera is mounted on a mechanism
which permits it to record the motion of the car as the car
travels on straightway BC. Determine (a) the speed of the car
in terms of b, θ , and θ, (b) the magnitude of the acceleration
in terms of b, θ , θ, and θ.
SOLUTION
(a)
We have
r=
b
cos θ
Then
r =
bθ sin θ
cos 2 θ
We have
v 2 = vr2 + vθ2 = (r)2 + (rθ) 2
2
or
 bθ sin θ   bθ 2
= 
 + 

2
 cos θ   cos θ 
 b2θ 2
b2θ 2  sin 2θ
=
+
1
 =
2 
2
4
cos θ  cos θ
 cos θ
bθ
v=±
cos 2θ
For the position of the car shown, θ is decreasing; thus, the negative root is chosen.
v=−
bθ

cos 2θ
v=−
bθ

cos 2θ
Alternative solution.
From the diagram
or
r = −v sin θ
bθ sin θ
= −v sin θ
cos 2θ
or



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PROBLEM 11.167 (Continued)
(b)
For rectilinear motion
a=
dv
dt
Using the answer from Part a
v=−
Then
a=
bθ
cos 2θ
d 
bθ
 −
dt  cos 2θ
= −b



θ cos 2θ − θ(−2θ cos θ sin θ )
cos 4θ
a=−
or
b
(θ + 2θ 2 tan θ ) 
cos 2θ
Alternative solution
From above
Then
Now
where
and
b
bθ sin θ
r =
cos θ
cos 2 θ
(θ sin θ + θ 2 cos θ )(cos 2θ ) − (θ sin θ )(−2θ cos θ sin θ )
r = b
cos 4θ
 θ sin θ θ 2 (1 + sin 2 θ ) 
= b
+

2
cos3θ
 cos θ

r=
a 2 = ar2 + aθ2
 θ sin θ θ 2 (1 + sin 2 θ )  bθ 2
+
ar = r − rθ 2 = b 
−
2
cos 2θ
 cos θ
 cos θ
b  
2θ 2 sin 2 θ 
=
+
θ
sin
θ


cos θ 
cos 2θ 
b sin θ 
ar =
(θ + 2θ 2 tan θ )
2
cos θ
aθ = rθ + 2rθ =
=
Then
bθ
bθ 2 sin θ
+2
cos θ
cos 2θ
b cos θ 
(θ + 2θ tan θ )
cos 2θ
a=±
b
(θ + 2θ 2 tan θ )[(sin θ ) 2 + (cos θ )2 ]1/ 2
2
cos θ
For the position of the car shown, θ is negative; for a to be positive, the negative root is chosen.
b
a=−
(θ + 2θ 2 tan θ ) 
cos 2 θ
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PROBLEM 11.168
After taking off, a helicopter climbs in
a straight line at a constant angle β . Its
flight is tracked by radar from Point A.
Determine the speed of the helicopter
in terms of d, β , θ , and θ.
SOLUTION
From the diagram
r
d
=
sin (180° − β ) sin ( β − θ )
or
d sin β = r (sin β cos θ − cos β sin θ )
tan β
tan β cos θ − sin θ
or
r=d
Then
r = d tan β
−(− tan β sin θ − cos θ ) 
θ
(tan β cos θ − sin θ )2
tan β sin θ + cos θ
= dθ tan β
(tan β cos θ − sin θ ) 2
From the diagram
vr = v cos ( β − θ )
where
vr = r
Then
dθ tan β
tan β sin θ + cos θ
= v(cos β cos θ + sin β sin θ )
(tan β cos θ − sin θ )2
= v cos β (tan β sin θ + cos θ )
v=
or
dθ tan β sec β

(tan β cosθ − sin θ )2
Alternative solution.
We have
v 2 = vr2 + vθ2 = (r) 2 + ( rθ) 2
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PROBLEM 11.168 (Continued)
Using the expressions for r and r from above

tan β sin θ + cos θ 
v =  dθ tan β

(tan β cos θ − sin θ ) 2 

2
1/ 2
or
 (tan β sin θ + cos θ )2

dθ tan β
v=±
+ 1

2
(tan β cos θ − sin θ )  (tan β cos θ − sin θ )

1/ 2


tan 2 β + 1
dθ tan β
=±

2
(tan β cos θ − sin θ )  (tan β cos θ − sin θ ) 
Note that as θ increases, the helicopter moves in the indicated direction. Thus, the positive root is chosen.
v=
dθ tan β sec β

(tan β cos θ − sin θ )2
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PROBLEM 11.169
At the bottom of a loop in the vertical plane, an
airplane has a horizontal velocity of 315 mi/h
and is speeding up at a rate of 10 ft/s2. The
radius of curvature of the loop is 1 mi. The
plane is being tracked by radar at O. What are
r , θ and θ for this
the recorded values of r, 
instant?
SOLUTION
Geometry. The polar coordinates are
r = (2400) 2 + (1800)2 = 3000 ft
Velocity Analysis.
 1800 
 = 36.87°
 2400 
θ = tan −1 
v = 315 mi/h = 462 ft/s
vr = 462 cos θ = 369.6 ft/s
vθ = −462sin θ = −277.2 ft/s
vr = r
vθ = rθ
θ =
r = 370 ft/s 
vθ
277.2
=−
r
3000
θ = −0.0924 rad/s 
Acceleration analysis.
at = 10 ft/s 2
an =
v2
ρ
=
(462) 2
= 40.425 ft/s 2
5280
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PROBLEM 11.169 (Continued)
ar = at cos θ + an sin θ = 10 cos 36.87° + 40.425 sin 36.87° = 32.255 ft/s 2
aθ = − at sin θ + an cos θ = −10 sin 36.87° + 40.425 cos 36.87° = 26.34 ft/s 2
ar = 
r − rθ 2 
r = ar + rθ 2

r = 32.255 + (3000)( −0.0924) 2
aθ = rθ + 2rθ
a
2rθ
θ = θ −
r
r
26.34 (2)(369.6)(−0.0924)
=
−
3000
3000

r = 57.9 ft/s 2 
θ = 0.0315 rad/s 2 
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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 ω. At the instant when β = 60°,
r and θ. Express your answers in terms
determine (a) r and θ, (b) 
of d and ω.
SOLUTION
Looking at d and β as polar coordinates with d = 0,
vβ = d β = d ω,
vd = d = 0
aβ = d β + 2d β = 0,
r = d 3 for angles shown.
Geometry analysis:
(a)
Velocity analysis:
ad = d − d β 2 = −d ω 2
Sketch the directions of v, er and eθ.
vr = r = v ⋅ er = d ω cos120°
1
r = − d ω 
2
vθ = rθ = v ⋅ eθ = d ω cos 30°
θ =
(b)
Acceleration analysis:
3
dω cos 30° dω 2
=
r
d 3
θ =
1
ω
2
Sketch the directions of a, er and eθ.
ar = a ⋅ e r = a cos150° = −
3
dω 2
2
3

r − rθ2 = −
d ω2
2

r =−
3
3
1 
dω 2 + rθ 2 = −
dω 2 + d 3  ω 
2
2
2 
2
r = −
3
dω 2 
4
1
aθ = a ⋅ eθ = d ω 2 cos120° = − d ω 2
2
a = rθ + 2rθ
θ
θ =
1
(aθ − 2rθ) =
r
1
3d
 1
 1
 1  
2
 − d ω − (2)  − d ω  ω  
2
2

 2  

θ = 0 
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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 θ = 60° to
the position θ = 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,
a high-speed camera is positioned at Point A. The camera
is mounted on a mechanism which permits it to record the
motion of the car as the car travels on straightway BC.
Determine (a) the speed of the car in terms of b, θ , and
θ, (b) the magnitude of the acceleration in terms of b, θ ,
θ, and θ.
SOLUTION
From the diagram:
Δr12 = 25 tan 60° − 25 tan 35°
= 25.796 m
Now
vave =
=
Δr12
Δt12
25.796 m
0.5 s
= 51.592 m/s
or
vave = 185.7 km/h 
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PROBLEM 11.172
For the helicopter of Problem 11.168, it was found that when the helicopter was at B, the distance and the
angle of elevation of the helicopter were r = 3000 ft and θ = 20°, respectively. Four seconds later, the radar
station sighted the helicopter at r = 3320 ft and θ = 23.1°. Determine the average speed and the angle of
climb β of the helicopter during the 4-s interval.
PROBLEM 11.168 After taking off, a helicopter climbs in a straight line at a constant angle β . Its flight is
tracked by radar from Point A. Determine the speed of the helicopter in terms of d, β , θ , and θ.
SOLUTION
We have
r0 = 3000 ft
r4 = 3320 ft
θ0 = 20°
θ 4 = 23.1°
From the diagram:
Δr 2 = 30002 + 33202
− 2(3000)(3320) cos (23.1° − 20°)
or
Δr = 362.70 ft
Now
vave =
Δr
Δt
362.70 ft
=
4s
= 90.675 ft/s
vave = 61.8 mi/h 
or
Also,
or
Δr cos β = r4 cos θ 4 − r0 cos θ0
cos β =
3320 cos 23.1° − 3000 cos 20°
362.70
or
β = 49.7° 
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PROBLEM 11.173
A particle moves along the spiral shown; determine the magnitude of the velocity
of the particle in terms of b, θ , and θ.
SOLUTION
Hyperbolic spiral.
r=
b
θ
dr
b dθ
b
=− 2
= − 2 θ
dt
θ dt
θ
b 
b
vr = r = − 2 θ
vθ = rθ = θ
r =
θ
θ
2
 1  1
v = vr2 + vθ2 = bθ  − 2  +  
 θ  θ 
bθ
= 2 1+θ 2
2
θ
v=
b
θ2
1 + θ 2 θ 
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PROBLEM 11.174
A particle moves along the spiral shown; determine the magnitude of the velocity
of the particle in terms of b, θ , and θ.
SOLUTION
Logarithmic spiral.
r = ebθ
r =
dr
dθ
= bebθ
= bebθ θ
dt
dt
vr = r = bebθ θ vθ = rθ = ebθ θ
v=
vr2
+ vθ = e θ b 2 + 1
2
bθ
v = ebθ 1 + b 2 θ 
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PROBLEM 11.175
A particle moves along the spiral shown. Knowing that θ is constant and
denoting this constant by ω , determine the magnitude of the acceleration of
the particle in terms of b, θ , and ω.
SOLUTION
b
Hyperbolic spiral.
r=
From Problem 11.173
r = −
θ

r =−
b
θ2
θ
b  2b  2
θ + 3θ
2
θ
θ
b
2b
b
ar = 
r − rθ 2 = − 2 θ + 3 θ 2 − θ 2
θ
θ
θ
b
b
b
 b 
aθ = rθ + 2rθ = θ + 2  − 2 θ θ = θ − 2 2 θ 2
θ
θ
θ
 θ 
Since θ = ω = constant, θ = 0, and we write:
b 2 bω 2
2
−
ω
ω = 3 (2 − θ 2 )
θ
θ3
θ
2
b
bω
aθ = −2 2 ω 2 = − 3 (2θ )
θ
θ
ar = +
2b
a = ar2 + aθ2 =
bω 2
θ
3
(2 − θ 2 ) 2 + (2θ ) 2 =
bω 2
θ
3
4 − 4θ 2 + θ 4 + 4θ 2
a=
bω 2
θ3
4 +θ 4 
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PROBLEM 11.176
A particle moves along the spiral shown. Knowing that θ is constant and
denoting this constant by ω , determine the magnitude of the acceleration of
the particle in terms of b, θ , and ω.
SOLUTION
Logarithmic spiral.
r = ebθ
dr
= bebθ θ
dt
r = bebθ θ + b 2 ebθ θ 2 = bebθ (θ + bθ 2 )
r =
ar = r − rθ 2 = bebθ (θ + bθ 2 ) − ebθ θ 2
a = rθ + 2rθ = ebθ θ + 2(bebθ θ)θ
θ
Since θ = ω = constant, θ = θ , and we write
ar = bebθ (bω 2 ) − ebθ ω 2 = ebθ (b 2 − 1)ω 2
aθ = 2bebθ ω 2
a = ar2 + aθ2 = ebθ ω 2 (b2 − 1)2 + (2b) 2
= ebθ ω 2 b4 − 2b 2 + 1 + 4b 2 = ebθ ω 2 b 4 + 2b 2 + 1
= ebθ ω 2 (b2 + 1) 2 = ebθ ω 2 (b2 + 1)
a = (1 + b 2 )ω 2 ebθ 
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PROBLEM 11.177
The motion of a particle on the surface of a right circular cylinder is
defined by the relations R = A, θ = 2π t , and z = B sin 2π nt , where A and B
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
θ = 2π t
z = B sin 2π nt
θ = 2π
z = 2π n B cos 2π nt
 = 0
R
θ = 0

z = −4π 2 n 2 B sin 2π nt
Velocity (Eq. 11.49)
v = R e R + Rθeθ + zk
v=
+ A(2π )eθ + 2π n B cos 2π nt k
v = 2π A2 + n 2 B 2 cos 2 2π nt 
Acceleration (Eq. 11.50)
 - Rθ 2 )e + ( Rθ + 2 Rθ)e + 
a = (R
zk
R
θ
a = −4π 2 Aek − 4π 2 n 2 B sin 2π nt k
a = 4π 2 A2 + n4 B 2 sin 2 2π nt 
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PROBLEM 11.178
Show that r = hφ sin θ knowing that at the instant shown, step AB of
the step exerciser is rotating counterclockwise at a constant rate φ.
SOLUTION
From the diagram
r 2 = d 2 + h 2 − 2dh cos φ
Then
Now
or
2rr = 2dhφ sin φ
r
d
=
sin φ sin θ
r=
d sin φ
sin θ
Substituting for r in the expression for r
 d sin φ 

 sin θ  r = dhφ sin φ


or
r = hφ sin θ
Q.E.D.

Alternative solution.
First note
α = 180° − (φ + θ )
Now
v = v r + vθ = re r + rθeθ
With B as the origin
vP = dφ
( 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 α
Then
Now
or
substituting
r = dφ sin α
h
d
=
sin α sin θ
d sin α = h sin θ
r = hφ sin θ
Q.E.D.
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PROBLEM 11.179
The three-dimensional motion of a particle is defined by the relations R = A(1 − e −t ), θ = 2π 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
 = − Ae−t
R
θ = 2π t
z = B (1 − e−t )
θ = 2π
z = Be−t
θ = 0

z = − Be−t
Velocity (Eq. 11.49)
v = R e R + Rθeθ + zk
v = Ae −t e R + 2π A(1 − e −t )eθ + Be−t k
(a)
When
t = 0: e−t = e0 = 1;
(b)
When
t = ∞ : e− t = e −∞ = 0
v = A2 + B 2 
v = Ae R + Bk
v = 2π Aeθ
v = 2π A 
Acceleration (Eq. 11.50)
 − Rθ1 )e + ( Rθ + 2 Rθ)e + 
a = (R
zk
R
θ
= [ − Ae−t − A(1 − e −t )4π 2 ]e R + [0 + 2 Ae−t (2π )]eθ − Be −t k
(a)
When
t = 0: e −t = e0 = 1
a = − Ae R + 4π Aeθ − B k
a = A2 + (4π A) 2 + B 2
(b)
When
a = (1 + 16π 2 ) A2 + B 2 
t = ∞ : e −t = e−∞ = 0
a = −4π 2 Ae R
a = 4π 2 A 
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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 motion of a particle is defined by the position vector r =
(Rt cos ωnt)i + ctj + (Rt sin ωnt)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 ωn t )i + ctj + ( Rt sin ωn t )k
Then
v=
dr
= R(cos ωn t − ωn t sin ωn t )i + cj + R(sin ωn t + ωn t cos ωn t )k
dt
and
a=
dv
dt
(
)
= R −ωn sin ωn t − ωn sin ωn t − ωn2 t cos ωn t i
+R
(
ωn cos ωn t + ωn cos ωn t − ωn2 t sin ωn t
)k
= ωn R[−(2sin ωn t + ωn t cos ωn t )i + (2cos ωn t − ωn t sin ωn t )k ]
It then follows that the vector ( v × a) is perpendicular to the osculating plane.
i
j
k
( v × a) = ωn R R(cos ωn t − ωn t sin ωn t ) c R(sin ωn t + ωn t cos ωn t )
−(2sin ωn t + ωn t cos ωn t ) 0 (2cos ωn t − ωn t sin ωn t )
= ωn R{c(2 cos ωn t − ωn t sin ωn t )i + R[ −(sin ωn t + ωn t cos ωn t )(2sin ωn t + ωn t cos ωn t )
− (cos ωn t − ωn t sin ωn t )(2 cos ωn t − ωn t sin ωn t )] j + c(2sin ωn t + ωn t cos ωn t )k
(
)
= ωn R c(2 cos ωn t − ωn t sin ωn t )i − R 2 + ωn2 t 2 j + c(2sin ωn t + ωn t cos ωn t )k 


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PROBLEM 11.180* (Continued)
The angle α formed by the vector ( v × a) and the y axis is found from
cos α =
Where
( v × a) ⋅ j
| ( v × a) || j |
|j| =1
(
( v × a) ⋅ j = −ωn R 2 2 + ωn2 t 2
)
(
|( v × a) | = ωn R c 2 (2 cos ωn t − ωn t sin ωn t ) 2 + R 2 2 + ωn2 t 2

)
2
1/ 2
+ c 2 (2sin ωn t + ωn t cos ωn t )2 
(
)
2 1/2
(
) 
= ωn R c 2 4 + ωn2 t 2 + R 2 2 + ωn2t 2

Then
(
)
ω R c ( 4 + ω t ) + R ( 2 + ω t ) 


−R ( 2 + ω t )
=
c 4 + ω t + R 2 + ω t 
) (
) 
 (
−ωn R 2 2 + ωn2t 2
cos α =
2
n
2 2
n
2
2 2
n
2
1/2
2 2
n
2
2 2
n
2
2 2
n
2 1/2
The angle β that the osculating plane forms with y axis (see the above diagram) is equal to
β = α − 90°
Then
cos α = cos ( β + 90°) = −sin β
− sin β =
Then
or
tan β =
(
)
) + R ( 2 + ω t ) 
− R 2 + ωn2t 2
(
c 2 4 + ω 2t 2
n

(
R 2 + ωn2 t 2
c
2
2 2
n
2
1/ 2
)
4 + ωn2 t 2
(
 R 2 + ωn2t 2
β = tan 
 c 4 + ω 2t 2
n

−1
)  


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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 = π /2 s.
SOLUTION
Given:
)
(
r = ( At cos t )i + A t 2 + 1 j + ( Bt sin t )k
r − ft, 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
3t
= 3(cos t − t sin t )i +
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(2sin t + t cos t )i + 2
j + (2 cos t − t sin t )k
(t + 1)3/2
and
(a)
t2 +1
At t = 0:
v = (3 ft/s)i
a = (3 ft/s 2 ) j + (2 ft/s 2 )k
Then
and
Then
v × a = 3i × (3j + 2k )
= 3(−2 j + 3k )
| v × a | = 3 (−2) 2 + (3) 2 = 3 13
eb =
3( −2 j + 3k )
3 13
cos θ x = 0
or
θ x = 90°
=
cos θ y = −
1
13
2
(−2 j + 3k )
13
θ y = 123.7°
cos θ z =
3
13
θ z = 33.7°

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PROBLEM 11.181* (Continued)
(b)
At t =
Then
π
2
s:

 3π
  3π
v = −
ft/s  i + 
ft/s  j + (1 ft/s)k


 2
  π2 +4


 π
24

a = −(6 ft/s 2 )i +  2
ft/s 2  j −  ft/s 2  k
3/2

 (π + 4)
 2
i
3π
v×a = −
2
−6
j
3π
2
(π + 4)1/ 2
24
(π 2 + 4)3/ 2
k
1
−
π
2

 
3π 2
24
3π 2
6
i
= −
+
−
+


2
1/2
4
(π 2 + 4)3/ 2  
 2(π + 4)


36π
18π
k
+ − 2
+ 2
3/2
1/2 
(π + 4) 
 (π + 4)
= −4.43984i − 13.40220 j + 12.99459k
and
Then
or

 j

| v × a | = [(−4.43984) 2 + ( −13.40220)2 + (12.99459) 2 ]1/ 2
= 19.18829
1
(−4.43984i − 13.40220 j + 12.99459k )
19.1829
4.43984
13.40220
12.99459
cos θ x = −
cos θ y = −
cos θ z =
19.18829
19.18829
19.18829
eb =
θ x = 103.4°
θ y = 134.3°
θ z = 47.4°

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PROBLEM 11.182
The motion of a particle is defined by the relation x = 2t 3 − 15t 2 + 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
a=
= 12t − 30
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 > 0.
For
1 s ≤ t ≤ 2.5 s, v ≤ 0.
At
t = 0,
At
t = 1 s, x1 = (2)(1)3 − (15)(1) 2 + (24)(1) + 4 = 15 m
x0 = 4 m.
Distance traveled over interval: x1 − x0 = 11 m
For
1 s ≤ 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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PROBLEM 11.183
A particle starting from rest at x = 1 m is accelerated so that its velocity doubles in magnitude between
x = 2 m and x = 8 m. Knowing that the acceleration of the particle is defined by the relation a = k[ x − (A/x)],
determine the values of the constants A and k if the particle has a velocity of 29 m/s when x = 16 m.
SOLUTION
We have
v

When x = 1 ft, v = 0:
v
0
dv
A

= a = kx− 
dx
x

vdv =

x
1
A

k  x −  dx
x

1 2
1
v = k  x 2 − A ln
2
2
or
x

x
1
1
1
= k  x 2 − A ln x − 
2
2
1 2
1
1
3

v2 = k  (2) 2 − A ln 2 −  = k  − A ln 2 
2
2
2
2




At x = 2 ft:
1 2
1
1
v8 = k  (8) 2 − A ln 8 −  = k (31.5 − A ln 8)
2
2
2
x = 8 ft:
Now
1 2
v
2 8
1 2
v
2 2
v8
= 2:
v2
= (2) 2 =
k (31.5 − A ln 8)
k ( 32 − A ln 2 )
6 − 4 A ln 2 = 31.5 − A ln 8
1
25.5 = A(ln 8 − 4 ln 2) = A(ln 8 − ln 24 ) = A ln  
2
A=
When x = 16 m, v = 29 m/s:
25.5
ln 12
A = −36.8 m 2 
1
1
25.5
1
(29) 2 = k  (16) 2 −
ln(16) − 
1
2
2
2
ln ( 2 )


1

420.5k = k 128 + 102 −  = 230.5k
2

= 230.5k
k = 1.832 s −2 
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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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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
or
48
111.366
=
sin α sin 155°
α = 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
at t = 3 min, B is (48 km/h) ( 60
) = 5.6 km southwest of the crossing.
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°
or
rB/A = 2.96 km 
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PROBLEM 11.186
Slider block B starts from rest and moves to the right with a constant
acceleration of 1 ft/s 2. Determine (a) the relative acceleration of
portion C of the cable with respect to slider block A, (b) the velocity
of portion C of the cable after 2 s.
SOLUTION
Let d be the distance between the left and right supports.
Constraint of entire cable: xB + ( xB − x A ) + 2(d − x A ) = constant
2vB − 3v A = 0
aA =
and
2
2
aB = (1) = 0.667 ft/s 2
3
3
Constraint of Point C:
a A = 0.667 ft/s 2
or
2(d − x A ) + yC/ A = constant
−2v A + vC/ A = 0
(a)
2aB − 3a A = 0
− 2a A + aC/ A = 0
and
aC/ A = 2a A = 2(0.667) = 1.333 ft/s 2
aC/ A = 1.333 ft/s 2

Velocity vectors after 2s: v A = (0.667)(2) = 1.333 ft/s
vC/ A = (1.333)(2) = 2.666 ft/s
v C = v A + v C/ A
Sketch the vector addition.
vC2 = v A2 + vC2 / A = (1.333) 2 + (2.666) 2 = 8.8889(ft/s) 2
vC = 2.981 ft/s
tan θ =
(b)
vC/ A
vA
=
2.666
= 2,
1.333
θ = 63.4°
vC = 2.98 ft/s
63.4° 
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PROBLEM 11.187
Collar A starts from rest at t = 0 and moves downward with a constant
acceleration of 7 in./s 2 . Collar B moves upward with a constant acceleration,
and its initial velocity is 8 in./s. Knowing that collar B moves through 20 in.
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 − yB ) = constant
Then
−2v A − vB + 4vC = 0
(1)
and
−2a A − aB + 4aC = 0
(2)
(v A )0 = 0
Given:
(a A ) = 7 in./s 2
( v B )0 = 8 in./s
a B = constant


At t = 2 s
(a)
y − (yB )0 = 20 in.
1
aB t 2
2
y B = ( y B ) 0 + ( vB ) 0 t +
We have
At t = 2 s:
−20 in. = (−8 in./s)(2 s) +
aB = −4 in./s 2
1
aB (2 s) 2
2
or
a B = 2 in./s 2 
Then, substituting into Eq. (2)
−2(7 in./s 2 ) − (−2 in./s 2 ) + 4aC = 0
aC = 3 in./s 2
or
aC = 3 in./s 2 
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PROBLEM 11.187 (Continued)
(b)
Substituting into Eq. (1) at t = 0
−2(0) − (−8 in./s) + 4(vC )0 = 0 or (vC )0 = −2 in./s 
vC = (vC )0 + aC t
Now
(c)
When vC = 0:
0 = (−2 in./s) + (3 in./s 2 )t
or
t=
2
3
t = 0.667 s 
yC = ( yC )0 + (vC )0 t +
We have
At t =
2
s
3
s:
2
yC − ( yC )0 = ( −2 in./s) 
3
= −0.667 in.
1
aC t 2
2
 1
2
s  + (3 in./s 2 ) 
 2
3
or

s

2
y C − (y C )0 = 0.667 in. 
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PROBLEM 11.188
A golfer hits a ball with an initial velocity of magnitude v0 at an
angle α 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 α = 31°, (b) a five-iron with α = 27°.
SOLUTION
The horizontal and vertical motions are
x
t cos α
1 2
1
y = (v0 sin α )t − gt = x tan α − gt 2
2
2
x = (v0 cos α )t
v0 =
or
(1)
2( x tan α − y)
g
or
t2 =
At the landing Point C:
yC = 0,
And
xC = (v0 cos α )t =
(2)
t =
2v0 sin α
g
2v02 sin α cos α
g
(3)
(a) α = 31°
To clear tree A:
x A = 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 s 2 ,
9.81
t A = 1.1084 s
30
= 31.58 m/s
1.1084cos 31°
xB = 100 m,
yB = 14 m
From (2),
(t B ) 2 =
2(100 tan 31° − 14)
= 9.3957 s 2 ,
9.81
From (1),
(v0 ) B =
100
= 38.06 m/s
3.0652 cos 31°
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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PROBLEM 11.188 (Continued)
(b)
α = 27°
By a similar calculation,
t A = 0.81846 s,
t B = 2.7447 s,
v0 = 41.138 m/s
xC = 139.56 m,
(v0 ) A = 41.138 m/s,
(v0 ) B = 40.890 m/s,
v0 = 41.1 m/s 
d = 29.6 m 
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PROBLEM 11.189
As the truck shown begins to back up with a constant acceleration
of 4 ft/s 2 , the outer section B of its boom starts to retract with a
constant acceleration of 1.6 ft/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 = 4 ft/s 2
For the boom,
a B/ A = 1.6 ft/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°
= 42 + 1.62 − 2(4)(1.6) cos 50°
aB = 3.214 ft/s 2
Law of sines:
sin α =
aB/ A sin 50°
aB
α = 22.4°,
(b)
=
1.6 sin 50°
= 0.38131
3.214
a B = 3.21 ft/s 2
22.4° 
v B = (vB )0 + aBt = 0 + (3.214)(2)
v B = 6.43 ft/s 2
22.4° 
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PROBLEM 11.190
A motorist traveling along a straight portion of a
highway is decreasing the speed of his automobile at
a constant rate before exiting from the highway onto a
circular exit ramp with a radius of 560-ft. He continues
to decelerate at the same constant rate so that 10 s after
entering the ramp, his speed has decreased to 20 mi/h,
a speed which he then maintains. Knowing that at this
constant speed the total acceleration of the automobile
is equal to one-quarter of its value prior to entering the
ramp, determine the maximum value of the total
acceleration of the automobile.
SOLUTION
First note
v10 = 20 mi/h =
88
ft/s
3
While the car is on the straight portion of the highway.
a = astraight = at
and for the circular exit ramp
a = at2 + an2
where
an =
v2
ρ
By observation, amax occurs when v is maximum, which is at t = 0 when the car first enters the ramp.
For uniformly decelerated motion
v = v0 + at t
and at t = 10 s:
v = constant  a = an =
a=
Then
or
astraight
2
v10
ρ
1
ast.
4
88
v 2 ( 3 ft/s )
1
= at  at = 10 =
ρ
4
560 ft
2
at = −6.1460 ft/s 2
(The car is decelerating; hence the minus sign.)
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PROBLEM 11.190 (Continued)
Then at t = 10 s:
or
Then at t = 0:
88
ft/s = v0 + (−6.1460 ft/s 2 )(10 s)
3
v0 = 90.793 ft/s
amax =
at2
 v2
+ 0
ρ




2
1/2
2

 (90.793 ft/s) 2  

2 2
= (−6.1460 ft/s ) + 
 
560 ft

 

or
amax = 15.95 ft/s 2 
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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 = 30 ft and yB = −18 ft.
Horizontal motion:
Vertical motion:
vx = v0 cos 25°
(1)
x = v0 t cos 25°
(2)
v y = v0 sin 25° − gt
y = v0t sin 25° −
(3)
1 2
gt
2
(4)
At Point B, Eq. (2) gives
v0 t B =
xB
30
=
= 33.101 ft
cos 25° cos 25°
Substituting into Eq. (4),
1
−18 = (33.101)(sin 25°) − (32.2)t B2
2
tB = 1.40958 s
(a)
Speed of the belt.
v0 =
v0t B
33.101
=
= 23.483
1.40958
tB


v0 = 23.4 ft/s 
Eqs. (1) and (3) give
vx = 23.483cos 25° = 21.283 ft/s
v y = (23.483) sin 25° − (32.2)(1.40958) = −35.464 ft/s
tan θ
−v y
vx
= 1.66632
θ = 59.03°
v = 41.36 ft/s
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PROBLEM 11.191 (Continued)
Components of acceleration.
a = 32.2 ft/s 2
at = 32.2sin θ
an = 32.2cos θ = 32.2 cos 59.03° = 16.57 ft/s 2
(b)
Radius of curvature at B.
an =
ρ=
v2
ρ
v 2 (41.36) 2
=
an
16.57
ρ = 103.2 ft 
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PROBLEM 11.192
The end Point B of a boom is originally 5 m from fixed Point A
when the driver starts to retract the boom with a constant radial
acceleration of 
r = −1.0 m/s 2 and lower it with a constant
angular acceleration θ = −0.5 rad/s 2 . 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
r = r0 + 
rt = 0 − 1.0t
r = r0 + r0 t +
r = 5 − (0.5)(2)2 = 3 m
r = (−1.0)(2) = −2 m/s
At t = 2 s,
θ 0 = 60° =
Angular motion.
π
3
rad, θ0 = 0, θ = −0.5 rad/s 2
1
π
+ 0 − 0.25t 2
2
3
θ = θ0 + θt = 0 − 0.5t
θ = θ0 + θ0 + θt 2 =
θ=
At t = 2 s,
π
+ 0 − (0.25)(2) 2 = 0.047198 rad = 2.70°
3

θ = −(0.5)(2) = −1.0 rad/s
Unit vectors er and eθ .
(a)
Velocity of Point B at t = 2 s.
v B = re r + rθeθ
= −(2 m/s)e r + (3 m)(−1.0 rad/s)eθ


v B = (−2.00 m/s)er + ( −3.00 m/s)eθ 

tan α =
v=
vθ −3.0
=
= 1.5
vr −2.0
vr2
2
α = 56.31°
2

2
+ vθ = (−2) + (−3) = 3.6055 m/s
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PROBLEM 11.192 (Continued)
Direction of velocity.
et =
v −2er − 3eθ
=
= −0.55470er − 0.83205eθ
3.6055
v
θ + α = 2.70 + 56.31° = 59.01°

(b)
v B = 3.61 m/s

59.0° 
Acceleration of Point B at t = 2 s.
a B = (
r − rθ 2 )er + ( rθ + 2rθ)eθ
= [−1.0 − (3)(−1) 2 ]e r + [(3)(−0.5) + (2)(−1.0)(−0.5)]eθ
a B = ( −4.00 m/s 2 )er + (2.50 m/s 2 )eθ 
tan β =
aθ
2.50
=
= −0.625
ar −4.00
β = −32.00°
a = ar2 + aθ2 = ( −4) 2 + (2.5) 2 = 4.7170 m/s 2
θ + β = 2.70° − 32.00° = −29.30°
a B = 4.72 m/s 2
Tangential component:
29.3° 
at = (a ⋅ et )et
at = (−4er + 2.5eθ ) ⋅ (−0.55470er − 0.83205eθ )et
= [(−4)(−0.55470) + (2.5)( −0.83205)]et
= (0.138675 m/s 2 )et = 0.1389 m/s 2
Normal component:
59.0°
a n = a − at
a n = −4er + 2.5eθ − (0.138675)(−0.55470e r − 0.83205eθ )
= (−3.9231 m/s 2 )er + (2.6154 m/s 2 )eθ
an = (3.9231) 2 + (2.6154) 2 = 4.7149 m/s 2
(c)
Radius of curvature of the path.
an =
ρ=
v2
ρ
v 2 (3.6055 m/s)2
=
an
4.7149 m/s
ρ = 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 = 500 ft, r = −105 ft/s,

r = −10 ft/s 2 , θ = 25°, θ = 0.07 rad/s, θ = 0.06 rad/s 2 .
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 = 500 ft, θ = 25°)
v = vr e r + vθ eθ = re r + rθeθ
= ( −105 ft/s)e r + (500 ft)(0.07 rad/s)eθ
v = ( −105 ft/s)e r + (35 ft/s)eθ 
Direction of velocity:
v = ( −105cos 25° − 35cos 65°)i + (35sin 65° − 105sin 25°) j
= ( −109.95 ft/s)i + ( −12.654 ft/s) j
v y −12.654
α = 6.565°
tan α =
=
vx −109.95
v = (105) 2 + (35)2 = 110.68 ft/s
v = 110.7 ft/s
6.57° 
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PROBLEM 11.193 (Continued)
(b)
Acceleration of the skier.
a = ar er + aθ eθ = (
r − rθ 2 )er + (rθ + 2rθ)eθ
ar = −10 − (500)(0.07)2 = −12.45 ft/s 2
aθ = (500)(0.06) + (2)(−105)(0.07) = 15.30 ft/s 2
a = (−12.45 ft/s 2 )er + (15.30 ft/s 2 )eθ 
a = (−12.45)(i cos 25° + j sin 25°) + (15.30)(−i cos 65° + j sin 65°)
= (−17.750 ft/s 2 )i + (8.6049 ft/s 2 ) j
ay
8.6049
tan β =
=
β = −25.9°
ax −17.750
a = (12.45) 2 + (15.30)2 = 19.725 ft/s 2
a = 19.73 ft/s 2
(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)




Vertical motion:


x0 = 0
x0 = 109.95 ft/s


x = x0 + x0 t = 109.95t 
(Uniformly accelerated motion)
y0 = 0
y 0 = 12.654 ft/s

y = 32.2 ft/s

(from Part a )
(from Part a) 
2
y = y0 + y 0 t +
1 2

yt = 12.654t − 16.1t 2 
2
At the landing point,
x = d cos 30°
y = 10 + d sin 30° or
(1)
y − 10 = d sin 30°
(2)
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PROBLEM 11.193 (Continued)
Multiply Eq. (1) by sin 30° and Eq. (2) by cos 30° and subtract
x sin 30° − ( y − 10) cos 30° = 0
(109.95t )sin 30° − (12.654t + 16.1t 2 − 10) cos 30° = 0
−13.943t 2 + 44.016t + 8.6603 = 0
t = −0.1858 s and 3.3427 s
Reject the negative root.
x = (109.95 ft/s)(3.3427 s) = 367.53 ft
d=
x
cos 30°
d = 424 ft 
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