PROBLEM 11.CQ1
A bus travels the 100 miles between A and B at 50 mi/h and then another
100 miles between B and C at 70 mi/h. The average speed of the bus for the
entire 200-mile trip is:
(a) more than 60 mi/h
(b) equal to 60 mi/h
(c) less than 60 mi/h
SOLUTION
The time required for the bus to travel from A to B is 2 h and from B to C is 100/70 = 1.43 h, so the total time
is 3.43 h and the average speed is 200/3.43 = 58 mph.
Answer: (c) 
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3
PROBLEM 11CQ2
Two cars A and B race each other down a straight
road. The position of each car as a function of time
is shown. Which of the following statements are
true (more than one answer can be correct)?
(a) At time t2 both cars have traveled the same
distance
(b) At time t1 both cars have the same speed
(c) Both cars have the same speed at some time t < t1
(d) Both cars have the same acceleration at some
time t < t1
(e) Both cars have the same acceleration at some
time t1 < t < t2
SOLUTION
The speed is the slope of the curve, so answer c) is true.
The acceleration is the second derivative of the position. Since A’s position increases linearly the second
derivative will always be zero. The second derivative of curve B is zero at the pont of inflection which occurs
between t1 and t2.
Answers: (c) and (e) 
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4
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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5
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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6
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
v1 = 20cos 2 − 30sin 2 = −35.6
a1 = −40sin 2 − 60 cos 2 = −11.40
x1 = 102.9 mm 
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)
vmax = −36.1 mm/s 
= −36.056
Note that we could have also used
vmax = 202 + 302 = 36.056
by combining the sine and cosine terms.
For amax we can take the derivative and set equal to zero or just combine the sine and cosine terms.
amax = 402 + 602 = 72.1 mm/s 2
amax = 72.1 mm/s 2 
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7
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
x0 = 0 mm 
v0 = 960 mm/s
v0 = 960 mm/s
a0 = −9216 mm/s 2
(b) At t = 0.3 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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8
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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9
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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10
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
t = 1.000 s and t = 4.00 s 
12t − 30 = 0
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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11
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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12
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
−10dt
0
0 − v0 = −10t f
tf =
v0 77.5
=
10
10
t f = 7.75 s 
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13
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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14
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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15
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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16
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
1
 1

x =  t 4 − 8t  −  (2) 4 − 8(2) 
2
 2

1
x = t 4 − 8t − 8 + 16
2
t
2
x=
1 4
t − 8t + 8 
2
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17
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 =
t
1.8
cos kt
k
0
1.8
(cos kt − 1) = 0.6cos kt − 0.6
3
v = 0.6cos kt m/s
Velocity:
t
t
x − x0 =  0 v dt = 0.6  0 cos kt dt =
x−0=
0.6
sin kt
k
t
0
0.6
(sin kt − 0) = 0.2sin kt
3
x = 0.2sin kt m
Position:
When t = 0.5 s,
k = 3 rad/s
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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18
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
v = −453 mm/s 
x = 0.12sin1.5 + 0.16 cos1.5 = 0.1310 m
x = 131.0 mm 
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19
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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20
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 ) 
t = 1.366 × 10−3 s 
or
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21
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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22
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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23
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
vmax = 1.004 m/s 
or
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24
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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25
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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26
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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27
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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28
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)
x = 42.0 ft 
v = 12.5 + 4e −2.4 = 12.863 ft/s
v = 12.86 ft/s 
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29
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:
v = 29.3 m/s 
2(−9) = −19t
t = 0.947 s 
or
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30
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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31
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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32
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
x = 145.2 m 
or
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33
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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34
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)
t3 − t1 = 6.17 s 
or
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35
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)
2
From Eq. (1)
t=
When x = 6 mi:
t=
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 
or
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36
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
32.2
(1 +
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:
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 
or
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37
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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38
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 

ve = 36.7 × 103 ft/s 
or
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39
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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40
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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41
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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42
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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43
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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44
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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45
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)
v 2 = v12 + 2a( y − y1 )
We have
At
y = ymax , v = 0
Then
0 = (252 ft/s) 2 + 2( −32.2 ft/s 2 )( ymax − 89.6) ft
ymax = 1076 ft 
or
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46
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
t D = 2.06 s 
or
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47
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:
vmax = 12.96 m/s 
x = x1 + v0 (t − t1 )
for
35 m < x ≤ 100 m
100 m = 35 m + (12.9628 m/s)(t2 − 5.4) s
t2 = 10.41 s 
or
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48
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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49
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) +
(vB )0 = 50 m/s
1
aB (8 s) 2
2
aB = 3.13 m/s 2 
or
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50
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
vM = 28.6 mi/h 
or
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51
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
a At 2 = 0 + 35.2t + (1.8)t 2
2
2
(1)
(2)
Motion of auto B:
vB = (vB )0 + aB t = 52.8 − 1.2t
x B = ( xB ) 0 + ( vB ) 0 t +
(a)
1
1
aB t 2 = 75 + 52.8t + (−1.2)t 2
2
2
(3)
(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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52
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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53
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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54
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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55
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 2
gt = 100t − 4.905t 2
2
(1)
(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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56
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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57
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

d = 906 ft 
or
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58
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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59
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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60
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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61
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
aB =
or
24
ft/s 2
3
a B = 8.00 ft/s 2 
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62
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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63
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
(3)
v A = 200 mm/s

vC = 600 mm/s

xB + ( xB − xC ) = constant
2vB − vC = 0
2(300 mm/s) − vC = 0
or
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64
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
vC/A = 400 mm/s
or

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65
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
aA = −
40
mm/s 2
3
a A = 13.33 mm/s 2
or

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66
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 +
a D = 13.33 mm/s 2

v B = 70.0 mm/s

1
aB t 2
2
xB − ( xB )0 = (150 mm/s)(4 s)
+
1
(−20.0 mm/s 2 )(4 s) 2
2
x B − (x B )0 = 440 mm
or

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67
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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68
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:
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. 
or
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69
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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70
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
where aC = −4 in./s 2
= 0 + 4t
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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71
PROBLEM 11.55 (Continued)
(a)
Time when vB is zero.
3 − 2t = 0
Motion of block D:
From (3),
vD =
(b)
t = 1.500 s 
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 = 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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72
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
t = 1.000 s 
ΔyC = (6)(1) − (3)(1)2 = 3 in.
ΔyC = 3.00 in. 
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73
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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74
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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75
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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76
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 +
or
a A = 2.00 mm/s 2 
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
y A − (y A )0 = 175.0 mm 
or
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77
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):
2(aD − 110) + 2( aC + 60) + aC = 0
3aC + 2aD = 100
or
Eq. (4):
−(aD − 110) − ( aC + 60) + 2aD = 0
− aC + aD = −50
or
(5)
(6)
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78
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:
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 
or
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79
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
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
− v A − vB − 2 vD = 0
(3)
and
− a A − aB + 2aD = 0
(4)
t =0
Given: At
v=0
( y A )0 = ( yB )0 = ( yC )0
All accelerations constant.
At t = 2 s
yC /A = 280 mm
vB /A = 80 mm/s
When
y A − ( y A )0 = 160 mm
yB − ( yB )0 = 320 mm
aB > 10 mm/s 2
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80
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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81
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:
−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 
or
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82
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
v0 = −14 ft/s
Change in v = area under a−t curve.
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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83
PROBLEM 11.61 (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
(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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84
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
v0 = −14 ft/s
Change in v = area under a−t curve.
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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85
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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86
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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87
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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88
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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89
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
v1 = 5(22 − t1 )
or
(22 − t1 )[5(22 − t1 )] = 640
t1 = 10.69 s and t1 = 33.3 s
10 s < t1 < 22 s 
We have
t1 = 10.69 s 
Between 26 s and 41 s:
Using similar triangles
41 − t2
15
=
20
300
t2 = 40.0 s 
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90
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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91
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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92
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 = (0.4032 + 40.0346 + 4.3203) s
(b)
We have
ainitial =
=
ttotal = 44.8 s 
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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93
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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94
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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95
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)
vt1 = v0 + A1
We have
 12 + 2 
1.8 m/s = 6 m/s − (t1 s) 
m/s2

 2 
t1 = 0.6 s 
(b)
v1.4 = vt1 + A2
We have
v1.4 = 1.8 m/s − (1.4 − 0.6) s × 2 m/s 2
v1.4 = 0.20 m/s 
Now x1.4 = A3 + A4 , where A3 is most easily determined using
integration. Thus,
a=
for 0 < t < t1:
−2 − (−12)
50
t − 12 = t − 12
0.6
3
dv
50
= a = t − 12
dt
3
Now
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96
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
x1.4 = 2.84 m 
or
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97
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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98
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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99
PROBLEM 11.71 (Continued)
Runner B: (vmax ) B = 8.8106, (t3 ) B = 149.45 s
and
26.765 s
t B = 25.2 + 26.765 = 51.965 s
Reject the larger root. Then total time
t B = 52.0 s 
Velocity of A at t = 51.965 s:
v1 = 8.6957 − (0.1)(51.965 − 25) = 5.999 m/s
Velocity of A at t = 51.279 s:
v2 = 8.6957 − (0.1)(52.279 − 25) = 5.968 m/s
Over 51.965 s ≤ t ≤ 52.965 s, runner A covers a distance 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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100
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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101
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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102
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)]
xB/A = 77.5 ft 
or
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103
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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104
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
t = 5.67 s 
Since 1.525 s is less than 2 s,
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105