Solutions for Problems in Dynamics (15th Edition) by Russell Hibbeler

1
12–1.
Starting from rest, a particle moving in a straight line has an
acceleration of a = (2t - 6) m>s2, where t is in seconds.What
is the particle’s velocity when t = 6 s, and what is its position
when t = 11 s?
SOLUTION
a = 2t - 6
dv = a dt
L
v
0
dv =
L
t
0
(2t - 6) dt
v = t2
- 6t
ds = v dt
L
s
0
ds =
L
t
0
(t2
- 6t) dt
s =
t3
3
- 3t2
When t = 6 s,
v = 0 Ans.
When t = 11 s,
s = 80.7 m Ans.
Ans:
v = 0
s = 80.7 m
s
m
t
b
9
8
@
g
m
a
i
l
.
c
o
m
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2
12–2.
SOLUTION
1S
+ 2 s = s0 + v0 t +
1
2
ac t2
= 0 + 12(10) +
1
2
(-2)(10)2
= 20 ft Ans.
If a particle has an initial velocity of v0 = 12 fts to the
right, at s0 = 0, determine its position when t = 10 s, if
a = 2 fts2
to the left.
Ans:
s = 20 ft

3
12–3.
A particle travels along a straight line with a velocity
v = (12 - 3t2
) ms, where t is in seconds.When t = 1 s, the
particle is located 10 m to the left of the origin. Determine
the acceleration when t = 4 s, the displacement from
t = 0 to t = 10 s, and the distance the particle travels during
this time period.
SOLUTION
v = 12 - 3t2
(1)
a =
dv
dt
= -6t t = 4 = -24 ms2
Ans.
L
s
-10
ds =
L
t
1
v dt =
L
t
1
(12 - 3t2
)dt
s + 10 = 12t - t3
- 11
s = 12t - t3
- 21
s t = 0 = -21
s t = 10 = -901
∆s = -901 - (-21) = -880 m Ans.
From Eq. (1):
v = 0 when t = 2s
s t = 2 = 12(2) - (2)3
- 21 = -5
sT = (21 - 5) + (901 - 5) = 912 m Ans.
Ans:
a = -24 ms2
∆s = -880 m
sT = 912 m
A particle travels along a straight line with a velocity
v = (12 - 3t2
) ms, where t is in seconds.When t = 1 s, the
particle is located 10 m to the left of the origin. Determine
the acceleration when t = 4 s, the displacement from
t = 0 to t = 10 s, and the distance the particle travels during
this time period.

4
*12–4.
SOLUTION
Velocity: To determine the constant acceleration , set , ,
and and apply Eq. 12–6.
Using the result , the velocity function can be obtained by applying
Eq. 12–6.
Ans.
v = A29.17s - 27.7 fts
v2
= 32
+ 2(4.583) (s - 4)
( :
+ ) v2
= v2
0 + 2ac(s - s0)
ac = 4.583 fts2
ac = 4.583 fts2
82
= 32
+ 2ac (10 - 4)
( :
+ ) v2
= v2
0 + 2ac(s - s0)
v = 8 fts
s = 10 ft
v0 = 3 fts
s0 = 4 ft
ac
A particle travels along a straight line with a constant
acceleration. When , and when ,
. Determine the velocity as a function of position.
v = 8 fts
s = 10 ft
v = 3 fts
s = 4 ft
A
Ans:
v = (19.17s - 27.7)fts

5
12–5.
The velocity of a particle traveling in a straight line is given
by v = (6t - 3t2) ms, where t is in seconds. If s = 0 when
t = 0, determine the particle’s deceleration and position
when t = 3 s. How far has the particle traveled during the
3-s time interval, and what is its average speed?
SOLUTION
ds
dt
= v
ds = (6t - 3t2
) dt
L
s
0
ds =
L
t
0
(6t - 3t2
)dt
s = 3t2
- t3
At t = 3 s, s = 3(3)2
- 33
= 0 Ans.
when v = 0; 0 = 6t - 3t2
t = 0 and 2 s
At t = 2 s, s = 3(2)2
- 23
= 4 m
From the figure sT = 2(4) = 8 m Ans.
vsp =
sT
△t
=
8
3
= 2.67 m/s Ans.
a =
dv
dt
= 6 - 6t
At t = 3 s, a = 6 - 6(3) = -12 m/s2
Ans.
Ans:
s = 0
sT = 8 m
vsp = 2.67 m/s
a = -12 m/s2

6
12–6.
A particle moving along a straight line is subjected to a
deceleration , where is in . If it has a
velocity and a position when ,
determine its velocity and position when .
t = 4 s
t = 0
s = 10 m
v = 8 ms
ms
v
a = (-2v3
) ms2
SOLUTION
Velocity: The velocity of the particle can be related to its position by applying
Eq. 12–3.
[1]
Position: The position of the particle can be related to the time by applying Eq.12–1.
When ,
Choose the root greater than Ans.
Substitute into Eq. [1] yields
Ans.
v =
8
16(11.94) - 159
= 0.250 ms
s = 11.94 m
10 m s = 11.94 m = 11.9 m
8s2
- 159s + 758 = 0
8(4) = 8s2
- 159s + 790
t = 4 s
8t = 8s2
- 159s + 790
L
t
0
dt =
L
s
10m
1
8
(16s - 159) ds
dt =
ds
v
v =
8
16s - 159
s - 10 =
1
2v
-
1
16
L
s
10m
ds =
L
v
8ms
-
dv
2v2
ds =
vdv
a
Ans:
s = 11.9 m
v = 0.250 ms

7
12–7.
A particle moves along a straight line such that its position
is e
h
t
e
n
i
m
r
e
t
e
D
.
y
b
d
e
n
i
f
e
d
velocity, average velocity, and the average speed of the
particle when .
t = 3 s
s = (2t3
+ 3t2
- 12t - 10) m
SOLUTION
Ans.
Ans.
Ans.
(vsp)avg =
sT
¢t
=
(17 - 10) + (17 + 35)
3 - 0
= 19.7 m s
vavg =
¢s
¢t
=
35 - (-10)
3 - 0
= 15 ms
s|t = 3 = 35
s|t = 1 = -17
s|t = 0 = -10
v = 0 at t = 1, t = -2
v|t = 3 = 6(3)2
+ 6(3) - 12 = 60 ms
v =
ds
dt
= 6t2
+ 6t - 12
s = 2t3
+ 3t2
- 12t - 10
Ans:
v|t = 3s = 60 ms
vavg = 15 ms
(vsp)avg = 19.7 ms

8
*12–8.
A particle is moving along a straight line such that its
position is defined by , where t is in
seconds. Determine (a) the displacement of the particle
during the time interval from to , (b) the
average velocity of the particle during this time interval,
and (c) the acceleration when .
t = 1 s
t = 5 s
t = 1 s
s = (10t2
+ 20) mm
SOLUTION
(a)
Ans.
(b)
Ans.
(c) Ans.
a =
d2
s
dt2
= 20 mm s2
(for all t)
vavg =
¢s
¢t
=
240
4
= 60 mms
¢t = 5 - 1 = 4 s
¢s = 270 - 30 = 240 mm
s|5 s = 10(5)2
+ 20 = 270 mm
s|1 s = 10(1)2
+ 20 = 30 mm
s = 10t2
+ 20
Ans:
∆s = 240 mm
vavg = 60 mms
a = 20 mms2

9
12–9.
A particle moves along a straight path with an acceleration
of a = (5s) ms2
, where s is in meters. Determine the
particle’s velocity when s = 2 m, if it is released from rest
when s = 1 m.
SOLUTION
a ds = v dv
5
s
ds = v dv
5
L
2
1
ds
s
=
L
v
0
v dv
5 (ln s) `
2
1
=
v2
2
v = 2.63 m/s Ans.
Ans:
v = 2.63 m/s

10
12–10.
SOLUTION
Ans.
v = 1.29 ms
0.8351 =
1
2
v2
L
2
1
5 ds
A3s
1
3 + s
5
2
B
=
L
v
0
v d
a ds = v d
a =
5
A3s
1
3 + s
5
2
B
A particle moves along a straight line with an acceleration
of , where s is in meters.
Determine the particle’s velocity when , if it starts
from rest when .Use a numerical method to evaluate
the integral.
s = 1m
s = 2 m
a = 5(3s13
+ s52
) ms2
v
v
Ans:
v = 1.29 ms

11
12–11.
A particle travels along a straight-line path such that in 4 s
it moves from an initial position sA = -8 m to a position
sB = +3 m. Then in another 5 s it moves from sB to
sC = -6 m. Determine the particle’s average velocity and
average speed during the 9-s time interval.
SOLUTION
Average Velocity: The displacement from A to C is ∆s = sC - SA = -6 - (-8)
= 2 m.
vavg =
∆s
∆t
=
2
4 + 5
= 0.222 ms Ans.
Average Speed: The distances traveled from A to B and B to C are sASB = 8 + 3
= 11.0 m and sBSC = 3 + 6 = 9.00 m, respectively.Then, the total distance traveled
is sTot = sASB + sBSC = 11.0 + 9.00 = 20.0 m.
(vsp)avg =
sTot
∆t
=
20.0
4 + 5
= 2.22 ms Ans.
Ans:
vavg = 0.222 ms
(vsp)avg = 2.22 ms

12
*12–12.
The speed of a particle traveling along a straight line within
a liquid is measured as a function of its position
as v = (100 - s) mms, where s is in millimeters. Determine
(a) the particle’s deceleration when it is located at point A,
where sA = 75 mm, (b) the distance the particle travels
before it stops, and (c) the time needed to stop the particle.
SOLUTION
(a) a ds = v dv
a = v
dv
ds
= (100 - s)(-1) = s - 100
When s = sA = 75 mm, a = 75 - 100 = -25 mm/s2
Ans.
(b) v = 100 - s
0 = 100 - s s = 100 mm Ans.
(c) v =
ds
dt
dt =
ds
v
=
ds
100 - s
L
t
0
dt =
L
100
0
ds
100 - s
t = -ln(100 - s)|100
0 S ∞ Ans.
Ans:
a = -25 mm/s2
s = 100 mm
t = -ln(100 - s)|100
0 S ∞

13
12–13.
SOLUTION
Stopping Distance: For normal driver, the car moves a distance of
before he or she reacts and decelerates the car. The
stopping distance can be obtained using Eq. 12–6 with and .
Ans.
For a drunk driver, the car moves a distance of before he
or she reacts and decelerates the car. The stopping distance can be obtained using
Eq. 12–6 with and .
Ans.
d = 616 ft
02
= 442
+ 2(-2)(d - 132)
A :
+ B v2
= v2
0 + 2ac (s - s0)
v = 0
s0 = d¿ = 132 ft
d¿ = vt = 44(3) = 132 ft
d = 517 ft
02
= 442
+ 2(-2)(d - 33.0)
A :
+ B v2
= v2
0 + 2ac (s - s0)
v = 0
s0 = d¿ = 33.0 ft
d¿ = vt = 44(0.75) = 33.0 ft
d
v1 44 ft/s
Ans:
Normal: d = 517 ft
Drunk: d = 616 ft
Tests reveal that a driver takes about 0.75 s before he or she
can react to a situation to avoid a collision. It takes about 3 s
for a driver having 0.1% alcohol in his system to do the
same. If such drivers are traveling on a straight road at
30 mph (44 fts) and their cars can decelerate at 2 fts2
,
determine the shortest stopping distance d for each from
the moment they see the pedestrians. Moral: If you must
drink, please don’t drive!

14
12–14.
The sports car travels along the straight road such
that v = 32100-s ms, where s is in meters. Determine
the time for the car to reach s = 60 m. How much time does
it take to stop?
s
v
SOLUTION
s–t Function. This function can be determined by integrating ds = vdt using the
initial condition s = 0 at t = 0.
ds = 32100 - s dt
L
t
o
dt =
1
3 L
s
o
ds
2100 - s
t = -
2
3
(2100 - s)3
s
o
t =
2
3
(10 - 2100 - s)
When s = 60 m,
t =
2
3
(10 - 2100 - 60) = 2.4503s = 2.45s Ans.
When the car stops, v = 0. Then
0 = 32100 - s s = 100 m
Thus, the required time is
t =
2
3
(10 - 2100 - 100) = 6.6667s = 6.67s Ans.
Ans:
t = 2.45s
t = 6.67s

15
12–15.
A ball is released from the bottom of an elevator which is
traveling upward with a velocity of . If the ball strikes
the bottom of the elevator shaft in 3 s, determine the height
of the elevator from the bottom of the shaft at the instant
the ball is released.Also, find the velocity of the ball when it
strikes the bottom of the shaft.
6fts
SOLUTION
Kinematics: When the ball is released, its velocity will be the same as the elevator at
the instant of release. Thus, . Also, , , , and
.
Ans.
Ans.
= -90.6 fts = 90.6 fts T
v = 6 + (-32.2)(3)
A + c B v = v0 + act
h = 127 ft
-h = 0 + 6(3) +
1
2
(-32.2)A32
B
A + c B s = s0 + v0t +
1
2
ac t2
ac = -32.2 fts2
s = -h
s0 = 0
t = 3 s
v0 = 6 fts
Ans:
h = 127 ft
v = 90.6 fts T

16
SOLUTION
Distance Traveled: The distance traveled by the particle can be determined by
applying Eq. 12–3.
When Ans.
Time: The time required for the particle to stop can be determined by applying
Eq. 12–2.
When Ans.
v = 0, t = 3.266 - 1.333a0
1
2 b = 3.27 s
t = -1.333av
1
2 b 6 ms
v
= a3.266 - 1.333v
1
2 b s
L
t
0
dt = -
L
v
6 ms
dv
1.5v
1
2
dt =
dv
a
v = 0, s = -0.4444a0
3
2 b + 6.532 = 6.53 m
= a -0.4444v
3
2 + 6.532b m
s =
L
v
6 ms
-0.6667 v
1
2 dv
L
s
0
ds =
L
v
6 ms
v
-1.5v
1
2
dv
ds =
vdv
a
Ans:
s = 6.53 m
t = 3.27 s
*12–16.
A particle is moving along a straight line with an initial
velocity of 6 ms when it is subjected to a deceleration of
a = (-1.5v12) ms2, where v is in ms. Determine how far it
travels before it stops. How much time does this take?

17
12–17.
SOLUTION
For B:
(1)
For A:
(2)
Require the moment of closest approach.
Worst case without collision would occur when .
At , from Eqs. (1) and (2):
Ans.
d = 16.9 ft
157.5 = d + 140.625
60(0.75) + 60(3.75 - 0.75) - 7.5(3.75 - 0.75)2
= d + 60(3.75) - 6(3.75)2
t = 3.75 s
sA = sB
t = 3.75 s
60 - 12t = 60 - 15(t - 0.75)
vA = vB
sA = 60(0.75) + 60(t - 0.75) -
1
2
(15) (t - 0.75)2
, [t 7 0.74]
( :
+ ) s = s0 + v0 t +
1
2
act2
vA = 60 - 15(t - 0.75), [t 7 0.75]
( :
+ ) v = v0 + act
sB = d + 60t -
1
2
(12) t2
( :
+ ) s = s0 + v0 t +
1
2
act2
vB = 60 - 12 t
( :
+ ) v = v0 + ac t
Car B is traveling a distance d ahead of car A. Both cars are
traveling at when the driver of B suddenly applies the
brakes, causing his car to decelerate at . It takes the
driver of car A 0.75 s to react (this is the normal reaction
time for drivers).When he applies his brakes, he decelerates
at . Determine the minimum distance d between the
cars so as to avoid a collision.
15 fts2
12 fts2
60 fts
d
A B
Ans:
d = 16.9 ft

18
12–18.
Car A starts from rest at and travels along a straight
road with a constant acceleration of until it reaches a
speed of . Afterwards it maintains this speed. Also,
when , car B located 6000 ft down the road is traveling
towards A at a constant speed of . Determine the
distance traveled by car A when they pass each other.
60 fts
t = 0
80 fts
6 fts2
t = 0
SOLUTION
Distance Traveled: Time for car A to achives can be obtained by
applying Eq. 12–4.
The distance car A travels for this part of motion can be determined by applying
Eq. 12–6.
For the second part of motion, car A travels with a constant velocity of
and the distance traveled in ( is the total time) is
Car B travels in the opposite direction with a constant velocity of and
the distance traveled in is
It is required that
The distance traveled by car A is
Ans.
sA = s1 + s2 = 533.33 + 80(46.67 - 13.33) = 3200 ft
t1 = 46.67 s
533.33 + 80(t1 - 13.33) + 60t1 = 6000
s1 + s2 + s3 = 6000
A :
+ B s3 = vt1 = 60t1
t1
v = 60 fts
A :
+ B s2 = vt¿ = 80(t1 - 13.33)
t1
t¿ = (t1 - 13.33) s
v = 80 fts
s1 = 533.33 ft
802
= 0 + 2(6)(s1 - 0)
A :
+ B v2
= v0
2
+ 2ac (s - s0)
t = 13.33 s
80 = 0 + 6t
A :
+ B v = v0 + ac t
y = 80 fts
A B
6000 ft
60 ft/s
Ans:
sA = 3200 ft

19
12–19.
A particle moves along a straight path with an acceleration
of a = (kt3 + 4) fts2, where t is in seconds. Determine the
constant k, knowing that v = 12 fts when t = 1 s, and that
v = -2 fts when t = 2 s.
SOLUTION
a =
dv
dt
(kt3
+ 4)dt = dv
L
2
1
(kt3
+ 4)dt =
L
-2
12
dv
1
4
kt4
+ 4t `
2
1
= v `
-2
12
k = -4.80 ft/s5
Ans.
Ans:
k = -4.80 ft/s5

20
*12–20.
The velocity of a particle traveling along a straight line is
, where is in seconds. If when
, determine the position of the particle when .
What is the total distance traveled during the time interval
to ?Also,what is the acceleration when ?
t = 2 s
t = 4 s
t = 0
t = 4 s
t = 0
s = 4 ft
t
v = (3t2
- 6t) fts
SOLUTION
Position: The position of the particle can be determined by integrating the kinematic
equation using the initial condition when Thus,
When
Ans.
The velocity of the particle changes direction at the instant when it is momentarily
brought to rest.Thus,
and
The position of the particle at and 2 s is
Using the above result, the path of the particle shown in Fig. a is plotted. From this
figure,
Ans.
Acceleration:
When
Ans.
a ƒt=2 s = 6122 - 6 = 6 fts2
:
t = 2 s,
a = 16t - 62 fts2
a =
dv
dt
=
d
dt
(3t2
- 6t)
A +
: B
sTot = 4 + 20 = 24 ft
s|2 s
= 23
- 3A22
B + 4 = 0
s|0 s
= 0 - 3A02
B + 4 = 4 ft
t = 0
t = 2 s
t = 0
t(3t - 6) = 0
v = 3t2
- 6t = 0
s|t = 4 s = 43
- 3(42
) + 4 = 20 ft
t = 4 s,
s = At3
- 3t2
+ 4B ft
s 2
s
4 ft
= (t 3
- 3t2
) 2
t
0
L
s
4 ft
ds =
L
t
0
A3t
2
- 6tBdt
ds = v dt
A +
: B
t = 0 s.
s = 4 ft
ds = v dt
Ans:
s t = 4 s = 20 ft
sTot = 24 ft
at = 2 s = 6 fts2 S

21
12–21.
A freight train travels at where t is the
elapsed time in seconds. Determine the distance traveled in
three seconds, and the acceleration at this time.
v = 6011 - e-t
2 fts,
SOLUTION
Ans.
At
Ans.
a = 60e-3
= 2.99 fts2
t = 3 s
a =
dv
dt
= 60(e-t
)
s = 123 ft
s = 60(t + e-t
)|0
3
L
s
0
ds =
L
v dt =
L
3
0
6011 - e-t
2dt
v = 60(1 - e-t
)
s v
Ans:
s = 123 ft
a = 2.99 fts2

22
12–22.
The acceleration of the boat is defined by
a = (1.5 v12) ms. Determine its speed when t = 4 s if it has
a speed of 3 ms when t = 0.
v
SOLUTION
V-t Function: This function can be determined by integrating dv = adt using the
initial condition v = 3 m/s at t = 0,
dv =
3
2
v
1
2 dt
L
t
o
dt =
2
3 L
v
3 m/s
dv
v
1
2
t =
4
3
v
1
2
`
v
3 m/s
t =
4
3
av
1
2
- 23b
v = a
3
4
t + 23b
2
m/s
When t = 4 s,
v = c
3
4
(4) + 23d
2
= 22.39 m/s = 22.4 m/s Ans.
Ans:
v = 22.4 ms

23
12–23.
acceleration is defined as a = (-2v) ms2, where v is in
meters per second. If v = 20 ms when s = 0 and t = 0,
determine the particle’s position, velocity, and acceleration
as functions of time.
SOLUTION
a = -2v
dv
dt
= -2v
L
v
20
dv
v
=
L
t
0
-2 dt
ln
v
20
= -2t
v = (20e-2t
) ms Ans.
a =
dv
dt
= (-40e-2t
) ms2
Ans.
L
s
0
ds = v dt =
L
t
0
(20e-2t
)dt
s = -10e-2t
t
0 = -10(e-2t
- 1)
s = 10(1 - e-2t
) m Ans.
Ans:
v = (20e -2t
) ms
a = (-40e -2t
) ms2
s = 10(1 - e-2t
) m

24
*12–24.
SOLUTION
Position:
The particle achieves its maximum height when .Thus,
Ans.
=
1
2k
ln¢1 +
k
g
v0
2
≤
hmax =
1
2k
ln¢
g + kv0
2
g
≤
v = 0
s =
1
2k
ln¢
g + kv0
2
g + kv2 ≤
s|s
0 = - c
1
2k
lnAg + kv2
B d 2
v
v0
L
s
0
ds =
L
v
v0
-
vdv
g + kv2
A + c B ds =
v dv
a
When a particle is projected vertically upward with an
initial velocity of , it experiences an acceleration
where g is the acceleration due to gravity,
k is a constant, and is the velocity of the particle.
Determine the maximum height reached by the particle.
v
a = -(g + kv2
),
v0
Ans:
hmax = 1
2kln¢1 + k
g v0
2
≤

25
12–25.
If the effects of atmospheric resistance are accounted for, a
falling body has an acceleration defined by the equation
, where is in and the
positive direction is downward. If the body is released from
rest at a very high altitude, determine (a) the velocity when
, and (b) the body’s terminal or maximum attainable
velocity (as ).
t : q
t = 5 s
ms
v
a = 9.81[1 - v2
(10-4
)] ms2
SOLUTION
Velocity: The velocity of the particle can be related to the time by applying Eq. 12–2.
(1)
a) When , then, from Eq. (1)
Ans.
b) If , .Then, from Eq. (1)
Ans.
vmax = 100 ms
e0.1962t
- 1
e0.1962t
+ 1
: 1
t : q
v =
100[e0.1962(5)
- 1]
e0.1962(5)
+ 1
= 45.5 ms
t = 5 s
v =
100(e0.1962t
- 1)
e0.1962t
+ 1
9.81t = 50lna
1 + 0.01v
1 - 0.01v
b
t =
1
9.81
c
L
v
0
dv
2(1 + 0.01v)
+
L
v
0
dv
2(1 - 0.01v)
d
L
t
0
dt =
L
v
0
dv
9.81[1 - (0.01v)2
]
(+ T) dt =
dv
a
Ans:
(a) v = 45.5 ms
(b) v max = 100 ms
If the effects of atmospheric resistance are accounted for, a
freely falling body has an acceleration defined by the
equation, a = 9.81[1 - v2
(10-4
)] ms2
, where v is in ms
and the positive direction is downward. If the body is
released from rest at a very high altitude, determine (a) the
velocity when t = 5 s, and (b) the body’s terminal or
maximum attainable velocity (as t S ∞).

26
12–26.
A ball is thrown with an upward velocity of 5 ms
from the top of a 10-m-high building. One second later
another ball is thrown upward from the ground with a
velocity of 10 ms. Determine the height from the ground
where the two balls pass each other.
Ans:
h = 4.54 m
SOLUTION
Kinematics: First, we will consider the motion of ball A with (vA)0 5 5 ms,
(sA)0 5 0, sA 5 (h 2 10)m, tA 5 t′, and ac 5 29.81 ms2.Thus,
(+ c ) sA = (sA)0 + (vA)0 tA +
1
2
act 2
A
h - 10 = 0 + 5t′ +
1
2
(-9.81)(t′)2
h = 5t′ - 4.905(t′)2
+ 10 (1)
Motion of ball B is with (vB)0 5 10 ms, (sB)0 5 0, sB 5 h, tB 5 t′ 2 1 and
ac 5 29.81 ms2.Thus,
(+ c ) sB = (sB)0 + (vB)0 tB +
1
2
act 2
B
h = 0 + 10(t′ - 1) +
1
2
(-9.81)(t′ - 1)2
h = 19.81t′ - 4.905(t′)2
- 14.905 (2)
Solving Eqs. (1) and (2) yields
h = 4.54 m Ans.
t′ = 1.68 s

27
12–27.
When a particle falls through the air, its initial acceleration
diminishes until it is zero, and thereafter it falls at a
constant or terminal velocity . If this variation of the
acceleration can be expressed as
determine the time needed for the velocity to become
Initially the particle falls from rest.
v = vf2.
a = 1gv2
f21v2
f - v2
2,
vf
a = g
SOLUTION
Ans.
t = 0.549 a
vf
g
b
t =
vf
2g
ln ¢
vf + vf2
vf - vf2
≤
t =
vf
2g
ln ¢
vf + v
vf - v
≤
1
2vf
ln ¢
vf + v
vf - v
≤ `
y
0
=
g
v2
f
t
L
v
0
dy
v2
f - v2¿
=
g
v2
f L
t
0
dt
dv
dt
= a = ¢
g
v2
f
≤ Av2
f - v2
B
Ans:
t = 0.549a
vf
g
b

28
*12–28.
A train is initially traveling along a straight track at a speed
of 90 kmh. For 6 s it is subjected to a constant deceleration
of 0.5 ms2, and then for the next 5 s it has a constant
deceleration ac. Determine ac so that the train stops at the
end of the 11-s time period.
SOLUTION
First stage of motion:
v = v0 + act
=
90(10)3
3600
+ (-0.5)(6)
= 22 m/s
Second stage of motion:
v = v0 + act
0 = 22 + ac(5)
ac = -4.40 m/s2
Ans.
Ans:
ac = -4.40 m/s2

29
12–29.
Two cars A and B start from rest at a stop line.
Car A has a constant acceleration of aA = 8 ms2, while
car B has an acceleration of aB = (2t32) ms2, where t is in
seconds. Determine the distance between the cars when A
reaches a velocity of vA = 120 kmh.
SOLUTION
For Car A:
1 +
S2 v2
= v0
2
+ 2ac(s -s0)
c
120(10)3
3600
d
2
= 0 + 2(8)(sA - 0)
sA = 69.44 m
1 +
S2 v = v0 + act

120(10)3
3600
= 0 + 8t t = 4.1667 s
For Car B:
1 +
S2
dv
dt
= a
dv = 2t
3
2
dt
3
vB
0
dv = 3
t
0
2t
3
2
dt
vB =
4
5
t
5
2
1 +
S2
ds
dt
= v
ds =
4
5
t
5
2
dt
3
sB
0
ds =
4
53
4.1667
0
t
5
2
dt
sB = 33.75 m
The distance between cars A and B is sAB = 69.44 - 33.75 = 35.7 m Ans.
Ans:
sAB = 35.7 m

30
12–30.
A sphere is fired downward into a medium with an initial
speed of . If it experiences a deceleration of
where t is in seconds, determine the
distance traveled before it stops.
a = (-6t) ms2
,
27 ms
SOLUTION
Velocity: at .Applying Eq. 12–2, we have
(1)
At , from Eq. (1)
Distance Traveled: at . Using the result and applying
Eq. 12–1, we have
(2)
At , from Eq. (2)
Ans.
s = 27(3.00) - 3.003
= 54.0 m
t = 3.00 s
s = A27t - t3
B m
L
s
0
ds =
L
t
0
A27 - 3t2
B dt
A + T B ds = vdt
v = 27 - 3t2
t0 = 0 s
s0 = 0 m
0 = 27 - 3t2
t = 3.00 s
v = 0
v = A27 - 3t2
B ms
L
v
27
dv =
L
t
0
-6tdt
A + T B dv = adt
t0 = 0 s
v0 = 27 ms
Ans:
s = 54.0 m

31
12–31.
position is given by , where t is in seconds.
Determine the distance traveled from to , the
average velocity, and the average speed of the particle
during this time interval.
t = 5 s
t = 0
s = (4t - t2
) ft
SOLUTION
Total Distance Traveled: The velocity of the particle can be determined by applying
Eq. 12–1.
The times when the particle stops are
The position of the particle at , 2 s and 5 s are
From the particle’s path, the total distance is
Ans.
Average Velocity: The displacement of the particle from to 5 s is
.
Ans.
Average Speed:
Ans.
ysp avg =
sTot
¢t
=
13.0
5
= 2.60 ft s
yavg =
¢s
¢t
=
-5
5
= -1.00 fts
¢s = s|t = 5 t - s|t = 0 s = -5.00 - 0 = -5.00 ft
t = 0 s
stot = 4.00 + 9.00 = 13.0 ft
s|t = 5 s = 4(5) - A52
B = -5.00 ft
s|t = 2 s = 4(2) - A22
B = 4.00 ft
s|t = 0 s = 4(0) - A02
B = 0
t = 0 s
4 - 2t = 0 t = 2 s
y =
ds
dt
= 4 - 2t
Ans:
Stot = 13.0 ft
vavg = - 1.00 fts
(vsp)avg = 2.60 fts

32
Ans:
t = 2 s
d = 20.4 m
*12–32.
Ball A is thrown vertically upward from the top of a 30-m-
high building with an initial velocity of 5 . At the same
instant another ball B is thrown upward from the ground
with an initial velocity of 20 . Determine the height from
the ground and the time at which they pass.
ms
ms
SOLUTION
Origin at roof:
Ball A:
Ball B:
Solving,
Ans.
Distance from ground,
Ans.
Also, origin at ground,
Require
sB = 20.4 m
t = 2 s
30 + 5t +
1
2
(-9.81)t2
= 20t +
1
2
(-9.81)t2
sA = sB
sB = 0 + 20t +
1
2
(-9.81)t2
sA = 30 + 5t +
1
2
(-9.81)t2
s = s0 + v0 t +
1
2
ac t2
d = (30 - 9.62) = 20.4 m
s = 9.62 m
t = 2 s
-s = -30 + 20t -
1
2
(9.81)t2
A + c B s = s0 + v0 t +
1
2
ac t2
-s = 0 + 5t -
1
2
(9.81)t2
A + c B s = s0 + v0 t +
1
2
ac t2

33
12–33.
As a body is projected to a high altitude above the earth’s
surface, the variation of the acceleration of gravity with
respect to altitude y must be taken into account. Neglecting
air resistance, this acceleration is determined from the
formula , where is the constant
gravitational acceleration at sea level, R is the radius of the
earth, and the positive direction is measured upward. If
and , determine the minimum
initial velocity (escape velocity) at which a projectile should
be shot vertically from the earth’s surface so that it does not
fall back to the earth. Hint: This requires that as
y : q.
v = 0
R = 6356 km
g0 = 9.81 ms2
g0
a = -g0[R2
(R + y)2
]
SOLUTION
Ans.
= 11167 ms = 11.2 kms
= 22(9.81)(6356)(10)3
v = 22g0 R
v2
2
2
0
y
=
g0 R2
R + y
2
q
0
L
0
y
v dv = -g0R
2
L
q
0
dy
(R + y)
2
v dv = a dy
Ans:
v = 11.2 kms

34
12–34.
SOLUTION
F
From Prob. 12–33,
Since
then
Thus
When , ,
Ans.
v = -3016 ms = 3.02 kms T
v = -6356(103
)
A
2(9.81)(500)(103
)
6356(6356 + 500)(106
)
y = 0
y0 = 500 km
v = -R
A
2g0 (y0 - y)
(R + y)(R + y0)
g0 R2
[
1
R + y
-
1
R + y0
] =
v2
2
g0 R2
c
1
R + y
d
y
y0
=
v2
2
-g0 R2
L
y
y0
dy
(R + y)2
=
L
v
0
v dv
a dy = v dv
(+ c) a = -g0
R2
(R + y)2
Accounting for the variation of gravitational acceleration
a with respect to altitude y (see Prob. 12–33), derive an
equation that relates the velocity of a freely falling particle
to its altitude. Assume that the particle is released from
rest at an altitude y0 from the earth’s surface. With what
velocity does the particle strike the earth if it is released
from rest at an altitude y0 = 500 km? Use the numerical
data in Prob. 12–33.
Ans:
v = -R
B
2g0(y0 - y)
(R + y)(R + y0)
vimp = 3.02 kms
Ans.

35
12–35.
A freight train starts from rest and travels with a constant
acceleration of . After a time it maintains a
constant speed so that when it has traveled 2000 ft.
Determine the time and draw the –t graph for the motion.
v
t¿
t = 160 s
t¿
0.5 fts2
SOLUTION
Total Distance Traveled: The distance for part one of the motion can be related to
time by applying Eq. 12–5 with and .
The velocity at time t can be obtained by applying Eq. 12–4 with .
(1)
The time for the second stage of motion is and the train is traveling at
a constant velocity of (Eq. (1)).Thus, the distance for this part of motion is
If the total distance traveled is , then
Choose a root that is less than 160 s, then
Ans.
t¿ = 27.34 s = 27.3 s
0.25(t¿)2
- 80t¿ + 2000 = 0
2000 = 0.25(t¿)2
+ 80t¿ - 0.5(t¿)2
sTot = s1 + s2
sTot = 2000
A :
+ B s2 = vt2 = 0.5t¿(160 - t¿) = 80t¿ - 0.5(t¿)2
v = 0.5t¿
t2 = 160 - t¿
A :
+ B v = v0 + act = 0 + 0.5t = 0.5t
v0 = 0
s1 = 0 + 0 +
1
2
(0.5)(t¿)2
= 0.25(t¿)2
A :
+ B s = s0 + v0 t +
1
2
ac t2
v0 = 0
s0 = 0
t = t¿
v–t Graph: The equation for the velocity is given by Eq. (1).When t = t¿ = 27.34 s,
v = 0.5(27.34) = 13.7 fts.
Ans:
t′ = 27.3 s.
When t = 27.3 s, v = 13.7 fts.

36
*12–36.
The s–t graph for a train has been experimentally
determined. From the data, construct the v–t and a–t graphs
for the motion; 0 … t … 40 s. For 0 … t 6 30 s, the curve is
s = (0.4t2) m, and then it becomes straight for t 7 30 s.
SOLUTION
0 … t 6 30:
s = 0.4t2
v =
ds
dt
= 0.8t
a =
dv
dt
= 0.8 ms2
30 6 t … 40:
s - 360 = a
600 - 360
40 - 30
b(t - 30)

s = 24t - 360

v =
ds
dt
= 24 ms

a =
dv
dt
= 0
t (s)
s (m)
600
360
30 40
Ans:
v =
ds
dt
= 0.8t
a =
dv
dt
= 0.8 ms2
s = 24t - 360
v =
ds
dt
= 24 ms
a =
dv
dt
= 0

37
12–37.
Two rockets start from rest at the same elevation. Rocket A
accelerates vertically at 20 ms2 for 12 s and then maintains
a constant speed. Rocket B accelerates at 15 ms2 until
reaching a constant speed of 150 ms. Construct the a–t, v–t,
and s–t graphs for each rocket until t = 20 s. What is the
distance between the rockets when t = 20 s?
SOLUTION
For rocket A:
For t 6 12 s
+ c vA = (vA)0 + aA t
vA = 0 + 20 t
vA = 20 t
+ c sA = (sA)0 + (vA)0 t +
1
2
aA t2
sA = 0 + 0 +
1
2
(20) t2
sA = 10 t2
When t = 12 s, vA = 240 ms
sA = 1440 m
For t 7 12 s
vA = 240 ms
sA = 1440 + 240(t - 12)
For rocket B:
For t 6 10 s
+ c vB = (vB)0 + aB t
vB = 0 + 15 t
vB = 15 t
+ c sB = (sB)0 + (vB)0 t +
1
2
aB t2
sB = 0 + 0 +
1
2
(15) t2
sB = 7.5 t2
When t = 10 s, vB = 150 ms
sB = 750 m
For t 7 10 s
vB = 150 ms
sB = 750 + 150(t - 10)
When t = 20 s, sA = 3360 m, sB = 2250 m
∆s = 1110 m = 1.11 km Ans.
Ans:
∆s = 1.11 km

38
12–38.
A particle starts from and travels along a straight line
with a velocity , where is in
seconds. Construct the and graphs for the time
interval .
0 … t … 4 s
a-t
v-t
t
v = (t2
- 4t + 3) m s
s = 0
SOLUTION
a–t Graph:
Thus,
The graph is shown in Fig. a.
Graph: The slope of the graph is zero when .Thus,
The velocity of the particle at , 2 s, and 4 s are
The graph is shown in Fig. b.
v-t
v|t=4 s = 42
- 4(4) + 3 = 3 ms
v|t=2 s = 22
- 4(2) + 3 = -1 ms
v|t=0 s = 02
- 4(0) + 3 = 3 ms
t = 0 s
t = 2 s
a = 2t - 4 = 0
a =
dv
dt
= 0
v-t
v–t
a-t
a|t=4 s = 2(4) - 4 = 4 ms2
a|t=2 = 0
a|t=0 = 2(0) - 4 = -4 ms2
a = (2t - 4)ms2
a =
dv
dt
=
d
dt
1t2
- 4t + 32
Ans:
at=0 = -4 ms2
at=2 s = 0
at=4 s = 4 ms2
vt=0 = 3 ms
vt=2 s = -1 ms
vt=4 s = 3 ms

39
12–39.
SOLUTION
If the position of a particle is defined by
the , , and graphs for .
0 … t … 10 s
a-t
v-t
s-t
s = [2 sin (p5)t + 4] , where t is in seconds, construct
m
Ans:
s = 2 sin a
p
5
tb + 4
v =
2p
5
cos a
p
5
tb
a = -
2p2
25
sin a
p
5
tb

40
*12–40.
An airplane starts from rest, travels 5000 ft down a runway,
and after uniform acceleration, takes off with a speed of
It then climbs in a straight line with a uniform
acceleration of until it reaches a constant speed of
Draw the s–t, v–t, and a–t graphs that describe
the motion.
220 mih.
3 fts2
162 mih.
SOLUTION
t = 28.4 s
322.67 = 237.6 + 3 t
v3 = v2 + act
s = 12 943.34 ft
(322.67)2
= (237.6)2
+ 2(3)(s - 5000)
v2
3 = v2
2 + 2ac(s3 - s2)
v3 = 220
mi
h
(1h) 5280 ft
(3600 s)(1 mi)
= 322.67 fts
t = 42.09 = 42.1 s
237.6 = 0 + 5.64538 t
v2 = v1 + act
ac = 5.64538 fts2
(237.6)2
= 02
+ 2(ac)(5000 - 0)
v2
2 = v2
1 + 2 ac(s2 - s1)
v2 = 162
mi
h
(1h) 5280 ft
(3600 s)(1 mi)
= 237.6 fts
v1 = 0
Ans:
0 … t 6 42.1 s, a = 5.65 fts2
vt = 42.1 s = 238 fts s t = 42.1 s = 5000 ft
42.1 s 6 t … 70.4 s, a = 3 fts2
vt =70.4 s = 323 fts s t = 70.4 s = 12943 ft

41
12–41.
SOLUTION
Thus,
Thus,
When t = 2.145 s, v = vmax = 10.7 fts
and h = 11.4 ft.
Ans.
t = t1 + t2 = 7.48 s
t2 = 5.345 s
t1 = 2.138 s
vmax = 10.69 fts
h = 11.429 ft
10 h = 160 - 4h
v2
max = 160 - 4h
0 = v2
max + 2(-2)(40 - h)
v2
max = 10h
v2
max = 0 + 2(5)(h - 0)
+ c v2
= v2
1 + 2 ac(s - s1)
+ c 40 - h = 0 + vmaxt2 -
1
2
(2) t2
2
h = 0 + 0 +
1
2
(5)(t2
1) = 2.5 t2
1
+ c s2 = s1 + v1t1+
1
2
act2
1
t1 = 0.4 t2
0 = vmax - 2 t2
+ c v3 = v2 + ac t
vmax = 0 + 5 t1
+ c v2 = v1 + act1
The elevator starts from rest at the first floor of the
building. It can accelerate at and then decelerate at
Determine the shortest time it takes to reach a floor
40 ft above the ground. The elevator starts from rest and
then stops. Draw the a–t, v–t, and s–t graphs for the motion.
2 fts2
.
5 fts2
40 ft
Ans:
t = 7.48 s. When t = 2.14 s,
v = vmax = 10.7 fts
h = 11.4 ft

42
12–42.
A car starting from rest moves along a straight track with an
acceleration as shown. Determine the time t for the car to
reach a speed of 50 ms and construct the v–t graph that
describes the motion until the time t.
10 t
t (s)
8
a (ms2
)
SOLUTION
For 0 … t … 10 s,
a =
8
10
t
dv = a dt
3
v
0
dv = 3
t
0
8
10
t dt
v =
8
20
t2
At t = 10 s,
v =
8
20
(10)2
= 40 ms
For t 7 10 s,
a = 8
dv = a dt
3
v
40
dv = 3
t
10
8 dt
v - 40 = 8t - 80
v = 8t - 40 When v = 50 ms
t =
50 + 40
8
= 11.25 s Ans.
Ans:
t = 11.25 s

43
SOLUTION
v–t Graph: The v–t function can be determined by integrating dv = a dt.
For 0 … t 6 10 s, a = 0. Using the initial condition v = 300 fts at t = 0,
L
300 fts
dv =
L
t
0
0 dt
v - 300 = 0
v = 300 fts Ans.
For 10 s 6 t 6 20 s,
a - (-20)
t - 10
=
-10 - (-20)
20 - 10
, a = (t - 30) fts2
. Using the
initial condition v = 300 fts at t = 10 s,
L
300 fts
dv =
L
t
10 s
(t - 30) dt
v - 300 = a
1
2
t2
- 30tb `
10 s
t
v = e
1
2
t2
- 30t + 550f fts Ans.
At t = 20 s,
v `
t=20 s
=
1
2
(202
) - 30(20) + 550 = 150 fts
For 20 s 6 t 6 t′, a = -10 fts. Using the initial condition v = 150 fts at t = 20 s,
L
150 fts
dv =
L
t
20 s
- 10 dt
v - 150 = (-10t) `
t
20 s
v = (-10t + 350) fts Ans.
It is required that at t = t′, v = 0.Thus
0 = -10 t′ + 350
t′ = 35 s Ans.
Using these results, the v9t graph shown in Fig. a can be plotted s-t Graph.
The s9t function can be determined by integrating ds = v dt. For 0 … t 6 10 s, the
initial condition is s = 0 at t = 0.
L
s
0
ds =
L
t
0
300 dt
s = {300 t} ft Ans.
At = 10 s,
s 0 t=10 s = 300(10) = 3000 ft
12–43.
The motion of a jet plane just after landing on a runway is
described by the a–t graph. Determine the time t′ when the
jet plane stops. Construct the v–t and s–t graphs for the
motion. Here s = 0, and v = 300 fts when t = 0.
t (s)
10
a (ft/s2
)
10
20 t¿
20
v
v
v

44
12–43. Continued
For 10 s 6 t 6 20 s, the initial condition is s = 3000 ft at t = 10 s.
L
s
3000 ft
ds =
L
t
10 s
a
1
2
t2
- 30t + 550bdt
s - 3000 = a
1
6
t3
- 15t2
+ 550tb `
t
10 s
s = e
1
6
t3
- 15t2
+ 550t - 1167f ft Ans.
At t = 20 s,
s =
1
6
(203
) - 15(202
) + 550(20) - 1167 = 5167 ft
For 20 s 6 t … 35 s, the initial condition is s = 5167 ft at t = 20 s.
L
s
5167 ft
ds =
L
t
20 s
(-10t + 350) dt
s - 5167 = (-5t2
+ 350t) `
t
20 s
s = 5 -5t2
+ 350t + 1676 ft Ans.
At t = 35 s,
s `
t = 35 s
= -5(352
) + 350(35) + 167 = 6292 ft
Using these results, the s-t graph shown in Fig. b can be plotted.
Ans:
t′ = 35 s
For 0 … t 6 10 s,
s = {300t} ft
v = 300 fts
For 10 s 6 t 6 20 s,
s = e
1
6
t3
- 15t2
+ 550t - 1167f ft
v = e
1
2
t2
- 30t + 550f fts
For 20 s 6 t … 35 s,
s = 5 -5t2
+ 350t + 1676 ft
v = (-10t + 350) fts

45
*12–44.
The v–t graph for a particle moving through an electric field
from one plate to another has the shape shown in the figure.
The acceleration and deceleration that occur are constant
and both have a magnitude of If the plates are
spaced 200 mm apart, determine the maximum velocity
and the time for the particle to travel from one plate to
the other. Also draw the s–t graph. When the
particle is at s = 100 mm.
t = t¿2
t¿
vmax
4 ms2
.
SOLUTION
Ans.
Ans.
When ,
When ,
s = 0.2 m
t = 0.447 s
s = - 2 t2
+ 1.788 t - 0.2
L
s
0.1
ds =
L
t
0.2235
1-4t+1.7882 dt
v = -4 t+1.788
L
v
0.894
ds = -
L
t
0.2235
4 dt
s = 0.1 m
t =
0.44721
2
= 0.2236 = 0.224 s
s = 2 t2
s = 0 + 0 +
1
2
(4)(t)2
s = s0 + v0 t +
1
2
ac t2
t¿ = 0.44721 s = 0.447 s
0.89442 = 0 + 4(
t¿
2
)
v = v0 + ac t¿
vmax = 0.89442 ms = 0.894 ms
v2
max = 0 + 2(4)(0.1 - 0)
v2
= v2
0 + 2 ac(s - s0)
s
2
= 100 mm = 0.1 m
ac = 4 m/s2
t¿/2 t¿
t
v
smax
vmax
s
Ans:
t=
= 0.447 s
s = 0.2 m

46
Ans:
When t = 0.1 s,
s = 0.5 m and a changes from
100 ms2
to -100 ms2
. When t = 0.2 s,
s = 1 m.
12–45.
SOLUTION
For ,
When ,
For ,
When ,
s = 1 m
s = 0.5 m and a changes
t = 0.2 s
When ,
t = 0.1 s
s = 1 m.
from 100 m/s2
s = - 50 t2
+ 20 t - 1
s - 0.5 = (-50 t2
+ 20 t - 1.5)
L
s
0.5
ds =
L
t
0.1
1-100t + 202dt
ds = v dt
a =
dv
dt
= - 100
v = -100 t + 20
0.1 s 6 t 6 0.2 s
s = 0.5 m
t = 0.1 s
s = 50 t2
L
s
0
ds =
L
t
0
100 t dt
ds = v dt
a =
dv
dt
= 100
v = 100 t
0 6 t 6 0.1 s
t¿/2 t¿
t
v
smax
vmax
s
The v–t graph for a particle moving through an electric field
from one plate to another has the shape shown in the figure,
where t¿ = 0.2 s and vmax = 10 ms. Draw the s–t and a–t graphs
for the particle. When t = t¿2 the particle is at s = 0.5 m.
to -100 m/s2. When t = 0.2 s,

47
12–46.
The a–s graph for a rocket moving along a straight track has
been experimentally determined. If the rocket starts at s = 0
when v = 0, determine its speed when it is at
s = 75 ft, and 125 ft, respectively. Use Simpson’s rule with
n = 100 to evaluate v at s = 125 ft.
s (ft)
a (ft/s2
)
100
5
a 5 6(s 10)5/3
Ans:
v `
s=75 ft
= 27.4 fts
v `
s=125 ft
= 37.4 fts
SOLUTION
0 … s 6 100
L
0
v dv =
L
s
0
5 ds
1
2
v2
= 5 s
v = 210 s
At s = 75 ft, v = 2750 = 27.4 fts Ans.
At s = 100 ft, v = 31.623
v dv = ads
L
31.623
v dv =
L
125
100
35 + 6(2s - 10)53
4 ds
1
2
v2
`
31.623
= 201.0324
v = 37.4 fts Ans.
v
v
v

48
12–47.
SOLUTION
Graph: For the time interval , the initial condition is at .
When ,
The initial condition is at .
When ,
The v–t graph is shown in Fig. a.
Graph: For the time interval , the initial condition is when
.
When ,
For the time interval , the initial condition is when
.
When ,
The s–t graph is shown in Fig. b.
s t=14 s =
2
3
A143
B - 9A142
B + 108(14) - 340.2 = 1237 m
t = 14 s
s = a
2
3
t3
- 9t2
+ 108t - 340.2b m
L
s
388.8 m
ds =
L
t
9 s
A2t2
- 18t + 108Bdt
A + c B ds = vdt
t = 9 s
s = 388.8 m
9 s 6 t … 14 s
s t=9 s =
8
5
A952
B = 388.8 m
t = 9 s
s =
8
5
t52
L
s
0
ds =
L
t
0
4t32
dt
A + c B ds = vdt
t = 0
s = 0
0 … t 6 9 s
s t
v t=14 s = 2A142
B - 18(14) + 108 = 248 ms
t = 14 s
v = A2t2
- 18t + 108B ms
L
v
108 ms
dv =
L
t
9 s
(4t - 18)dt
A + c B dv = adt
t = 9 s
v = 108 ms
v t=9 s = 4A932
B = 108 ms
t = 9 s
v = A4t32
Bms
L
v
0
dv =
L
t
0
6t12
dt
A + c B dv = adt
s = 0
v = 0
0 … t 6 9 s
v t
The rocket has an acceleration described by the graph. If it
starts from rest, construct the and graphs for the
motion for the time interval .
0 … t … 14s
s-t
v-t
t(s)
a(m/s2
)
38
18
9 14
a2
36t
a 4t 18

49
t (s)
600
360
30 40
s (m)
s 24t 360
s 0.4t2
5
5
*12–48.
Thes–tgraphforatrainhasbeendeterminedexperimentally.
From the data, construct the v–t and a–t graphs for the
motion.
SOLUTION
v–t Graph: The velocity in terms of time t can be obtained by applying v =
ds
dt
.
For time interval 0 s … t … 30 s,
v =
ds
dt
= 0.8t
When t = 30 s, v = 0.8(30) = 24.0 ms
For time interval 30 s 6 t … 40 s,
v =
ds
dt
= 24.0 m/s
a–t Graph: The acceleration in terms of time t can be obtained by applying a =
dv
dt
.
For time interval 0 s … t 6 30 s and 30 s 6 t … 40 s, a =
dv
dt
= 0.800 ms2
and
a =
dv
dt
= 0, respectively.

50
12–49.
The jet car is originally traveling at a velocity of 10 ms
when it is subjected to the acceleration shown. Determine
the car’s maximum velocity and the time t′ when it stops.
When t = 0, s = 0.
6
15
4
t (s)
a (m/s2
)
t¿
Ans:
vmax = 100 ms
t′ = 40 s
SOLUTION
V9t Function: The v-t function can be determined by integrating dv = a dt. For
0 … t 6 15 s, a = 6 ms2
. Using the initial condition v = 10 ms at t = 0,
L
10 ms
dv =
L
t
0
6dt
v - 10 = 6t
v = {6t + 10} ms
The maximum velocity occurs when t = 15 s.Then
vmax = 6(15) + 10 = 100 ms Ans.
For 15 s 6 t … t′, a = -4 ms, Using the initial condition v = 100 ms at t = 15 s,
L
100 ms
dv =
L
t
15 s
- 4dt
v - 100 = (-4t) `
t
15 s
v = {-4t + 160} ms
It is required that v = 0 at t = t′.Then
0 = -4t′ + 160 t′ = 40 s Ans.
v
v
Free sample finished here!
Solution Manual for chapter 1 has 236 pages. Last solved problem is #232.

Solutions for Problems in Dynamics (15th Edition) by Russell Hibbeler

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Solutions for Problems in Dynamics (15th Edition) by Russell Hibbeler