Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition

Name: Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition
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Author: Lucy Gray

Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition provides the perfect textbook solutions, giving you the help you need to succeed in your studies.

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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
(t 2 - 6t) dt
s = t 3
3 - 3t 2
When t = 6 s,
v = 0 Ans.
When t = 11 s,
s = 80.7 m Ans.
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?
Ans:
v = 0
s = 80.7 m
1
12–1.

12–2.
SOLUTION
When ,
Ans.
Ans.
Since then
Ans.d = 6 - = 6 m
v Z 0
s = 639.5 m
v = >s
t =
s = 1 t - 1
2 t2 + t +
L
s
ds = L
t
0
(t - t + dt
v = t - t +
L
v
dv = L
t
0
(4 t - 1) dt
The acceleration of a particle as it moves along a straight
line is given by where t is in seconds. If
and when determine the
particle’s velocity and position when Also,
determine the total distance the particle travels during this
time period.
t =
t = 0,v = >ss =
a = 1 t - 12 m>s ,2
5
5 m2 m
5 s.
3
3
4 5
2
4 5)
5
5 5 2
5 s
625 m
39.5 2 37.5
4
v = 625 m>s
s = 639.5 m
d = 637.5 m
Ans:
2

12–3.
The velocity of a particle traveling in a straight line is given
by v = (6t - 3t2) m>s, 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
v = 6t - 3t2
a = dv
dt = 6 - 6t
At t = 3 s
a = -12 m>s2 Ans.
ds = v dt
L
s
0
ds = L
t
0
(6t - 3t 2 )dt
s = 3t2 - t3
At t = 3 s
s = 0 Ans.
Since v = 0 = 6t - 3t2
, when t = 0 and t = 2 s.
when t = 2 s, s = 3(2) 2 - (2) 3 = 4 m
sT = 4 + 4 = 8 m Ans.
(vsp )avg = sT
t = 8
3 = 2.67 m>s Ans.
a = -12 m>s 2
s = 0
s T =
(vsp) avg = 2.67 m>s
= 8 m
Ans:
3

*12–4.
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 st = 1 s
s = (10t2 + 20) mm
SOLUTION
(a)
Ans.
(b)
Ans.
(c) Ans.a = d2s
dt2 = 20 mm s2 (for all t)
vavg = ¢s
¢t = 240
4 = 60 mm>s
¢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 mm>s
a = 20 mm>s2
4

12–5.
A particle moves along a straight line such that its position
is defined by s = (t2 - 6t + 5) m. Determine the average
velocity, the average speed, and the acceleration of the
particle when t = 6 s.
Solution
s = t 2 - 6t + 5
v = ds
dt = 2t - 6
a = dv
dt = 2
v = 0 when t = 3
s t = 0 = 5
s t = 3 = -4
s t = 6 = 5
vavg = ∆s
∆t = 0
6 = 0 Ans.
(vsp )avg = sT
∆t = 9 + 9
6 = 3 m>s Ans.
a t = 6 = 2 m>s2 Ans.
Ans:
vavg = 0
(vsp ) avg = 3 m>s
a t = 6 s = 2 m>s2
5

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*12–8.
A particle travels along a straight line with a velocity
v = (12 - 3t2
) m>s, 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 m>s2 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 m>s2
∆s = -880 m
sT = 912 m
8

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12–9.
SOLUTION
Motion of car A:
Motion of car B:
The distance between cars A and B is
Ans.sBA = |s B - s A| = ` vAvB
a A
- v2
A
2a A
` = ` 2vAvB - v2
A
2a A
`
sB = vBt = vB a vA
aA
b = vAvB
aA
sA = v2
A
2aA
0 = vA
2 + 2( - a A)(s A - 0)
v2 = v0
2 + 2ac(s - s0)
0 = vA - a At t = vA
a A
v = v0 + act
When two cars A and B are next to one another, they are
traveling in the same direction with speeds and
respectively. If B maintains its constant speed, while A
begins to decelerate at determine the distance d
between the cars at the instant A stops.
a A ,
vB ,vA
A B
d
SBA = ` 2vAvB - v2
A
2aA
`
Ans:
9

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12–10.
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. T hen in another 5 s it moves from sB to
sC = -6 m. Determine the particle’s a verage 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 m>s 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 m>s Ans.
Ans:
vavg = 0.222 m>s
(vsp ) avg = 2.22 m>s
10

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12–13.
The acceleration of a particle as it moves along a straight
line is given by a = (2t - 1) m>s2
, where t is in seconds. If
s = 1 m and v = 2 m>s when t = 0, determine the
particle’s velocity and position when t = 6 s. Also,
determine the total distance the particle travels during this
time period.
Solution
a = 2t - 1
dv = a dt
L
v
2
dv = L
t
0
(2t - 1)dt
v = t2 - t + 2
dx = v dt
L
s
t
ds = L
t
0
(t 2 - t + 2)dt
s = 1
3 t3 - 1
2 t2 + 2t + 1
When t = 6 s
v = 32 m>s Ans.
s = 67 m Ans.
Since v 0 for 0 … t … 6 s, then
d = 67 - 1 = 66 m Ans.
Ans:
v = 32 m>s
s = 67 m
d = 66 m
13

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12–14.
A train starts from rest at station A and accelerates at
for 60 s. Afterwards it travels with a constant
velocity for 15 min. It then decelerates at 1 until it is
brought to rest at station B. Determine the distance
between the stations.
m>s2
0.5 m>s2
SOLUTION
Kinematics: For stage (1) motion, and Thus,
For stage (2) motion, and Thus,
For stage (3) motion, and Thus,
Ans.= 28 350 m = 28.4 km
s3 = 27 900 + 30(30) + 1
2(-1)(302
)
s = s0 + v0t + 1
2act2+
:
t = 30 s
0 = 30 + (-1)t
v = v0 + actA +
: B
ac = -1 m>s2 .v0 = 30 m>s, v = 0, s0 = 27 900 m
s2 = 900 + 30(900) + 0 = 27 900 m
s = s0 + v0t + 1
2 act2
A +
: B
t = 15(60) = 900 s.ac = 0s0 = 900 m,v0 = 30 m>s,
v1 = 0 + 0.5(60) = 30 m>s
v = v0 + actA +
: B
s1 = 0 + 0 + 1
2(0.5)(602
) = 900 m
s = s0 + v0t + 1
2 act2
A +
: B
ac = 0.5 m>s2
.t = 60 s,s0 = 0,v0 = 0,
Ans:
s = 28.4 km
14

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12–15.
A particle is moving along a straight line such that its
velocity is defined as , where s is in meters.
If when , determine the velocity and
acceleration as functions of time.
t = 0s = 2 m
v = (-4s2
) m>s
SOLUTION
Ans.
Ans.a = dv
dt = 16(2)(8t + 1)(8)
(8t + 1)4 = 256
(8t + 1)3 m>s2
v = -4a 2
8t + 1 b
2
= - 16
(8t + 1)2 m>s
s = 2
8t + 1
t = 1
4 (s- 1 - 0.5)
-s- 1 | s
2 = -4t|0
t
L
s
2
s- 2 ds = L
t
0
-4 dt
ds
dt = -4s2
v = -4s2
v = 16
(8t + 1)2 m>s
a = 256
(8t + 1)3 m>s 2
Ans:
15

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Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition

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