Engineering Mechanics: Dynamics 14th Edition Solution Manual

Name: Engineering Mechanics: Dynamics 14th Edition Solution Manual
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Find all the textbook answers you need with Engineering Mechanics: Dynamics 14th Edition Solution Manual, featuring clear, step-by-step solutions.

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1
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.
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?
Ans:
s = 80.7 m

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 ft>s to the
right, at s0 = 0, determine its position when t = 10 s, if
a = 2 ft>s2 to the left.
Ans:
s = 20 ft

3
12–3.
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

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 = A 29.17s - 27.7 ft>s
v2 = 3 2 + 2(4.583) (s - 4)
( :+ ) v2 = v2
0 + 2a c (s - s0)
a c = 4.583 ft>s 2
a c = 4.583 ft>s 2
8 2 = 3 2 + 2a c (10 - 4)
( :+ ) v2 = v2
0 + 2a c (s - s0)
v = 8 ft>ss = 10 ft
v0 = 3 ft>ss0 = 4 fta c
A particle travels along a straight line with a constant
acceleration. When , and when ,
. Determine the velocity as a function of position.v = 8 ft>s
s = 10 ftv = 3 ft>ss = 4 ft
A
Ans:
v = ( 19.17s - 27.7 ) ft>s

5
12–5.
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.
Ans:
sT = 8 m
vavg = 2.67 m>s

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6
12–6.
SOLUTION
Position: The position of the particle when is
Ans.
Total DistanceTraveled: 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 , 1 s and 5 s are
From the particle’s path, the total distance is
Ans.stot = 10.5 + 48.0 + 10.5 = 69.0 ft
s t = 5 s = 1.5(53
) - 13.5(5 2
) + 22.5(5) = -37.5 ft
s t = 1 s = 1.5(13
) - 13.5(1 2
) + 22.5(1) = 10.5 ft
s t = 0 s = 1.5(03
) - 13.5(0 2
) + 22.5(0) = 0
t = 0 s
t = 1 s and t = 5 s
4.50t2 - 27.0t + 22.5 = 0
v = ds
dt = 4.50t2 - 27.0t + 22.5
s|t = 6s = 1.5(6 3
) - 13.5(62
) + 22.5(6) = -27.0 ft
t = 6 s
The position of a particle along a straight line is given by
, where t is in seconds.
Determine the position of the particle when and the
total distance it travels during the 6-s time interval. Hint:
Plot the path to determine the total distance traveled.
t = 6 s
s = (1.5t3 - 13.5t2 + 22.5t) ft
Ans:
s t = 6 s = -27.0 ft
stot = 69.0 ft

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7
12–7.
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

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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 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

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9
12–9.
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

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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. 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

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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
0 2 = 44 2 + 2(-2)(d - 132)
A :+ B v2 = v2
0 + 2a c (s - s0)
v = 0s0 = d¿ = 132 ft
d¿ = vt = 44(3) = 132 ft
d = 517 ft
0 2 = 44 2 + 2(-2)(d - 33.0)
A :+ B v2 = v2
0 + 2a c (s - s0)
v = 0s0 = d¿ = 33.0 ft
d¿ = vt = 44(0.75) = 33.0 ft
Tests reveal that a normal driver takes about 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 ) and their cars can decelerate at ,
determine the shortest stopping distance d for each from
the moment they see the pedestrians. Moral: If you must
drink, please don’t drive!
2 ft>s 2
ft>s
0.75 s
d
v1 44 ft/s
Ans:
Normal: d = 517 ft
drunk: d = 616 ft

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14
12–14.
The position of a particle along a straight-line path is
defined by s = (t3 - 6t2 - 15t + 7) ft, where t is in seconds.
Determine the total distance traveled when t = 10 s. What
are the particle’s average velocity, average speed, and the
instantaneous velocity and acceleration at this time?
Solution
s = t 3 - 6t2 - 15t + 7
v = ds
dt = 3t 2 - 12t - 15
When t = 10 s,
v = 165 ft>s Ans.
a = dv
dt = 6t - 12
When t = 10 s,
a = 48 ft>s2 Ans.
When v = 0,
0 = 3t2 - 12t - 15
The positive root is
t = 5 s
When t = 0, s = 7 ft
When t = 5 s, s = -93 ft
When t = 10 s, s = 257 ft
Total distance traveled
sT = 7 + 93 + 93 + 257 = 450 ft Ans.
vavg = ∆s
∆t = 257 - 7
10 - 0 = 25.0 ft>s Ans.
(vsp )avg = sT
∆t = 450
10 = 45.0 ft>s Ans.
Ans:
v = 165 ft>s
a = 48 ft>s2
sT = 450 ft
vavg = 25.0 ft>s
(vsp ) avg = 45.0 ft>s

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15
12–15.
SOLUTION
Ans.
Ans.s = 1
k B ¢2kt + ¢ 1
n2
0
≤ ≤
1
2
- 1
n0
R
s =
2a2kt + a 1
n2
0
b b
1
2
2k
3
t
0
L
s
0
ds = L
t
0
dt
a2kt + a 1
v2
0
b b
1
2
ds = n dt
n = a2kt + a 1
n2
0
b b
- 1
2
- 1
21n- 2 - n- 2
0 2 = - kt
L
n
n0
n- 3 dn = L
t
0
- k dt
a = dn
dt = - kn3
A particle is moving with a velocity of when and
If it is subjected to a deceleration of
where k is a constant, determine its velocity and position as
functions of time.
a = -kv3 ,t = 0.
s = 0v0
Ans:
v = a2kt + 1
v2
0
b
- 1>2
s = 1
k c a2kt + 1
v2
0
b
1>2
- 1
v0
d

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Mechanical Engineering

Engineering Mechanics: Dynamics 14th Edition Solution Manual

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