SOLUTION MANUAL FOR ENGINEERING MECHANICS DYNAMICS 12 EDITION By HIBBELER

TEST BANK ENGINEERING MECHANICS DYNAMICS
12 EDITION By HIBBELER| LATEST QUESTIONS 100%
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TEST BANK FOR ENGINEERING MECHANICS DYNAMICS

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•12–1. A car starts from rest and with constant
acceleration achieves a velocity of 15 m>s when it travels a
distance of 200 m. Determine the acceleration of the car
and the time required.




Kinematics:

v0 = 0, v = 15 m>s, s0 = 0, and s = 200 m.

A:
+ B v2 = v0 2 + 2ac(s - s0)

152 = 02 + 2ac(200 - 0)

ac = 0.5625 m>s2 Ans.

A:
+ B v = v0 + act

15 = 0 + 0.5625t

t = 26.7 s Ans.



12–2. A train starts from rest at a station and travels with
a constant acceleration of 1 m>s2. Determine the velocity of
the train when t = 30 s and the distance traveled during
this time.




Kinematics:

ac = 1 m>s2, v0 = 0, s0 = 0, and t = 30 s.

A:
+ B v = v0 + act

= 0 + 1(30) = 30 m>s Ans.

A:
+ B 1 2
s = s0 + v0t + at
2 c

(1) A 302 B
1
= 0 + 0 +
2

= 450 m Ans.




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12–3. An elevator descends from rest with an acceleration
of 5 ft>s2 until it achieves a velocity of 15 ft>s. Determine the
time required and the distance traveled.




Kinematics:

ac = 5 ft>s2, v0 = 0, v = 15 ft>s, and s0 = 0.

A+TB v = v0 + act

15 = 0 + 5t

t = 3s Ans.

A+TB v2 = v0 2 + 2ac(s - s0)

152 = 02 + 2(5)(s - 0)

s = 22.5 ft Ans.



*12–4. A car is traveling at 15 m>s, when the traffic light
50 m ahead turns yellow. Determine the required constant
deceleration of the car and the time needed to stop the car
at the light.




Kinematics:

v0 = 0, s0 = 0, s = 50 m and v0 = 15 m>s.

A:
+ B v2 = v0 2 + 2ac(s - s0)

0 = 152 + 2ac(50 - 0)

ac = - 2.25 m>s2 = 2.25 m>s2 ; Ans.

A:
+ B v = v0 + act

0 = 15 + ( -2.25)t

t = 6.67 s Ans.




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•12–5. A particle is moving along a straight line with the
acceleration a = (12t – 3t1/2) ft>s2, where t is in seconds.
Determine the velocity and the position of the particle as a
function of time. When t = 0, v = 0 and s = 15 ft.


Velocity:

A:
+ B dv = a dt

A 12t - 3t1>2 B dt
v t
dv =
L0 L0


v冷0 = A 6t2 - 2t3>2 B 2
v t

0


v = A 6t2 - 2t3>2 B ft>s Ans.

Position: Using this result and the initial condition s = 15 ft at t = 0 s,

A:
+ B ds = v dt

A 6t2 - 2t3>2 B dt
s t
ds =
L15 ft L0

4 5>2 2 t
s冷15 ft = a 2t3 - t b
s

5 0


s = a2t3 - t + 15 b ft
4 5>2
Ans.
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12–6. A ball is released from the bottom of an elevator
which is traveling upward with a velocity of 6 ft>s. 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.


Kinematics: When the ball is released, its velocity will be the same as the elevator at
the instant of release. Thus, v0 = 6 ft>s. Also, t = 3 s, s0 = 0, s = - h, and
ac = - 32.2 ft>s2.

A+cB
1
s = s0 + v0t + a t2
2 c

( - 32.2) A 32 B
1
- h = 0 + 6(3) +
2
h = 127 ft Ans.

A+cB v = v0 + act

v = 6 + (- 32.2)(3)

= - 90.6 ft>s = 90.6 ft>s T Ans.




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12–7. A car has an initial speed of 25 m>s and a constant
deceleration of 3 m>s2. Determine the velocity of the car
when t = 4 s. What is the displacement of the car during the
4-s time interval? How much time is needed to stop the car?


v = v0 + act

v = 25 + (- 3)(4) = 13 m>s Ans.

1
¢s = s - s0 = v0 t + a t2
2 c

1
¢s = s - 0 = 25(4) + ( - 3)(4)2 = 76 m Ans.
2

v = v0 + ac t

0 = 25 + ( - 3)(t)

t = 8.33 s Ans.



*12–8. 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.


A:
+ B 1
s = s0 + v0 t + a t2
2 c

1
= 0 + 12(10) + ( - 2)(10)2
2

= 20 ft Ans.



•12–9. The acceleration of a particle traveling along a
straight line is a = k>v, where k is a constant. If s = 0,
v = v0 when t = 0, determine the velocity of the particle as
a function of time t.


Velocity:

A:
+ B dv
dt =
a
t v
dv
dt =
L0 Lv0 k>v
t v
1
dt = vdv
L0 Lv0 k
t
1 2 2v
t2 = v
0 2k v0

A v2 - v0 2 B
1
t =
2k

v = 22kt + v0 2 Ans.




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12–10. Car A starts from rest at t = 0 and travels along a 60 ft/s
straight road with a constant acceleration of 6 ft>s2 until it A B
reaches a speed of 80 ft>s. Afterwards it maintains this
speed. Also, when t = 0, car B located 6000 ft down the
road is traveling towards A at a constant speed of 60 ft>s.
Determine the distance traveled by car A when they pass 6000 ft
each other.




Distance Traveled: Time for car A to achives y = 80 ft>s can be obtained by
applying Eq. 12–4.

A:
+ B y = y0 + ac t
80 = 0 + 6t
t = 13.33 s

The distance car A travels for this part of motion can be determined by applying
Eq. 12–6.

A:
+ B y2 = y20 + 2ac (s - s0)
802 = 0 + 2(6)(s1 - 0)
s1 = 533.33 ft

For the second part of motion, car A travels with a constant velocity of y = 80 ft>s
and the distance traveled in t¿ = (t1 - 13.33) s (t1 is the total time) is

A:
+ B s2 = yt¿ = 80(t1 - 13.33)

Car B travels in the opposite direction with a constant velocity of y = 60 ft>s and
the distance traveled in t1 is

A:
+ B s3 = yt1 = 60t1

It is required that

s1 + s2 + s3 = 6000
533.33 + 80(t1 - 13.33) + 60t1 = 6000
t1 = 46.67 s

The distance traveled by car A is

sA = s1 + s2 = 533.33 + 80(46.67 - 13.33) = 3200 ft Ans.




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12–11. 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.


v = 12 - 3t2 (1)

dv
a = = - 6t 2 = - 24 m>s2 Ans.
dt t=4


A 12 - 3t2 B dt
s t t
ds = v dt =
L-10 L1 L1

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.




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*12–12. A sphere is fired downwards into a medium with
an initial speed of 27 m>s. If it experiences a deceleration of
a = ( - 6t) m>s2, where t is in seconds, determine the
distance traveled before it stops.


Velocity: y0 = 27 m>s at t0 = 0 s. Applying Eq. 12–2, we have

A+TB dy = adt
y t
dy = -6tdt
L27 L0

y = A 27 - 3t2 B m>s [1]

At y = 0, from Eq.[1]

0 = 27 - 3t2 t = 3.00 s

Distance Traveled: s0 = 0 m at t0 = 0 s. Using the result y = 27 - 3t2 and applying
Eq. 12–1, we have

A+TB ds = ydt

A 27 - 3t2 B dt
s t
ds =
L0 L0

s = A 27t - t3 B m [2]

At t = 3.00 s, from Eq. [2]

s = 27(3.00) - 3.003 = 54.0 m Ans.



•12–13. A particle travels along a straight line such that in
2 s it moves from an initial position sA = + 0.5 m to a
position sB = - 1.5 m. Then in another 4 s it moves from sB
to sC = + 2.5 m. Determine the particle’s average velocity
and average speed during the 6-s time interval.


¢s = (sC - sA) = 2 m

sT = (0.5 + 1.5 + 1.5 + 2.5) = 6 m

t = (2 + 4) = 6 s

¢s 2
vavg = = = 0.333 m>s Ans.
t 6

sT 6
(vsp)avg = = = 1 m>s Ans.
t 6




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12–14. 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.


Average Velocity: The displacement from A to C is ¢s = sC - SA = - 6 - ( -8)
= 2 m.

¢s 2
yavg = = = 0.222 m>s Ans.
¢t 4 + 5

Average Speed: The distances traveled from A to B and B to C are
sA : B = 8 + 3 = 11.0 m and sB : C = 3 + 6 = 9.00 m, respectively. Then, the
total distance traveled is sTot = sA : B + sB : C = 11.0 - 9.00 = 20.0 m.

A ysp B avg =
sTot 20.0
= = 2.22 m>s Ans.
¢t 4 + 5



12–15. Tests reveal that a normal driver takes about 0.75 s v1 ⫽ 44 ft/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 ft>s) and their cars can
decelerate at 2 ft>s2, determine the shortest stopping
distance d for each from the moment they see the d
pedestrians. Moral: If you must drink, please don’t drive!


Stopping Distance: For normal driver, the car moves a distance of
d¿ = yt = 44(0.75) = 33.0 ft before he or she reacts and decelerates the car. The
stopping distance can be obtained using Eq. 12–6 with s0 = d¿ = 33.0 ft and y = 0.

A:
+ B y2 = y20 + 2ac (s - s0)

02 = 442 + 2(- 2)(d - 33.0)

d = 517 ft Ans.

For a drunk driver, the car moves a distance of d¿ = yt = 44(3) = 132 ft before he
or she reacts and decelerates the car. The stopping distance can be obtained using
Eq. 12–6 with s0 = d¿ = 132 ft and y = 0.

A:
+ B y2 = y20 + 2ac (s - s0)

02 = 442 + 2(- 2)(d - 132)

d = 616 ft Ans.




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*12–16. As a train accelerates uniformly it passes
successive kilometer marks while traveling at velocities of
2 m>s and then 10 m>s. Determine the train’s velocity when
it passes the next kilometer mark and the time it takes to
travel the 2-km distance.

Kinematics: For the first kilometer of the journey, v0 = 2 m>s, v = 10 m>s, s0 = 0,
and s = 1000 m. Thus,

A:
+ B v2 = v0 2 + 2ac (s - s0)

102 = 22 + 2ac (1000 - 0)

ac = 0.048 m>s2

For the second kilometer, v0 = 10 m>s, s0 = 1000 m, s = 2000 m, and
0.048 m>s2. Thus,

A:
+ B v2 = v0 2 + 2ac (s - s0)

v2 = 102 + 2(0.048)(2000 - 1000)

v = 14 m>s Ans.

For the whole journey, v0 = 2 m>s, v = 14 m>s, and 0.048 m>s2. Thus,

A:
+ B v = v0 + act

14 = 2 + 0.048t

t = 250 s Ans.


•12–17. A ball is thrown with an upward velocity of 5 m>s
from the top of a 10-m high building. One second later
another ball is thrown vertically from the ground with a
velocity of 10 m>s. Determine the height from the ground
where the two balls pass each other.


Kinematics: First, we will consider the motion of ball A with (vA)0 = 5 m>s,
(sA)0 = 0, sA = (h - 10) m, tA = t¿ , and ac = - 9.81 m>s2. Thus,


A+cB
1
sA = (sA)0 + (vA)0 tA + actA 2
2
1
h - 10 = 0 + 5t¿ + ( -9.81)(t¿)2
2
h = 5t¿ - 4.905(t¿)2 + 10 (1)

Motion of ball B is with (vB)0 = 10 m>s, (sB)0 = 0, sB = h, tB = t¿ - 1 and
ac = - 9.81 m>s2. Thus,

A+cB
1
sB = (sB)0 + (vB)0 tB + ac tB 2
2
1
h = 0 + 10(t¿ - 1) + ( -9.81)(t¿ - 1)2
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 m

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