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Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 1 145 242 302 396 504 591 632 666 714 786

Engineering Mechanics - Dynamics

Chapter 12

Problem 12-1 A truck traveling along a straight road at speed v1, increases its speed to v2 in time t. If its acceleration is constant, determine the distancetraveled. Given: v1 = 20 Solution: a = v2 − v1 t 1 2 at 2 a = 1.852 m s
2

km hr

v2 = 120

km hr

t = 15 s

d = v1 t +

d = 291.67 m

Problem 12-2 A car starts from rest and reaches a speed v after traveling a distance d along a straight road. Determine its constant acceleration and the time of travel. Given: Solution: v = 2a d
2

v = 80

ft s
2

d = 500 ft

v a = 2d t = v aa = 6.4

ft s
2

v = at

t = 12.5 s

Problem 12-3 A baseball is thrown downward from a tower of height h with an initial speed v0. Determine the speed at which it hits the ground and the time of travel. Given: h = 50 ft Solution: v = v0 + 2g h
2

g = 32.2

ft s
2

v0 = 18

ft s

v = 59.5

ft s

1

Engineering Mechanics - Dynamics

Chapter 12

t =

v − v0 gt = 1.29 s

*Problem 12–4 Starting from rest, a particle moving in a straight line has an acceleration of a = (bt + c). What is the particle’s velocity at t1 and what is its position at t2? Given: Solution: a ( t) = b t + c v ( t1 ) = 0 m s ⌠ v ( t ) = ⎮ a ( t ) dt ⌡0 d ( t2 ) = 80.7 m
t

b = 2

m s
3

c = −6

m s
2

t1 = 6 s

t2 = 11 s

⌠ d ( t ) = ⎮ v ( t ) dt ⌡0

tProblem 12-5 Traveling with an initial speed v0 a car accelerates at rate a along a straight road. How long will it take to reach a speed vf ? Also, through what distance does the car travel during this time? Given: Solution: vf = v0 + a t
2 2

v0 = 70

km hr

a = 6000

km hr
2

vf = 120

km hr

t =

vf − v0 a
2 2

t = 30 s

vf = v0 + 2a s

vf − v0 s = 2a

s = 792 mProblem 12-6 where t is the elapsed time. Determine the distance A freight train travels at v = v0 1 − e traveled in time t1, and the acceleration at this time.

(

−bt

)

2

Engineering Mechanics - Dynamics

Chapter 12

Given: v0 = 60 b = 1 s ft s

t1 = 3 s Solution: v ( t) = v0 1 − e d ( t1 ) = 123.0 ft

(

−bt

)

a ( t) =

d v ( t) dt ft s
2

⌠ d ( t ) = ⎮ v ( t ) dt⌡0

t

a ( t1 ) = 2.99

Problem 12-7 The position of a particle along a straight line is given by sp = at3 + bt2 + ct. Determine its maximum acceleration and maximum velocity during the time interval t0 ≤ t ≤ tf. Given: a = 1 ft s Solution: sp = a t + b t + c t vp = d 2 sp = 3a t + 2b t + c dt d d vp = s = 6a t + 2b 2 p dt dt
2 3 2 3

b = −9

ft s
2

c = 15

ft s

t0 = 0 s

tf =10 s

ap =

Since the acceleration is linear in time then the maximum will occur at the start or at the end. We check both possibilities. amax = max ( 6a t0 + b , 6a tf + 2b) amax = 42 ft s
2

The maximum velocity can occur at the beginning, at the end, or where the acceleration is zero. We will check all three locations. tcr = −b 3a tcr = 3 s

3

Engineering Mechanics - DynamicsChapter 12

vmax = max 3a t0 + 2b t0 + c , 3a tf + 2b tf + c , 3a tcr + 2b tcr + c

(

2

2

2

)

vmax = 135

ft s

*Problem 12-8 From approximately what floor of a building must a car be dropped from an at-rest position so that it reaches a speed vf when it hits the ground? Each floor is a distance h higher than the one below it. (Note: You may want to remember this when travelingat speed vf ) Given: Solution: ac = g Number of floors Height of one floor vf = 0 + 2ac s N h = 12 ft N = H h N = 8.427 N = ceil ( N)
2

vf = 55 mph

h = 12 ft

g = 32.2
2

ft s
2

H =

vf

2ac

H = 101.124 ft

The car must be dropped from floor number N = 9

Problem 12–9 A particle moves along a straight line such that its position is defined by sp = at3 + bt2 + c....
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