Solucionario De Hibbeler
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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