Hibbeler Dynamics 12th Edition Solutions Chapter 12
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Table of Contents Chapter 12 1 Chapter 13 145 Chapter 14 242 Chapter 15 302 Chapter 16 396 Chapter 17 504 Chapter 18 591 Chapter 19 632 Chapter 20 666 Chapter 21 714 Chapter 22 786 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 distance traveled. Given: v1 20 km hr = v2 120 kmhr= t 15 s= Solution: a v2 v1− t = a 1.852 m s2 = d v1 t 1 2 a t2+= 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: v 80 ft s = d 500 ft= Solution: v2 2a d= a v 2 2d = a 6.4 ft s2 = v a t= t v a = 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= g 32.2 ft s2 = v0 18 fts= Solution: v v0 2 2g h+= v 59.5 ft s = 1 Engineering Mechanics - Dynamics Chapter 12 t v v0− g = t 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: b 2 m s3 = c 6− m s2 = t1 6 s= t2 11 s= Solution: a t( ) b t c+= v t( ) 0 t ta t( ) ⌠⎮⌡ d= d t( ) 0 t tv t( ) ⌠⎮⌡ d= v t1( ) 0 ms= d t2( ) 80.7 m= Problem 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: v0 70 km hr = a 6000 km hr2 = vf 120 kmhr= Solution: vf v0 a t+= t vf v0− a = t 30 s= vf 2 v0 2 2a s+= s vf 2 v0 2− 2a = s 792 m= Problem 12-6 A freight train travels at v v0 1 e b− t−( )= where t is the elapsed time.
Determine the distance traveled in time t1, and the acceleration at this time. 2 Engineering Mechanics - Dynamics Chapter 12 v0 60 ft s = b 1 s = t1 3 s= Solution: v t( ) v0 1 e b− t−( )= a t( ) t v t( )d d = d t( ) 0 t tv t( ) ⌠⎮⌡ d= d t1( ) 123.0 ft= a t1( ) 2.99 ft s2 = 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 s3 = b 9− ft s2 = c 15 ft s = t0 0 s= tf 10 s= Solution: sp a t 3 b t2+ c t+= vp t sp d d = 3a t2 2b t+ c+= ap t vp d d = 2t sp d d 2 = 6a t 2b+= 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 s2 = 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 Given: Engineering Mechanics - Dynamics Chapter 12 vmax max 3a t0 2 2b t0+ c+ 3a tf2 2b tf+ c+, 3a tcr2 2b tcr+ c+,( )= vmax 135 fts=.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?
Hibbeler 12th Edition Solution Manual
Each floor is a distance h higher than the one below it. (Note: You may want to remember this when traveling at speed vf ) Given: vf 55 mph= h 12 ft= g 32.2 ft s2 = Solution: ac g= vf2 0 2ac s+= H vf 2 2ac = H 101.124 ft= Number of floors N Height of one floor h 12 ft= N H h = N 8.427= N ceil N( )= 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. Determine the average velocity, the average speed, and the acceleration of the particle at time t1. Given: a 1 m s3 = b 3− m s2 = c 2 m= t0 0 s= t1 4 s= Solution: sp t( ) a t 3 b t2+ c+= vp t( ) t sp t( ) d d = ap t( ) t vp t( ) d d = Find the critical velocity where vp = 0. 4 Engineering Mechanics - Dynamics Chapter 12 t2 1.5 s= Given vp t2( ) 0= t2 Find t2( )= t2 2 s= vave sp t1( ) sp t0( )− t1 = vave 4 ms= vavespeed sp t2( ) sp t0( )− sp t1( ) sp t2( )−+ t1 = vavespeed 6 ms= a1 ap t1( )= a1 18 m s2 = Problem 12–10 A particle is moving along a straight line such that its acceleration is defined as a = −kv. If v = v0 when d = 0 and t = 0, determine the particle’s velocity as a function of position and the distance the particle moves before it stops.
Rc Hibbeler 12th Edition Solutions
Given: k 2 s = v0 20 ms= Solution: ap v( ) k− v= v s vd d k− v= v0 v v1 ⌠⎮⌡ d k− sp= Velocity as a function of position v v0 k sp−= Distance it travels before it stops 0 v0 k sp−= sp v0 k = sp 10 m= Problem 12-11 The acceleration of a particle as it moves along a straight line is given by a = bt + c. If s = s0 and v = v0 when t = 0, determine the particle’s velocity and position when t = t1. Also, determine the total distance the particle travels during this time period. Given: b 2 m s3 = c 1− m s2 = s0 1 m= v0 2 ms= t1 6 s= 5 Engineering Mechanics - Dynamics Chapter 12 v0 v v1 ⌠⎮⌡ d 0 t tb t c+( )⌠⎮⌡ d= v v0 b t2 2 + c t+= s0 s s1 ⌠⎮⌡ d 0 t tv0 b t2 2 + c t+⎛⎜⎝ ⎞⎟⎠ ⌠⎮ ⎮⌡ d= s s0 v0 t+ b6 t 3+ c 2 t2+= When t = t1 v1 v0 b t1 2 2 + c t1+= v1 32 ms= s1 s0 v0 t1+ b6 t1 3+ c 2 t1 2+= s1 67 m= The total distance traveled depends on whether the particle turned around or not. To tell we will plot the velocity and see if it is zero at any point in the interval t 0 0.01t1, t1.= v t( ) v0 b t 2 2 + c t+= If v never goes to zero then 0 2 4 6 0 20 40 v t( ) t d s1 s0−= d 66 m=.Problem 12–12 A particle, initially at the origin, moves along a straight line through a fluid medium such that its velocity is defined as v = b(1 − e−ct).
Determine the displacement of the particle during the time 0.