MEN 201 Problems For Chapter 12 with Solutions, Spring 2012

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 km hr = t 15 s = Solution: a v2 v1 − t = a 1.852 m s2 = d v1 t 1 + 2at2 = 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 = 2ad a v2 = 2d a 6.4 ft s2 = v = at 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 ft s = Solution: v v0 2 + 2gh = 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( ) bt + c = v t( ) 0 t t a t( ) ⌠ ⎮⌡ d = d t( ) 0 t t v t( ) ⌠ ⎮⌡ d = v t1 ( ) 0 m s = 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 km hr = Solution: vf v0 + at = t vf v0 − a = t 30 s = vf 2 v0 2 + 2as = 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 d v t( ) d = d t( ) 0 t t v 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 at3 + bt2 + ct = vp t d sp d = 3at2 + 2bt + c = ap t d vp d = 2t sp d d 2 = 6at + 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 6at0 + b 6atf + 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 3at0 2 + 2bt0 + c 3atf 2 + 2btf + c , 3atcr 2 + 2btcr + c , ( ) = 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 traveling at speed vf ) Given: vf = 55 mph h = 12 ft g 32.2 ft s2 = Solution: ac = g vf 2 0 + 2ac s = H vf 2 2ac = H 101.124 ft = Number of floors N Height of one floor h 12ft = 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( ) at3 + bt2 + 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 m s = vavespeed sp t2 ( ) sp t0 ( ) − sp t1 ( ) sp t2 ( ) − + t1 = vavespeed 6 m s = 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. Given: k 2 = s v0 20 m s = Solution: ap v ( ) k = − v v s v d d k − v = v0 v 1 v ⌠ ⎮⌡ d k − sp = Velocity as a function of position v v0 − ksp = Distance it travels before it stops 0 v0 − ksp = 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 m s = t1 6 s = 5 Engineering Mechanics - Dynamics Chapter 12 v0 v 1 v ⌠ ⎮⌡ d 0 t t bt + c ( ) ⌠ ⎮⌡ d = v v0 bt2 2 + + ct = s0 s 1 s ⌠ ⎮⌡ d 0 t t v0 bt2 2 + + ct ⎛⎜ ⎝ ⎞⎟ ⎠ ⌠ ⎮ ⎮ ⌡ d = s s0 + v0 t b + 6t3 c + 2t2 = When t = t1 v1 v0 bt1 2 2 + + ct1 = v1 32 m s = s1 s0 + v0 t1 b 6t1 3 + c 2t1 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 bt2 2 + + ct = 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 < t < t1. Given: b 1.8 m s = c 0.3 s = t1 3 s = 6 Solution: Engineering Mechanics - Dynamics Chapter 12 v t( ) b 1 e c − t ( − ) = sp t( ) 0 t t v t( ) ⌠ ⎮⌡ d = sp t1 ( ) 1.839 m = Problem 12–13 The velocity of a particle traveling in a straight line is given v = bt + ct2. If s = 0 when t = 0, determine the particle’s deceleration and position when t = t1. How far has the particle traveled during the time t1, and what is its average speed? Given: b 6 m s2 = c −3 m s3 = t0 = 0 s t1 3 s = Solution: v t( ) bt + ct2 = a t( ) t v t( ) d d = sp t( ) 0 t t v t( ) ⌠ ⎮⌡ d = Deceleration a1 a t1 ( ) = a1 −12 m s2 = Find the turning time t2 t2 = 1.5 s Given v t2 ( ) = 0 t2 Find t2 ( ) = t2 2 s = Total distance traveled d sp t1 ( ) sp t2 ( ) − sp t2 ( ) sp t0 ( ) − + = d 8 m = Average speed vavespeed d t1 − t0 = vavespeed 2.667 m s = Problem 12–14 A particle moves along a straight line such that its position is defined by s = bt2 + ct + d. Determine the average velocity, the average speed, and the acceleration of the particle when t = t1. Given: b 1 m s2 = c −6 m s = d = 5 m t0 = 0 s t1 6 s = Solution: sp t( ) bt2 + ct + d = v t( ) t sp t( ) d d = a t( ) t v t( ) d d = Find the critical time t2 = 2s Given v t2 ( ) = 0 t2 Find t2 ( ) = t2 3 s = vavevel sp t1 ( ) sp t0 ( ) − t1 = vavevel 0 m s = 7 Solution: Engineering Mechanics - Dynamics Chapter 12 vavespeed sp t1 ( ) sp t2 ( ) − sp t2 ( ) sp t0 ( ) − + t1 = vavespeed 3 m s = a1 a t1 ( ) = a1 2 m s2 = Problem 12–15 A particle is moving along a straight line such that when it is at the origin it has a velocity v0. If it begins to decelerate at the rate a = bv1/2 determine the particle’s position and velocity when t = t1. Given: v0 4 m s = b −1.5 m s3 = t1 = 2 s a v ( ) b v = Solution: a v ( ) b v = t d v d = v0 v v 1 v ⌠ ⎮ ⎮⌡ d 2 v v0 − ( ) = bt = v t( ) v0 1 + 2bt ⎛⎜⎝ ⎞⎟⎠ 2 = v t1 ( ) 0.25 m s = sp t( ) 0 t t v t( ) ⌠ ⎮⌡ d = sp t1 ( ) 3.5 m = *Problem 12-16 A particle travels to the right along a straight line with a velocity vp = a / (b + sp). Determine its deceleration when sp = sp1. Given: a 5 m2 s = b = 4 m sp1 2 m = Solution: vp a b + sp = ap vp dvp dsp = a b sp + a − b ( + sp )2 = a2 − b ( + sp )3 = ap1 a2 − b ( + sp1 )3 = ap1 −0.116 m s2 = 8 Engineering Mechanics - Dynamics Chapter 12 Problem 12–17 Two particles A and B start from rest at the origin s = 0 and move along a straight line such that aA = (at − b) and aB = (ct2 − d), where t is in seconds. Determine the distance between them at t and the total distance each has traveled in time t. Given: a 6 ft s3 = b 3 ft s2 = c 12 ft s3 = d 8 ft s2 = t 4 s = Solution: dvA dt at − b = vA at2 2 − bt ⎛⎜ ⎝ ⎞⎟ ⎠ = sA at3 6 bt2 − 2 ⎛⎜ ⎝ ⎞⎟ ⎠ = dvB dt ct2 − d = vB ct3 3 s − dt ⎛⎜ ⎝ ⎞⎟ ⎠ = sB ct4 12 s dt2 − 2 ⎛⎜ ⎝ ⎞⎟ ⎠ = Distance between A and B dAB at3 6 bt2 2 − ct4 − 12 s dt2 2 + = dAB 46.33 m = Total distance A and B has travelled. D at3 6 bt2 2 − ct4 + 12 s dt2 2 − = D 70.714 m = Problem 12–18 A car is to be hoisted by elevator to the fourth floor of a parking garage, which is at a height h above the ground. If the elevator can accelerate at a1, decelerate at a2, and reach a maximum speed v, determine the shortest time to make the lift, starting from rest and ending at rest. Given: h = 48 ft a1 0.6 ft s2 = a2 0.3 ft s2 = v 8 ft s = Solution: Assume that the elevator never reaches its maximum speed. Guesses t1 = 1 s t2 = 2 s vmax 1 ft s = h1 1 ft = Given vmax a1 t1 = 9 Engineering Mechanics - Dynamics Chapter 12 h1 1 2a1 t1 2 = 0 vmax a2 t2 − t1 ( ) − = h h1 vmax t2 − t1 ( ) + 1 2a2 t2 − t1 ( )2 − = t1 t2 vmax h1 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find t1 t2 , , vmax , h1 ( ) = t2 21.909 s = Since vmax 4.382 ft s = < v 8 ft s = then our original assumption is correct. Problem 12-19 A stone A is dropped from rest down a well, and at time t1 another stone B is dropped from rest. Determine the distance between the stones at a later time t2. Given: d = 80 ft t1 = 1 s t2 = 2 s g 32.2 ft s2 = Solution: aA = g vA = gt sA g 2t2 = aB = g vB g t ( − t1 ) = sB g 2 t ( − t1 )2 = At time t2 sA2 g 2t2 2 = sA2 64.4 ft = sB2 g 2 t2 − t1 ( )2 = sB2 16.1 ft = d sA2 − sB2 = d 48.3 ft = *Problem 12-20 A stone A is dropped from rest down a well, and at time t1 another stone B is dropped from rest. Determine the time interval between the instant A strikes the water and the instant B strikes the water. Also, at what speed do they strike the water? 10 Engineering Mechanics - Dynamics Chapter 12 Given: d = 80 ft t1 = 1 s g 32.2 ft s2 = Solution: aA = g vA = gt sA g 2t2 = aB = g vB g t ( − t1 ) = sB g 2 t ( − t1 )2 = Time to hit for each particle tA 2d g = tA 2.229 s = tB 2d g + t1 = tB 3.229 s = Δt tB − tA = Δt 1 s = Speed vA = gtA vB = vA vA 71.777 ft s = vB 71.777 ft s = Problem 12–21 A particle has an initial speed v0. If it experiences a deceleration a = bt, determine the distance traveled before it stops. Given: v0 27 m s = b −6 m s3 = Solution: a t( ) = bt v t( ) b t2 2 + v0 = sp t( ) b t3 6 + v0 t = t 2v0 −b = t = 3 s sp t( ) 54 m = Problem 12-22 The acceleration of a rocket traveling upward is given by ap = b + c sp. Determine the rocket’s velocity when sp = sp1 and the time needed to reach this altitude. Initially, vp = 0 and sp = 0 when t = 0. 11 Engineering Mechanics - Dynamics Chapter 12 Given: b 6 m s2 = c 0.02 1 s2 = sp1 2000 m = Solution: ap b + csp = vp dvp dsp = 0 vp vp vp ⌠ ⎮⌡ d 0 sp sp b ( + csp ) ⌠ ⎮⌡ d = vp 2 2 bsp c 2sp 2 + = vp dsp dt = 2bsp csp 2 + = vp1 2bsp1 csp1 2 + = vp1 322.49 m s = t 0 sp sp 1 2bsp csp 2 + ⌠⎮ ⎮ ⎮⌡ d = t1 0 sp1 sp 1 2bsp csp 2 + ⌠⎮ ⎮ ⎮⌡ d = t1 19.274 s = Problem 12-23 The acceleration of a rocket traveling upward is given by ap = b + c sp. Determine the time needed for the rocket to reach an altitute sp1. Initially, vp = 0 and sp = 0 when t = 0. Given: b 6 m s2 = c 0.02 1 s2 = sp1 100 m = Solution: ap b + csp = vp dvp dsp = 0 vp vp vp ⌠ ⎮⌡ d 0 sp sp b ( + csp ) ⌠ ⎮⌡ d = vp 2 2 bsp c 2sp 2 + = vp dsp dt = 2bsp csp 2 + = vp1 2bsp1 csp1 2 + = vp1 37.417 m s = 12 Engineering Mechanics - Dynamics Chapter 12 t 0 sp sp 1 2bsp csp 2 + ⌠⎮ ⎮ ⎮⌡ d = t1 0 sp1 sp 1 2bsp csp 2 + ⌠⎮ ⎮ ⎮⌡ d = t1 5.624 s = *Problem 12–24 A particle is moving with velocity v0 when s = 0 and t = 0. If it is subjected to a deceleration of a k = − v3 , where k is a constant, determine its velocity and position as functions of time. Solution: a dv = dt k − v3 = v0 v v v 3 − ⌠ ⎮ ⌡ d 0 t −k t ⌠ ⎮⌡ d = 1 − 2 v 2 − v0 − 2 − ( ) k − t = v t( ) 1 2kt 1 v0 2 + = ds vdt = 0 s 1 s ⌠ ⎮⌡ d 0 t t 1 2kt 1 v0 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + ⌠⎮ ⎮ ⎮ ⎮⌡ d = s t( ) 1 k 2kt 1 v0 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + 1 v0 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = Problem 12–25 A particle has an initial speed v0. If it experiences a deceleration a = bt, determine its velocity when it travels a distance s1. How much time does this take? Given: v0 27 m s = b −6 m s3 = s1 10 m = Solution: a t( ) = bt v t( ) b t2 2 + v0 = sp t( ) b t3 6 + v0 t = Guess t1 = 1 s Given sp t1 ( ) = s1 t1 Find t1 ( ) = t1 0.372 s = v t1 ( ) 26.6 m s = 13 Engineering Mechanics - Dynamics Chapter 12 Problem 12-26 Ball A is released from rest at height h1 at the same time that a second ball B is thrown upward from a distance h2 above the ground. If the balls pass one another at a height h3 determine the speed at which ball B was thrown upward. Given: h1 40 ft = h2 5 ft = h3 20 ft = g 32.2 ft s2 = Solution: For ball A: For ball B: aA = −g aB −g = vA g = − t vB g − t + vB0 = sA −g 2 ⎛⎜⎝ ⎞⎟⎠ t2 + h1 = sB −g 2 ⎛⎜⎝ ⎞⎟⎠ t2 + vB0 t + h2 = Guesses t = 1 s vB0 2 ft s = Given h3 −g 2 ⎛⎜⎝ ⎞⎟⎠ t2 + h1 = h3 −g 2 ⎛⎜⎝ ⎞⎟⎠ t2 + vB0 t + h2 = t ⎛⎜vB0 ⎝ ⎞⎟ ⎠ Find t vB0 ( , ) = t = 1.115 s vB0 31.403 ft s = Problem 12–27 A car starts from rest and moves along a straight line with an acceleration a = k s−1/3. Determine the car’s velocity and position at t = t1. Given: k 3 m4 s6 ⎛⎜ ⎜⎝ ⎞⎟ ⎟⎠ 1 3 = t1 6 s = 14 Engineering Mechanics - Dynamics Chapter 12 Solution: a v sp v d d = ksp 1 − 3 = 0 v v v ⌠ ⎮⌡ d v2 2 = 0 sp s ksp 1 − 3 ⌠ ⎮ ⎮ ⌡ d = 3 2ksp 2 3 = v 3ksp 1 3 = t sp d d = 3kt 0 sp sp sp 1 − 3 ⌠ ⎮ ⎮ ⌡ d = 3 2sp 2 3 = sp t( ) 2 3kt 3 ⎛⎜⎝ ⎞⎟⎠ 3 2 = sp t1 ( ) 41.6 m = v t( ) t sp t( ) d d = v t1 ( ) 10.39 m s = *Problem 12-28 The acceleration of a particle along a straight line is defined by ap = b t + c. At t = 0, sp = sp0 and vp = vp0. When t = t1 determine (a) the particle's position, (b) the total distance traveled, and (c) the velocity. Given: b 2 m s3 = c −9 m s2 = sp0 = 1 m vp0 10 m s = t1 9 s = Solution: ap bt + c = vp b 2 ⎛⎜⎝ ⎞⎟⎠ t2 + ct + vp0 = sp b 6 ⎛⎜⎝ ⎞⎟⎠ t3 c 2 ⎛⎜⎝ ⎞⎟⎠ t2 + + vp0t + sp0 = a ) The position sp1 b 6 ⎛⎜⎝ ⎞⎟⎠ t1 3 c 2 ⎛⎜⎝ ⎞⎟⎠ t1 2 + + vp0t1 + sp0 = sp1 −30.5 m = b ) The total distance traveled - find the turning times vp b 2 ⎛⎜⎝ ⎞⎟⎠ t2 + ct + vp0 = 0 = t2 −c c2 − 2bvp0 − b = t2 1.298 s = 15 Engineering Mechanics - Dynamics Chapter 12 t3 −c c2 − 2bvp0 + b = t3 7.702 s = sp2 b 6 ⎛⎜⎝ ⎞⎟⎠ t2 3 c 2 ⎛⎜⎝ ⎞⎟⎠ t2 2 + + vp0t2 + sp0 = sp2 7.127 m = sp3 b 6 ⎛⎜⎝ ⎞⎟⎠ t3 3 c 2t3 2 + + vp0t3 + sp0 = sp3 −36.627 m = d sp2 − sp0 sp2 − sp3 + sp1 − sp3 + = d 56.009 m = c ) The velocity vp1 b 2 ⎛⎜⎝ ⎞⎟⎠ t1 2 + ct1 + vp0 = vp1 10 m s = Problem 12–29 A particle is moving along a straight line such that its acceleration is defined as a = k s2. If v = v0 when s = sp0 and t = 0, determine the particle’s velocity as a function of position. Given: k 4 1 ms2 = v0 −100 m s = sp0 10 m = Solution: a v sp v d d = ksp 2 = v0 v v v ⌠ ⎮⌡ d sp0 sp sp ksp 2 ⌠ ⎮ ⌡ d = 1 2 v2 v0 2 − ( ) 1 3k sp 3 sp0 3 − ( ) = v v0 2 2 3k sp 3 sp0 3 − ( ) + = Problem 12–30 A car can have an acceleration and a deceleration a. If it starts from rest, and can have a maximum speed v, determine the shortest time it can travel a distance d at which point it stops. Given: a 5 m s2 = v 60 m s = d 1200 m = Solution: Assume that it can reach maximum speed Guesses t1 = 1 s t2 = 2 s t3 = 3 s d1 = 1 m d2 2 m = Given at1 = v 1 2at1 2 = d1 d2 d1 v t2 − t1 ( ) + = 16 Engineering Mechanics - Dynamics Chapter 12 d d2 v t3 − t2 ( ) + 1 2a t3 − t2 ( )2 − = 0 v a t3 − t2 ( ) − = t1 t2 t3 d1 d2 ⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠ Find t1 t2 , , t3 , d1 , d2 ( ) = t1 t2 t3 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ 12 20 32 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ s = d1 d2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 360 ⎛⎜⎝840 ⎞⎟⎠ m = t3 32 s = Problem 12-31 Determine the time required for a car to travel a distance d along a road if the car starts from rest, reaches a maximum speed at some intermediate point, and then stops at the end of the road. The car can accelerate at a1 and decelerate at a2. Given: d = 1 km a1 1.5 m s2 = a2 2 m s2 = Let t1 be the time at which it stops accelerating and t the total time. Solution: Guesses t1 = 1 s d1 = 1 m t = 3 s v1 1 m s = Given d1 a1 2 t1 2 = v1 = a1 t1 v1 a2 t ( − t1 ) = d d1 v1 t ( − t1 ) + 1 2a2 t ( − t1 )2 − = t1 t v1 d1 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find t1 t, , v1 , d1 ( ) = t1 = 27.603 s v1 41.404 m s = d1 571.429 m = t 48.305 s = *Problem 12-32 When two cars A and B are next to one another, they are traveling in the same direction with speeds vA0 and vB0 respectively. If B maintains its constant speed, while A begins to decelerate at the rate aA, determine the distance d between the cars at the instant A stops. 17 Engineering Mechanics - Dynamics Chapter 12 Solution: Motion of car A: −aA = constant 0 vA0 − aAt = sA vA0 t 1 − 2aAt2 = t vA0 aA = sA vA0 2 2aA = Motion of car B: aB = 0 vB = vB0 sB = vB0 t sB vB0 vA0 aA = The distance between cars A and B is d sB − sA = vB0 vA0 aA vA0 2 2aA − = 2vB0 vA0 vA0 2 − 2aA = d 2vB0 vA0 vA0 2 − 2aA = Problem 12-33 If the effects of atmospheric resistance are accounted for, a freely falling body has an acceleration defined by the equation a g 1 ( − cv2 ) = , where the positive direction is downward. If the body is released from rest at a very high altitude, determine (a) the velocity at time t1 and (b) the body’s terminal or maximum attainable velocity as t →∞. Given: t1 = 5 s g 9.81 m s2 = c 10 4 − s2 m2 = Solution: (a) a dv = dt g 1 ( − cv2 ) = Guess v1 1 m s = Given 0 v1 v 1 1 cv2 − ⌠ ⎮ ⎮ ⌡ d 0 t1 g t ⌠ ⎮⌡ d = v1 Find v1 ( ) = v1 45.461 m s = 18 Engineering Mechanics - Dynamics Chapter 12 (b) Terminal velocity means a = 0 0 g 1 cvterm 2 ( − ) = vterm 1 c = vterm 100 m s = Problem 12-34 As a body is projected to a high altitude above the earth’s surface, the variation of the acceleration of gravity with respect to altitude y must be taken into account. Neglecting air resistance, this acceleration is determined from the formula a = g − [R2/(R+y)2], where g is the constant gravitational acceleration at sea level, R is the radius of the earth, and the positive direction is measured upward. If g = 9.81 m/s2 and R = 6356 km, determine the minimum initial velocity (escape velocity) at which a projectile should be shot vertically from the earth’s surface so that it does not fall back to the earth. Hint: This requires that v = 0 as y → ∞. Solution: g 9.81 m s2 = R 6356 km = vdv = ady g − R2 R + y ( )2 dy = v 0 v v ⌠ ⎮⌡ d g − R2 0 ∞ y 1 R + y ( )2 ⌠ ⎮ ⎮ ⌡ d = v2 − 2 g − R = v 2gR = v 11.2 km s = Problem 12-35 Accounting for the variation of gravitational acceleration a with respect to altitude y (see Prob. 12-34), derive an equation that relates the velocity of a freely falling particle to its altitude. Assume that the particle is released from rest at an altitude y0 from the earth’s surface. With what velocity does the particle strike the earth if it is released from rest at an altitude y0. Use the numerical data in Prob. 12-34. Solution: g 9.81 m s2 = R = 6356 km y0 500 km = vdv = ady g − R2 R + y ( )2 dy = 19 Engineering Mechanics - Dynamics Chapter 12 0 v v v ⌠ ⎮⌡ d g − R2 y0 y y 1 R + y ( )2 ⌠ ⎮ ⎮ ⌡ d = v2 2 gR2 1 R y + 1 R + y0 − ⎛⎜⎝ ⎞⎟⎠ = gR2 y0 − y ( ) R + y ( ) R + y0 ( ) = v 2gR2 y0 − y ( ) R + y ( ) R + y0 ( ) = When it hits, y = 0 vearth 2gR y0 R + y0 = vearth 3.016 km s = *Problem 12-36 When a particle falls through the air, its initial acceleration a = g diminishes until it is zero, and thereafter it falls at a constant or terminal velocity vf. If this variation of the acceleration can be expressed as a = (g/vf 2)(v2 f − v2), determine the time needed for the velocity to become v < vf. Initially the particle falls from rest. Solution: dv dt = a g vf 2 vf 2 − v2 ( ) = 0 v v 1 vf 2 v2 − ⌠⎮ ⎮ ⎮⌡ d g vf 2 0 t 1 t ⌠ ⎮⌡ d = 1 2vf ln vf v + vf v − ⎛⎜ ⎝ ⎞⎟ ⎠ g vf 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t = t vf 2g ln vf v + vf v − ⎛⎜ ⎝ ⎞⎟ ⎠ = Problem 12-37 An airplane starts from rest, travels a distance d down a runway, and after uniform acceleration, takes off with a speed vr It then climbs in a straight line with a uniform acceleration aa until it reaches a constant speed va. Draw the s-t, v-t, and a-t graphs that describe the motion. Given: d = 5000 ft vr 162 mi hr = aa 3 ft s2 = va 220 mi hr = Solution: First find the acceleration and time on the runway and the time in the air ar vr 2 = 2d ar 5.645 ft s2 = tr vr ar = tr 42.088 s = 20 Engineering Mechanics - Dynamics Chapter 12 ta va vr − aa = ta 28.356 s = The equations of motion t1 0 0.01tr , .. tr = a1 t1 ( ) ar s2 ft = v1 t1 ( ) art1 s ft = s1 t1 ( ) 1 2art1 2 1 ft = t2 tr 1.01tr , tr + ta .. = a2 t2 ( ) aa s2 ft = v2 t2 ( ) artr aa t2 − tr ( ) + ⎡⎣ ⎤⎦ s ft = s2 t2 ( ) 1 2artr 2 artr t2 − tr ( ) + 1 2aa t2 − tr ( )2 + ⎡⎢⎣ ⎤⎥⎦ 1 ft = The plots 0 10 20 30 40 50 60 70 80 0 5000 1 .104 1.5.104 Time in seconds Distance in ft s1 t1 ( ) s2 t2 ( ) t1 t2 , 0 10 20 30 40 50 60 70 80 0 200 400 Time in seconds Velocity in ft/s v1 t1 ( ) v2 t2 ( ) t1 t2 , 21 Engineering Mechanics - Dynamics Chapter 12 0 10 20 30 40 50 60 70 80 0 2 4 6 Time in seconds Acceleration in ft/s^2 a1 t1 ( ) a2 t2 ( ) t1 t2 , Problem 12-38 The elevator starts from rest at the first floor of the building. It can accelerate at rate a1 and then decelerate at rate a2. Determine the shortest time it takes to reach a floor a distance d above the ground. The elevator starts from rest and then stops. Draw the a-t, v-t, and s-t graphs for the motion. Given: a1 5 ft s2 = a2 2 ft s2 = d 40 ft = Solution: Guesses t1 = 1 s t 2 s = d1 = 20 ft vmax 1 ft s = Given vmax = a1 t1 d1 1 2a1 t1 2 = vmax a2 t ( − t1 ) = d d1 vmax t ( − t1 ) + 1 2a2 t ( − t1 )2 − = t1 t d1 vmax ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find t1 t, , d1 , vmax ( ) = d1 = 11.429 ft t1 = 2.138 s vmax 10.69 ft s = t 7.483 s = The equations of motion ta 0 0.01t1 , .. t1 = td t1 1.01t1 , .. t = aa ta ( ) a1 s2 ft = ad td ( ) −a2 s2 ft = 22 Engineering Mechanics - Dynamics Chapter 12 va ta ( ) a1 ta s ft = vd td ( ) vmax a2 td − t1 ( ) − ⎡⎣ ⎤⎦ s ft = sa ta ( ) 1 2a1 ta 2 1 ft = sd td ( ) d1 vmax td − t1 ( ) + 1 2a2 td − t1 ( )2 − ⎡⎢⎣ ⎤⎥⎦ 1 ft = The plots 0 1 2 3 4 5 6 7 8 5 0 5 Time in seconds Acceleration in ft/s^2 aa ta ( ) ad td ( ) ta td , 0 1 2 3 4 5 6 7 8 0 5 10 15 Time in seconds Velocity in ft/s va ta ( ) vd td ( ) ta td , 23 Engineering Mechanics - Dynamics Chapter 12 0 1 2 3 4 5 6 7 8 0 20 40 Time in seconds Distance in ft sa ta ( ) sd td ( ) ta td , Problem 12–39 If the position of a particle is defined as s = bt + ct2, construct the s–t, v–t, and a–t graphs for 0 ≤ t ≤ T . Given: b = 5 ft c −3 ft = T = 10 s t 0 0.01T , .. T = Solution: sp t( ) bt + ct2 ( ) 1 ft = v t( ) b + 2ct ( ) s ft = a t( ) (2c ) s2 ft = 0 2 4 6 8 10 400 200 0 200 Time (s) Displacement (ft) sp t( ) t 0 2 4 6 8 10 100 50 0 50 Time (s) Velocity (ft/s) v t( ) t 24 Engineering Mechanics - Dynamics Chapter 12 0 2 4 6 8 10 6.01 6.005 6 5.995 5.99 Time (s) Acceleration (ft/s^2) a t( ) t *Problem 12-40 If the position of a particle is defined by sp = b sin(ct) + d, construct the s-t, v-t, and a-t graphs for 0 ≤ t ≤ T. Given: b = 2 m c π 5 1 s = d = 4 m T = 10 s t 0 0.01T , .. T = Solution: sp t( ) bsin ct ( ) + d ( ) 1 m = vp t( ) bccos ct ( ) s m = ap t( ) b − c2 sin ct ( ) s m2 = 0 2 4 6 8 10 2 4 6 Time in seconds Distance in m sp t( ) t 25 Engineering Mechanics - Dynamics Chapter 12 0 2 4 6 8 10 2 0 2 Time in seconds Velocity in m/s vp t( ) t 0 2 4 6 8 10 1 0 1 Time in seconds Acceleration in m/s^2 ap t( ) t Problem 12-41 The v-t graph for a particle moving through an electric field from one plate to another has the shape shown in the figure. The acceleration and deceleration that occur are constant and both have a magnitude a. If the plates are spaced smax apart, determine the maximum velocity vmax and the time tf for the particle to travel from one plate to the other. Also draw the s-t graph. When t = tf/2 the particle is at s = smax/2. Given: a 4 m s2 = smax 200 mm = Solution: smax 2 1 2a tf 2 ⎛⎜⎝ ⎞⎟⎠ 2 ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = 26 Engineering Mechanics - Dynamics Chapter 12 tf 4smax a = tf 0.447 s = vmax a tf 2 = vmax 0.894 m s = The plots t1 0 0.01tf , tf .. 2 = s1 t1 ( ) 1 2at1 2 1 m = t2 tf 2 1.01 tf 2 , .. tf = s2 t2 ( ) 1 2a tf 2 ⎛⎜⎝ ⎞⎟⎠ 2 a tf 2 t2 tf − 2 ⎛⎜⎝ ⎞⎟⎠ + 1 2a t2 tf − 2 ⎛⎜⎝ ⎞⎟⎠ 2 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 m = 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0 0.1 0.2 Time in seconds Distance in m s1 t1 ( ) s2 t2 ( ) t1 t2 , Problem 12-42 The v-t graph for a particle moving through an electric field from one plate to another has the shape shown in the figure, where tf and vmax are given. Draw the s-t and a-t graphs for the particle. When t = tf /2 the particle is at s = sc. Given: tf 0.2 s = vmax 10 m s = sc 0.5 m = Solution: a 2vmax tf = a 100 m s2 = 27 Engineering Mechanics - Dynamics Chapter 12 The plots t1 0 0.01tf , tf .. 2 = s1 t1 ( ) 1 2at1 2 1 m = a1 t1 ( ) a s2 m = t2 tf 2 1.01 tf 2 , .. tf = s2 t2 ( ) 1 2a tf 2 ⎛⎜⎝ ⎞⎟⎠ 2 a tf 2 t2 tf − 2 ⎛⎜⎝ ⎞⎟⎠ + 1 2a t2 tf − 2 ⎛⎜⎝ ⎞⎟⎠ 2 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 1 m = a2 t2 ( ) −a s2 m = 0 0.05 0.1 0.15 0.2 0 0.5 1 Time in seconds Distance in m s1 t1 ( ) s2 t2 ( ) t1 t2 , 0 0.05 0.1 0.15 0.2 100 0 100 Time in seconds Acceleration in m/s^2 a1 t1 ( ) a2 t2 ( ) t1 t2 , Problem 12–43 A car starting from rest moves along a straight track with an acceleration as shown. Determine the time t for the car to reach speed v. 28 Engineering Mechanics - Dynamics Chapter 12 Given: v 50 m s = a1 8 m s2 = t1 10 s = Solution: Assume that t > t1 Guess t 12 s = Given v a1 t1 t1 2 2 a1 t ( − t1 ) + = t = Find t( ) t 11.25 s = *Problem 12-44 A motorcycle starts from rest at s = 0 and travels along a straight road with the speed shown by the v-t graph. Determine the motorcycle's acceleration and position when t = t4 and t = t5. Given: v0 5 m s = t1 4 s = t2 10 s = t3 15 s = t4 8 s = t5 12 s = Solution: At t = t4 Because t1 < t4 < t2 then a4 dv = dt 0 = s4 1 2v0 t1 t4 − t1 ( )v0 + = s4 30 m = At t = t5 Because t2 < t5 < t3 then a5 v0 − t3 − t2 = a5 −1 m s2 = 29 Engineering Mechanics - Dynamics Chapter 12 s5 1 2t1 v0 v0 t2 − t1 ( ) + 1 2v0 t3 − t2 ( ) + 1 2 t3 t5 − t3 − t2 v0 t3 − t5 ( ) − = s5 48 m = Problem 12–45 From experimental data, the motion of a jet plane while traveling along a runway is defined by the v–t graph shown. Construct the s-t and a-t graphs for the motion. Given: v1 80 m s = t1 10 s = t2 40 s = Solution: k1 v1 t1 = k2 0 m s2 = τ1 0 0.01t1 , .. t1 = s1 τ1 ( ) 1 2k1τ1 2 ⎛⎜⎝ ⎞⎟⎠ m = a1 τ1 ( ) k1 s2 m = τ2 t1 1.01t1 , .. t2 = s2 τ2 ( ) v1τ2 1 2k1 t1 2 − ⎛⎜⎝ ⎞⎟⎠ m = a2 τ2 ( ) k2 s2 m = 30 Engineering Mechanics - Dynamics Chapter 12 0 5 10 15 20 25 30 35 40 0 1000 2000 3000 Time (s) Position (m) s1 τ1 ( ) s2 τ2 ( ) τ1 τ2 , 0 5 10 15 20 25 30 35 40 0 5 10 Time (s) Acceleration (m/s^2) a1 τ1 ( ) a2 τ2 ( ) τ1 τ2 , Problem 12–46 A car travels along a straight road with the speed shown by the v–t graph. Determine the total distance the car travels until it stops at t2. Also plot the s–t and a–t graphs. Given: t1 30 s = t2 48 s = v0 6 m s = Solution: k1 v0 t1 = 31 Engineering Mechanics - Dynamics Chapter 12 k2 v0 t2 − t1 = τ1 0 0.01t1 , .. t1 = s1 t( ) 1 ⎛⎜⎝2k1 t2 ⎞⎟⎠ = a1 t( ) = k1 a2 t( ) −k2 = τ2 t1 1.01t1 , .. t2 = s2 t( ) s1 t1 ( ) v0 + k2 t1 ( ) t ( − t1 ) + k2 2 t2 t1 ( − 2 ) − ⎡⎢⎣ ⎤⎥⎦ = d s2 t2 ( ) = d 144 m = 0 10 20 30 40 50 0 100 200 Time (s) Distance (m) s1 τ1 ( ) s2 τ2 ( ) τ1 τ2 , 0 10 20 30 40 50 0.4 0.2 0 0.2 Time (s) Acceleration (m/s^2) a1 τ1 ( ) a2 τ2 ( ) τ1 τ2 , 32 Engineering Mechanics - Dynamics Chapter 12 The v–t graph for the motion of a train as it moves from station A to station B is shown. Draw the a–t graph and determine the average speed and the distance between the stations. Given: t1 30 s = t2 90 s = t3 120 s = v1 40 ft s = Solution: τ1 0 0.01t1 , .. t1 = τ2 t1 1.01t1 , .. t2 = τ3 t2 1.01t2 , .. t3 = a1 t( ) v1 t1 s2 ft = a2 t( ) = 0 a3 t( ) v1 − t3 t2 − s2 ft = 0 20 40 60 80 100 120 5 .105 0 5 .105 1 .106 Time (s) Acceleration (ft/s^2) a1 τ1 ( ) a2 τ2 ( ) a3 τ3 ( ) τ1 τ2 , τ3 , d 1 2v1 t1 v1 t2 − t1 ( ) + 1 2v1 t3 − t2 ( ) + = d 3600ft = speed d t3 = speed 30 ft s = 33 Problem 12–47 Engineering Mechanics - Dynamics Chapter 12 *Problem 12–48 The s–t graph for a train has been experimentally determined. From the data, construct the v–t and a–t graphs for the motion; 0 ≤ t ≤ t2 . For 0 ≤ t ≤ t1 , the curve is a parabola, and then it becomes straight for t ≥ t1 . Given: t1 30 s = t2 40 s = s1 360 m = s2 600 m = Solution: k1 s1 t1 2 = k2 s2 s1 − t2 − t1 = τ1 0 0.01t1 , .. t1 = τ2 t1 1.01t1 , .. t2 = sp1 t( ) = k1 t2 v1 t( ) = 2k1 t a1 t( ) 2k1 = sp2 t( ) sp1 t1 ( ) k2 t ( − t1 ) + = v2 t( ) = k2 a2 t( ) 0 = 0 5 10 15 20 25 30 35 40 0 10 20 30 Time (s) Velocity (m/s) v1 τ1 ( ) v2 τ2 ( ) τ1 τ2 , 34 Engineering Mechanics - Dynamics Chapter 12 0 5 10 15 20 25 30 35 40 0 0.5 1 Time (s) Acceleration (m/s^2) a1 τ1 ( ) a2 τ2 ( ) τ1 τ2 , Problem 12-49 The v-t graph for the motion of a car as if moves along a straight road is shown. Draw the a-t graph and determine the maximum acceleration during the time interval 0 < t < t2. The car starts from rest at s = 0. Given: t1 10 s = t2 30 s = v1 40 ft s = v2 60 ft s = Solution: τ1 0 0.01t1 , .. t1 = a1 τ1 ( ) 2v1 t1 2 ⎛⎜ ⎜⎝ ⎞⎟ ⎟⎠ τ1 s2 ft = τ2 t1 1.01t1 , .. t2 = a2 τ2 ( ) v2 v1 − t2 t1 − s2 ft = 35 Engineering Mechanics - Dynamics Chapter 12 0 5 10 15 20 25 30 0 5 10 Time in seconds Acceleration in ft/s^2 a1 τ1 ( ) a2 τ2 ( ) τ1 τ2 , amax 2 v1 t1 2 ⎛⎜ ⎜⎝ ⎞⎟ ⎟⎠ t1 = amax 8 ft s2 = Problem 12-50 The v-t graph for the motion of a car as it moves along a straight road is shown. Draw the s-t graph and determine the average speed and the distance traveled for the time interval 0 < t < t2. The car starts from rest at s = 0. Given: t1 10 s = t2 30 s = v1 40 ft s = v2 60 ft s = Solution: The graph τ1 0 0.01t1 , .. t1 = s1 τ1 ( ) v1 t1 2 τ1 3 3 1 ft = τ2 t1 1.01t1 , .. t2 = s2 τ2 ( ) v1 t1 3 v1 τ2 − t1 ( ) + v2 v1 − t2 t1 − τ2 − t1 ( )2 2 + ⎡⎢ ⎢⎣ ⎤⎥ ⎥⎦ 1 ft = 36 Engineering Mechanics - Dynamics Chapter 12 0 5 10 15 20 25 30 0 500 1000 1500 Time in seconds Distance in ft s1 τ1 ( ) s2 τ2