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Engineering Mechanics: Dynamics 14th Edition Solution Manual

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    1 Solution a = 2 t - 6 d v = a dt L v 0 d v = L t 0 (2t - 6) dt v = t 2 - 6 t ds = v dt L s 0 d s = L t 0 (t 2 - 6t) dt s = t 3 3 - 3t 2 When t = 6 s , v = 0 Ans. When t = 11 s , s = 8 0.7 m Ans. 12–1. Starting from rest, a particle moving in a straight line has an acceleration of a = (2 t - 6) m > s 2 , where t is in seconds. What is the particle’s velocity when t = 6 s, and what is its position when t = 11 s? Ans: s = 80.7 m

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    2 12–2. SOLUTION 1 S + 2 s = s 0 + v 0 t + 1 2 a c t 2 = 0 + 12(10) + 1 2 ( - 2)(10) 2 = 20 ft Ans. If a particle has an initial velocity of v 0 = 12 ft > s to the right, at s 0 = 0, determine its position when t = 10 s, if a = 2 ft > s 2 to the left. Ans: s = 20 ft

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    3 12–3. A particle travels along a straight line with a velocity v = (12 - 3 t 2 ) 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. SOLUTION v = 12 - 3 t 2 (1) a = d v dt = - 6 t t = 4 = - 24 m > s 2 Ans. L s - 10 ds = L t 1 v dt = L t 1 (12 - 3 t 2 ) dt s + 10 = 12 t - t 3 - 11 s = 12 t - t 3 - 21 s t = 0 = - 21 s t = 10 = - 901 ∆ s = - 901 - ( - 21) = - 880 m Ans. From Eq. (1): v = 0 when t = 2 s s t = 2 = 12(2) - (2) 3 - 21 = - 5 s T = (21 - 5) + (901 - 5) = 912 m Ans. Ans: a = - 24 m > s 2 ∆ s = - 880 m s T = 912 m

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    4 *12–4. SOLUTION Velocity: To determine the constant acceleration , set , , and and apply Eq. 12–6. Using the result , the velocity function can be obtained by applying Eq. 12–6. Ans. v = A 2 9.17s - 27.7 ft > s v 2 = 3 2 + 2(4.583) ( s - 4) ( : + ) v 2 = v 2 0 + 2 a c ( s - s 0 ) a c = 4.583 ft > s 2 a c = 4.583 ft > s 2 8 2 = 3 2 + 2 a c (10 - 4) ( : + ) v 2 = v 2 0 + 2 a c ( s - s 0 ) v = 8 ft > s s = 10 ft v 0 = 3 ft > s s 0 = 4 ft a c A part i c l e trave l s a l ong a stra i g h t li ne w i t h a constant acceleration. When , and when , . Determine the velocity as a function of position. v = 8 ft > s s = 10 ft v = 3 ft > s s = 4 ft A Ans: v = ( 1 9.17s - 27.7 ) ft > s

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    5 12–5. The velocity of a particle traveling in a straight line is given by v = (6 t - 3 t 2 ) m > s, where t is in seconds. If s = 0 when t = 0, determine the particle’s deceleration and position when t = 3 s. How far has the particle traveled during the 3-s time interval, and what is its average speed? Solution v = 6 t - 3 t 2 a = d v dt = 6 - 6 t At t = 3 s a = - 12 m > s 2 Ans. ds = v dt L s 0 ds = L t 0 (6t - 3t 2 )dt s = 3 t 2 - t 3 At t = 3 s s = 0 Ans. Since v = 0 = 6 t - 3 t 2 , when t = 0 and t = 2 s. when t = 2 s, s = 3(2) 2 - (2) 3 = 4 m s T = 4 + 4 = 8 m Ans. ( v sp ) avg = s T t = 8 3 = 2.67 m > s Ans. Ans: s T = 8 m v avg = 2.67 m > s

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    6 12–6. SOLUTION Pos i t i on : The position of the particle when is Ans. Total D i stanceTraveled : The velocity of the particle can be determined by applying Eq. 12–1. The times when the particle stops are The position of the particle at , 1 s and 5 s are From the particle’s path, the total distance is Ans. s tot = 10.5 + 48.0 + 10.5 = 69.0 ft s t = 5 s = 1.5(5 3 ) - 13.5(5 2 ) + 22.5(5) = - 37.5 ft s t = 1 s = 1.5(1 3 ) - 13.5(1 2 ) + 22.5(1) = 10.5 ft s t = 0 s = 1.5(0 3 ) - 13.5(0 2 ) + 22.5(0) = 0 t = 0 s t = 1 s and t = 5 s 4.50 t 2 - 27.0 t + 22.5 = 0 v = ds dt = 4.50 t 2 - 27.0 t + 22.5 s | t = 6 s = 1.5(6 3 ) - 13.5(6 2 ) + 22.5(6) = - 27.0 ft t = 6 s The position of a particle along a straight line is given by , where t is in seconds. Determine the position of the particle when and the total distance it travels during the 6-s time interval. Hint: Plot the path to determine the total distance traveled. t = 6 s s = (1.5 t 3 - 13.5 t 2 + 22.5 t ) ft Ans: s  t = 6 s = - 27.0 ft s tot = 69.0 ft

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    7 12–7. A particle moves along a straight line such that its position is defined by s = ( t 2 - 6 t + 5) m. Determine the average velocity, the average speed, and the acceleration of the particle when t = 6 s. Solution s = t 2 - 6t + 5 v = ds dt = 2 t - 6 a = d v dt = 2 v = 0 when t = 3 s  t = 0 = 5 s  t = 3 = - 4 s  t = 6 = 5 v avg = ∆ s ∆ t = 0 6 = 0 Ans. ( v sp ) avg = s T ∆ t = 9 + 9 6 = 3 m > s Ans. a  t = 6 = 2 m > s 2 Ans. Ans: v avg = 0 ( v sp ) avg = 3 m > s a  t = 6 s = 2 m > s 2

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    8 *12–8. A particle is moving along a straight line such that its position is defined by , where t is in seconds. Determine (a) the displacement of the particle during the time interval from to , (b) the average velocity of the particle during this time interval, and (c) the acceleration when . t = 1 s t = 5 s t = 1 s s = (10 t 2 + 20) mm SOLUTION (a) Ans. (b) Ans. (c) Ans. a = d 2 s dt 2 = 20 mm s 2 (for all t ) v avg = ¢ s ¢ t = 240 4 = 60 mm > s ¢ t = 5 - 1 = 4 s ¢ s = 270 - 30 = 240 mm s | 5 s = 10(5) 2 + 20 = 270 mm s | 1 s = 10(1) 2 + 20 = 30 mm s = 10 t 2 + 20 Ans: ∆ s = 240 mm v avg = 60 mm > s a = 20 mm > s 2

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    9 12–9. The acceleration of a particle as it moves along a straight line is given by a = (2 t - 1) m > s 2 , where t is in seconds. If s = 1 m and v = 2 m > s when t = 0, determine the particle’s velocity and position when t = 6 s. Also, determine the total distance the particle travels during this time period. Solution a = 2t - 1 d v = a dt L v 2 d v = L t 0 (2t - 1)dt v = t 2 - t + 2 dx = v dt L s t ds = L t 0 (t 2 - t + 2)dt s = 1 3 t 3 - 1 2 t 2 + 2 t + 1 When t = 6 s v = 32 m > s Ans. s = 67 m Ans. Since v  0 for 0 … t … 6 s, then d = 67 - 1 = 66 m Ans. Ans: v = 32 m > s s = 67 m d = 66 m

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    10 12–10. SOLUTION Ans. v = 1.29 m > s 0.8351 = 1 2 v 2 L 2 1 5 ds A 3 s 1 3 + s 5 2 B = L v 0 v d a ds = v d a = 5 A 3 s 1 3 + s 5 2 B A particle moves along a straight line with an acceleration of , where s is in meters. Determine the particle’s velocity when , if it starts from rest when . Use a numerical method to evaluate the integral. s = 1 m s = 2 m a = 5 > (3 s 1 > 3 + s 5 > 2 ) m > s 2 v v Ans: v = 1.29 m > s

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    11 12–11. A particle travels along a straight-line path such that in 4 s it moves from an initial position s A = - 8 m to a position s B = + 3 m. T hen in another 5 s it moves from s B to s C = - 6 m. Determine the particle’s a verage velocity and average speed during the 9-s time interval. SOLUTION Average Velocity: The displacement from A to C is ∆ s = s C - S A = - 6 - ( - 8) = 2 m. v avg = ∆ s ∆ t = 2 4 + 5 = 0.222 m > s Ans. Average Speed: The distances traveled from A to B and B to C are s A S B = 8 + 3 = 11.0 m and s B S C = 3 + 6 = 9.00 m, respectively. Then, the total distance traveled is s Tot = s A S B + s B S C = 11.0 + 9.00 = 20.0 m. ( v sp ) avg = s Tot ∆ t = 20.0 4 + 5 = 2.22 m > s Ans. Ans: v avg = 0.222 m > s ( v sp ) avg = 2.22 m > s

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    12 *12–12. SOLUTION Ans. Ans. s = 0.792 km = 792 m (120) 2 = 70 2 + 2(6000)( s - 0) v 2 = v 1 2 + 2 a c ( s - s 1 ) t = 8.33(10 - 3 ) hr = 30 s 120 = 70 + 6000( t ) v = v 1 + a c t Traveling with an initial speed of a car accelerates at along a straight road. How long will it take to reach a speed of Also, through what distance does the car travel during this time? 120 km > h? 6000 km > h 2 70 km > h, Ans: t = 30 s s = 792 m

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    13 12–13. SOLUTION Stopping Distance: For normal driver, the car moves a distance of before he or she reacts and decelerates the car. The stopping distance can be obtained using Eq. 12–6 with and . Ans. For a drunk driver, the car moves a distance of before he or she reacts and decelerates the car. The stopping distance can be obtained using Eq. 12–6 with and . Ans. d = 616 ft 0 2 = 44 2 + 2( - 2)( d - 132) A : + B v 2 = v 2 0 + 2 a c ( s - s 0 ) v = 0 s 0 = d ¿ = 132 ft d ¿ = v t = 44(3) = 132 ft d = 517 ft 0 2 = 44 2 + 2( - 2)( d - 33.0) A : + B v 2 = v 2 0 + 2 a c ( s - s 0 ) v = 0 s 0 = d ¿ = 33.0 ft d ¿ = v t = 44(0.75) = 33.0 ft Tests reveal that a normal driver takes about 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 ) and their cars can decelerate at , determine the shortest stopping distance d for each from the moment they see the pedestrians. Moral : If you must drink, please don’t drive! 2 ft > s 2 ft > s 0.75 s d v 1 44 ft / s Ans: Normal: d = 517 ft d runk: d = 616 ft

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    14 12–14. The position of a particle along a straight-line path is defined by s = ( t 3 - 6 t 2 - 15 t + 7) ft, where t is in seconds. Determine the total distance traveled when t = 10 s. What are the particle’s average velocity, average speed, and the instantaneous velocity and acceleration at this time? Solution s = t 3 - 6 t 2 - 15 t + 7 v = ds dt = 3t 2 - 12t - 15 When t = 10 s, v = 165 ft > s Ans. a = d v dt = 6 t - 12 When t = 10 s, a = 48 ft > s 2 Ans. When v = 0, 0 = 3 t 2 - 12 t - 15 The positive root is t = 5 s When t = 0, s = 7 ft When t = 5 s, s = - 93 ft When t = 10 s, s = 257 ft Total distance traveled s T = 7 + 93 + 93 + 257 = 450 ft Ans. v avg = ∆ s ∆ t = 257 - 7 10 - 0 = 25.0 ft > s Ans. ( v sp ) avg = s T ∆ t = 450 10 = 45.0 ft > s Ans. Ans: v = 165 ft > s a = 48 ft > s 2 s T = 450 ft v avg = 25.0 ft > s ( v sp ) avg = 45.0 ft > s

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    15 12–15. SOLUTION Ans. Ans. s = 1 k B ¢ 2 kt + ¢ 1 n 2 0 ≤ ≤ 1 2 - 1 n 0 R s = 2 a 2 kt + a 1 n 2 0 b b 1 2 2 k 3 t 0 L s 0 ds = L t 0 dt a 2 kt + a 1 v 2 0 b b 1 2 ds = n dt n = a 2 kt + a 1 n 2 0 b b - 1 2 - 1 2 1 n - 2 - n - 2 0 2 = - kt L n n 0 n - 3 d n = L t 0 - k dt a = d n dt = - k n 3 A part i c l e i s mov i ng w i t h a ve l oc i ty of w h en an d If it is subjected to a deceleration of where k is a constant, determine its velocity and position as functions of time. a = - kv 3 , t = 0. s = 0 v 0 Ans: v = a 2 kt + 1 v 2 0 b - 1 > 2 s = 1 k c a 2 kt + 1 v 2 0 b 1 > 2 - 1 v 0 d

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