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Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition

Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition provides the perfect textbook solutions, giving you the help you need to succeed in your studies.

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    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. 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: v = 0 s = 80.7 m 1 12–1.

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    12–2. SOLUTION When , Ans. Ans. Since then Ans. d = 6 - = 6 m v Z 0 s = 639.5 m v = > s t = s = 1 t - 1 2 t 2 + t + L s ds = L t 0 ( t - t + dt v = t - t + L v d v = L t 0 (4 t - 1) dt The acceleration of a particle as it moves along a straight line is given by where t is in seconds. If and when determine the particle’s velocity and position when Also, determine the total distance the particle travels during this time period. t = t = 0, v = > s s = a = 1 t - 1 2 m > s , 2 5 5 m 2 m 5 s. 3 3 4 5 2 4 5) 5 5 5 2 5 s 625 m 39.5 2 37.5 4 v = 625 m > s s = 639.5 m d = 637.5 m Ans: 2

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    12– 3. 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. a = - 12 m > s 2 s = 0 s T = ( v sp ) avg = 2.67 m > s = 8 m Ans: 3

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    *12– 4. 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 4

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    12– 5. 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 5

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    12–6. stone A is dropped from rest down a well, and in 1 s another stone B is dropped from rest. Determine the distance between the stones another second later. SOLUTION Ans. ¢ s = - . 1 = . s B = s A = 0 + 0 + 1 2 (9.81)(1) 2 s A = s A = 0 + 0 + 1 2 ( . )(2) 2 + T s = s 1 + v 1 t + 1 2 a c t 2 A 9 81 19.62 m 4.91 m 19.62 4 9 14 71 m  s = 14.71 m Ans: 6

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    12–7. A bus starts from rest with a constant acceleration of 1 . Determine the time required for it to attain a speed of 25 and the distance traveled. m > s m > s 2 SOLUTION Kinematics : and . Ans. Ans. s = 312.5 m 25 2 = 0 + 2(1)( s - 0) v 2 = v 2 0 + 2 a c ( s - s 0 ) A + : B t = 25 s 25 = 0 + (1) t v = v 0 + a c t A + : B a c = 1 m > s 2 s 0 = 0, v = 25 m > s, v 0 = 0, t = 25 s s = 312.5 m Ans: 7

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    *12– 8. 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 8

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    12–9. SOLUTION Motion of car A : Motion of car B : The distance between cars A and B is Ans. s BA = | s B - s A | = ` v A v B a A - v 2 A 2 a A ` = ` 2 v A v B - v 2 A 2 a A ` s B = v B t = v B a v A a A b = v A v B aA s A = v 2 A 2 a A 0 = v A 2 + 2( - a A )( s A - 0) v 2 = v 0 2 + 2 a c ( s - s 0 ) 0 = v A - a A t t = v A a A v = v 0 + a c t W h en two cars A an d B are next to one anot h er, t h ey are traveling in the same direction with speeds and respectively. If B maintains its constant speed, while A begins to decelerate at determine the distance d between the cars at the instant A stops. a A , v B , v A A B d S BA = ` 2 v A v B - v 2 A 2 a A ` Ans: 9

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    12– 10. 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 10

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    12– 11. 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 11

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    *12– 12. 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 12

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    12– 13. 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 13

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    12– 14. A train s tart s from re s t at s tation A and accelerate s at for 60 s . Afterward s it tra v el s with a con s tant v elocity for 15 min. It then decelerate s at 1 until it i s brou g ht to re s t at s tation B . Determine the di s tance between the s tation s . m > s 2 0.5 m > s 2 SOLUTION Kinematics: For s ta g e (1) motion, and Thu s , For s ta g e (2) motion, and Thu s , For s ta g e (3) motion, and Thu s , An s. = 28 350 m = 28.4 km s 3 = 27 900 + 30(30) + 1 2 ( - 1)(30 2 ) s = s 0 + v 0 t + 1 2 a c t 2 + : t = 30 s 0 = 30 + ( - 1) t v = v 0 + a c t A + : B a c = - 1 m > s 2 . v 0 = 30 m > s , v = 0, s 0 = 27 900 m s 2 = 900 + 30(900) + 0 = 27 900 m s = s 0 + v 0 t + 1 2 a c t 2 A + : B t = 15(60) = 900 s . a c = 0 s 0 = 900 m, v 0 = 30 m > s , v 1 = 0 + 0.5(60) = 30 m > s v = v 0 + a c t A + : B s 1 = 0 + 0 + 1 2 (0.5)(60 2 ) = 900 m s = s 0 + v 0 t + 1 2 a c t 2 A + : B a c = 0.5 m > s 2 . t = 60 s , s 0 = 0, v 0 = 0, Ans: s = 28.4 km 14

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    12–15. A particle is moving along a straight line such that its velocity is defined as , where s is in meters. If when , determine the velocity and acceleration as functions of time. t = 0 s = 2 m v = ( - 4 s 2 ) m > s SOLUTION Ans. Ans. a = dv dt = 16(2)(8 t + 1)(8) (8 t + 1) 4 = 256 (8 t + 1) 3 m > s 2 v = - 4 a 2 8 t + 1 b 2 = - 16 (8 t + 1) 2 m > s s = 2 8 t + 1 t = 1 4 ( s - 1 - 0.5) - s - 1 | s 2 = - 4 t | 0 t L s 2 s - 2 ds = L t 0 - 4 dt ds dt = - 4 s 2 v = - 4 s 2 v = 16 (8 t + 1) 2 m > s a = 256 (8 t + 1) 3 m > s 2 Ans: 15

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