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# solution_pdf4 - reddy(ar38357 Kinematics B clancy(SCI411-2...

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reddy (ar38357) – Kinematics B – clancy – (SCI411-2) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An automobile accelerates from rest at 0 . 2 m / s 2 for 21 s. The speed is then held constant for 17 s, after which there is an ac- celeration of - 1 . 5 m / s 2 until the automobile stops. What total distance was traveled? Correct answer: 0 . 12138 km. Explanation: For the acceleration, Let : v 0 = 0 , a 1 = 0 . 2 m / s 2 , and t 1 = 21 s . The displacement is s 1 = v 0 t 0 + 1 2 a 1 t 2 1 = 1 2 a 1 t 2 1 , reaching a speed of v 1 = v 0 + a t 1 = a 1 t 1 . For the constant speed, Let : a 2 = 0 and t 2 = 17 s . The displacement is s 2 = v 1 t 2 = a 1 t 1 t 2 , maintaining a speed of v 2 = v 1 = a 1 t 1 . During the braking interval, Let : a 3 = - 1 . 5 m / s 2 and v f = 0 . v 2 3 = v 2 2 + 2 a 3 s 3 = 0 s 3 = - v 2 2 2 a 3 = - ( a 1 t 1 ) 2 2 a 3 , so s = s 1 + s 2 + s 3 = 1 2 a 1 t 2 1 + a 1 t 1 t 2 - a 2 1 t 2 2 a 3 = 1 2 (0 . 2 m / s 2 ) (21 s) 2 + (0 . 2 m / s 2 ) (21 s) (17 s) - (0 . 2 m / s 2 ) 2 (21 s) 2 2 ( - 1 . 5 m / s 2 ) = 121 . 38 m = 0 . 12138 km . 002 10.0 points Two students are on a balcony 20 . 9 m above the street. One student throws a ball ver- tically downward at 14 . 2 m / s; at the same instant, the other student throws a ball verti- cally upward at the same speed. The second ball just misses the balcony on the way down. How far apart are the balls 0 . 48 s after they are thrown? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 13 . 632 m. Explanation: Let : h = 20 . 9 m , v i, 1 = 14 . 2 m / s , v i, 2 = - 14 . 2 m / s , Δ t = 0 . 48 s , and g = 9 . 8 m / s 2 . For the first ball, Δ y 1 = v i, 1 Δ t - 1 2 g t ) 2 = ( - 14 . 2 m / s) (0 . 48 s) - 1 2 ( 9 . 8 m / s 2 ) (0 . 48 s) 2 = - 7 . 94496 m which is 7 . 94496 m below the balcony.

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reddy (ar38357) – Kinematics B – clancy – (SCI411-2) 2 For the second ball, Δ y 2 = v i, 2 Δ t - 1 2 g t ) 2 = (14 . 2 m / s) (0 . 48 s) - 1 2 ( 9 . 8 m / s 2 ) (0 . 48 s) 2 = 5 . 68704 m The distance between the balls is | Δ y 2 - Δ y 1 | = | 5 . 68704 m - ( - 7 . 94496 m) | = 13 . 632 m .
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