HW2 solution

# HW2 solution - choi(hc5878 hw2 Demkov(59910 This print-out...

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choi (hc5878) – hw2 – Demkov – (59910) 1 This print-out should have 46 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points If a race car completes a 3 . 7 mi oval track in 64 . 5 s, what is its average speed? Correct answer: 206 . 512 mi / h. Explanation: d = v t , so v = d t = 3 . 7 mi 64 . 5 s 60 s 1 min 60 min 1 h = 206 . 512 mi / h . 002 (part 2 of 2) 10.0 points Did the car accelerate? 1. No, the speed didn’t change. 2. Yes, the direction of the motion changed. correct 3. Yes, the speed changed. Explanation: 003 (part 1 of 2) 10.0 points The maker of a certain automobile advertises that the automobile takes 17 to accelerate from 14 . 5 mi/h to 57 . 9 mi/h. Assume con- stant acceleration. a) Compute the acceleration. Correct answer: 3 . 74431 ft / s 2 . Explanation: Dimensional Analysis for v : mi hr · 5280 ft 1 mi · 1 hr 3600 s = ft s v 0 = 14 . 5 mi / h = 21 . 2667 ft / s v f = 57 . 9 mi / h = 84 . 92 ft / s Solution Assume the constant acceleration is a . Then a ( t ) = a v = a integraldisplay v dt = integraldisplay a dt v = a integraldisplay dt = at + c v (0) = 21 . 2667 = c = v 0 = 21 . 2667, so v = at + 21 . 2667 v (17) = 84 . 92 = a = v v 0 t 3 . 74431 004 (part 2 of 2) 10.0 points b) How far does the car travel during the 17 seconds? Correct answer: 902 . 587 ft. Explanation: v = 3 . 74431 t + 21 . 2667 s = 3 . 74431 t + 21 . 2667 integraldisplay s dt = integraldisplay (3 . 74431 t + 21 . 2667) dt s = 3 . 74431 integraldisplay t dt + 21 . 2667 integraldisplay dt = 1 2 (3 . 74431) t 2 + 21 . 2667 + c 1 At t = 0, the car is at its initial position s 0 = 0, so s = 3 . 74431 t 2 2 + 21 . 2667 t and s (17) = 902 . 587. 005 (part 1 of 3) 10.0 points A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0 . 5 m / s 2 . A green car arrives at the position of the stop-light 5 s after the light had turned green. What is the lapse time of the blue car when the green car catches it if the green car main- tains the slowest constant speed necessary to catch up to the blue car? Correct answer: 10 s. Explanation:

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choi (hc5878) – hw2 – Demkov – (59910) 2 Let : x 0 = 0 m , stop-light position v 0 b = 0 m / s a b = 0 . 5 m / s 2 , and Δ t = 5 s . The position x b and x g of both cars are equal x b = x g when the blue car reaches the green car. The kinematic equations are x b = x 0 + v 0 b t + 1 2 a b t 2 = 1 2 a b t 2 , and (1) x g = x 0 + v g ( t Δ t ) = v g ( t Δ t ) . (2) For x = x b = x g , we have v g ( t Δ t ) = 1 2 a b t 2 , so (3) v g = a b t 2 2 ( t Δ t ) . (4) To minimize v g , set its derivative equal to zero, then d dt bracketleftbigg a t t 2 2 ( t Δ t ) bracketrightbigg = 0 4 ( t Δ t ) a b t 2 a b t 2 4 ( t Δ t ) 2 = 0 2 ( t Δ t ) t = 0 t 2 Δ t = 0 t = 2 Δ t (5) = 2 (5 s) = 10 s . 006 (part 2 of 3) 10.0 points What is the constant speed of the green car? Correct answer: 5 m / s. Explanation: Substitute t = 2 Δ t from Eq. 5 into Eq. 4 v g = a b t 2 2 ( t Δ t ) (4) = a b (2 Δ t ) 2 2 Δ t = 2 a b Δ t (6) = 2 (0 . 5 m / s 2 ) (5 s) = 5 m / s . Check to show x b = x g at t = 2 Δ t x b = 1 2 a b t 2 (1) = 1 2 a b (2 Δ t ) 2 = 2 a Δ t 2 = 2 (0 . 5 m / s 2 ) (5 s) 2 = 25 m , and x g = v g ( t Δ t ) (2) = v g (2 Δ t Δ t ) = v g Δ t = (5 m / s) (5 s) = 25 m .
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