hw2-solutions - bello(rtb473 – hw2 – Demkov –(59910 1...

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Unformatted text preview: bello (rtb473) – 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 2 . 5 mi oval track in 50 . 5 s, what is its average speed? Correct answer: 178 . 218 mi / h. Explanation: d = v t , so v = d t = 2 . 5 mi 50 . 5 s 60 s 1 min 60 min 1 h = 178 . 218 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 11 to accelerate from 15 . 4 mi/h to 48 . 8 mi/h. Assume con- stant acceleration. a) Compute the acceleration. Correct answer: 4 . 45333 ft / s 2 . Explanation: Dimensional Analysis for v : mi hr · 5280 ft 1 mi · 1 hr 3600 s = ft s v = 15 . 4 mi / h = 22 . 5867 ft / s v f = 48 . 8 mi / h = 71 . 5733 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) = 22 . 5867 = ⇒ c = v = 22 . 5867, so v = at + 22 . 5867 v (11) = 71 . 5733 = ⇒ a = v − v t ≈ 4 . 45333 004 (part 2 of 2) 10.0 points b) How far does the car travel during the 11 seconds? Correct answer: 517 . 88 ft. Explanation: v = 4 . 45333 t + 22 . 5867 s ′ = 4 . 45333 t + 22 . 5867 integraldisplay s ′ dt = integraldisplay (4 . 45333 t + 22 . 5867) dt s = 4 . 45333 integraldisplay t dt + 22 . 5867 integraldisplay dt = 1 2 (4 . 45333) t 2 + 22 . 5867 + c 1 At t = 0, the car is at its initial position s = 0, so s = 4 . 45333 t 2 2 + 22 . 5867 t and s (11) = 517 . 88. 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 6 . 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: 13 s. bello (rtb473) – hw2 – Demkov – (59910) 2 Explanation: Let : x = 0 m , stop-light position v b = 0 m / s a b = 0 . 5 m / s 2 , and Δ t = 6 . 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 + v b t + 1 2 a b t 2 = 1 2 a b t 2 , and (1) x g = x + 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 (6 . 5 s) = 13 s ....
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This note was uploaded on 11/10/2008 for the course PHY 301 taught by Professor Swinney during the Spring '07 term at University of Texas.

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hw2-solutions - bello(rtb473 – hw2 – Demkov –(59910 1...

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