This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: obennoskey (blo242) – hw2 – Demkov – (59910) 1 This printout should have 46 questions. Multiplechoice 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 . 6 mi oval track in 48 . 4 s, what is its average speed? Correct answer: 267 . 769 mi / h. Explanation: d = v t , so v = d t = 3 . 6 mi 48 . 4 s 60 s 1 min 60 min 1 h = 267 . 769 mi / h . 002 (part 2 of 2) 10.0 points Did the car accelerate? 1. Yes, the speed changed. 2. Yes, the direction of the motion changed. correct 3. No, the speed didn’t change. 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 45 . 3 mi/h. Assume con stant acceleration. a) Compute the acceleration. Correct answer: 2 . 65726 ft / s 2 . Explanation: Dimensional Analysis for v : mi hr · 5280 ft 1 mi · 1 hr 3600 s = ft s v = 14 . 5 mi / h = 21 . 2667 ft / s v f = 45 . 3 mi / h = 66 . 44 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 = 21 . 2667, so v = at + 21 . 2667 v (17) = 66 . 44 = ⇒ a = v − v t ≈ 2 . 65726 004 (part 2 of 2) 10.0 points b) How far does the car travel during the 17 seconds? Correct answer: 745 . 507 ft. Explanation: v = 2 . 65726 t + 21 . 2667 s ′ = 2 . 65726 t + 21 . 2667 integraldisplay s ′ dt = integraldisplay (2 . 65726 t + 21 . 2667) dt s = 2 . 65726 integraldisplay t dt + 21 . 2667 integraldisplay dt = 1 2 (2 . 65726) t 2 + 21 . 2667 + c 1 At t = 0, the car is at its initial position s = 0, so s = 2 . 65726 t 2 2 + 21 . 2667 t and s (17) = 745 . 507. 005 (part 1 of 3) 10.0 points A blue car pulls away from a red stoplight just after it has turned green with a constant acceleration of 0 . 8 m / s 2 . A green car arrives at the position of the stoplight 4 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: 8 s. Explanation: obennoskey (blo242) – hw2 – Demkov – (59910) 2 Let : x = 0 m , stoplight position v b = 0 m / s a b = 0 . 8 m / s 2 , and Δ t = 4 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 (4 s) = 8 s ....
View
Full
Document
This note was uploaded on 11/02/2008 for the course PHY 301 taught by Professor Demovk during the Spring '08 term at University of Texas at Dallas, Richardson.
 Spring '08
 DEMOVK

Click to edit the document details