SOL15 - Thevalingam, Donald Homework 15 Due: Jan 16 2007,...

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Unformatted text preview: Thevalingam, Donald Homework 15 Due: Jan 16 2007, midnight Inst: Eslami 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points How far will anobject move in9 s if itsaverage speed during that time is 40 m / s? Correct answer: 360 m. Explanation: If an object is moving during time t with average speed v then the distance traveled will be s = vt keywords: 002 (part 1 of 1) 10 points In order to qualify for the finals in a racing event, a race car must achieve an average speed of 252 km / h on a track with a total length of 1800 m. If a particular car covers the first half of the track at an average speed of 216 km / h, what minimum average speed must it have in the second half of the event in order to qualify? Correct answer: 302 . 4 km / h. Explanation: Let t t be the maximum time to complete the trip. t t = totaldistance neededaveragespeed = d v av The time spent to the first half, t 1 is t 1 = half distance v av 1 = d/ 2 v av 1 Thus, the maximum time that can be spent second half of the trip is t 2 = t t- t 1 and the required average speed on the second half is v av 2 = d/ 2 t 2 keywords: 003 (part 1 of 3) 10 points A ball is dropped from rest at point O . After falling for some time, it passes by a window of height 3 . 1 m and it does so during time 0 . 31 s. The acceleration of gravity is 9 . 8 m / s 2 . O A B 3 . 1 m x y The ball accelerates all the way down; let v A be its speed as it passes the windows top A and v B its speed as it passes the windows bottom B . How much did the ball speed up as it passed thewindow; i.e. , calculate v down = v B- v A ? Correct answer: 3 . 038 m / s. Explanation: Let : h = 3 . 1 m , t AB = 0 . 31 s , and g = 9 . 8 m / s 2 . O A B h t AB t y Assume: Down is positive. The ball falls under a constant acceleration g , so g = v t = v B- v A t and the change of its velocity during time t AB is simply ~v = ~v B- ~v A = g t AB , assuming the downward direction to be positive. v down = (9 . 8 m / s 2 )(0 . 31 s) = 3 . 038 m / s . Thevalingam, Donald Homework 15 Due: Jan 16 2007, midnight Inst: Eslami 2 004 (part 2 of 3) 10 points Calculate the speed v A at which the ball passes the windows top....
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This note was uploaded on 03/04/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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SOL15 - Thevalingam, Donald Homework 15 Due: Jan 16 2007,...

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