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: Thevalingam, Donald Homework 15 Due: Jan 16 2007, midnight Inst: Eslami 1 This printout should have 16 questions. Multiplechoice 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....
View
Full
Document
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.
 Spring '08
 Turner
 Physics, Work

Click to edit the document details