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Unformatted text preview: cerutti (cpc566) HW01 TSOI (58160) 1 This printout should have 24 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A student throws a baseball at a large gong 49 m away and hears the sound of the gong 1 . 47281 s later. The speed of sound in air is 330 m / s. What was the average speed of the base ball on its way to the gong? (For simplicity, assume its trajectory to be a straight line.) Correct answer: 37 m / s. Explanation: Let : d = 49 m , t = 1 . 47281 s , and v = 330 m / s . It takes t s = d v s = 49 m 330 m / s = 0 . 148485 s for the sound to travel back from the gong, so the flight time of the baseball was only t b = t tot t s = 1 . 47281 s . 148485 s = 1 . 32432 s and the baseballs average speed was v b = d t = 49 m 1 . 32432 s = 37 m / s . 002 10.0 points Consider three position curves between time points t A and t B . s B s A B A t A t B 3 2 1 t s Choose the correct relationship among quantities v 1 , v 2 , and v 3 . v = v A + v B 2 when a is a constant. 1. v 1 &gt; v 2 &gt; v 3 2. v 1 &lt; v 2 &lt; v 3 3. v 1 = v 2 = v 3 correct Explanation: The average velocity of an object is v = displacement time = s B s A t B t A . All three curves have exactly the same change in position s = s B s A in exactly the same time interval t = t B t A , so all three average velocities are equal: v 1 = v 2 = v 3 . 003 (part 1 of 3) 10.0 points Given: The acceleration of gravity on Earth is 9 . 8 m / s 2 . Consider a ball thrown up from the ground (the point O). It passes a window (the seg ment AB) in the time interval 0 . 205 s (see the figure). The points in the figure represent the sequential order, and are not drawn to scale. The distance AB = 1 . 18 m. O C B A 1 . 18 m b b b b b b b b b b b x y Find the average speed as the ball passes the window. Correct answer: 5 . 7561 m / s. Explanation: Basic Concepts: vectora = vectorv t cerutti (cpc566) HW01 TSOI (58160) 2 For constant acceleration v = s t = v i + v f 2 v = v + a t. For the acceleration of gravity ( a = g ) v = v g t. Solution: The average velocity is given by v = s t =  AB  t = 1 . 18 m . 205 s = 5 . 7561 m / s . Alternative Part 1: d = y B y A = v A t 1 2 g t 2 v A = d + 1 2 g t 2 t = 1 . 18 m + 1 2 (9 . 8 m / s 2 ) (0 . 205 s) 2 (0 . 205 s) = 6 . 7606 m / s v B = v A g t = 6 . 7606 m / s (9 . 8 m / s 2 ) (0 . 205 s) = 4 . 7516 m / s v = v A + v B 2 = 6 . 7606 m / s + 4 . 7516 m / s 2 = 5 . 7561 m / s . 004 (part 2 of 3) 10.0 points What is the magnitude of the decrease of the velocity from A to B? Correct answer: 2 . 009 m / s. Explanation: For a constant acceleration vectora = vectorv t , =  vectorv  =  vectora t  = g t....
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This note was uploaded on 05/04/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

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