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Unformatted text preview: Version One Homework 2 Heinz 81204 Sep 02, 2004 1 This printout should have 22 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. Serway CP 02 06 02:01, trigonometry, numeric, > 1 min, nor mal. 001 (part 1 of 5) 2 points The position versus time for a certain object moving along the xaxis is shown. The objects initial position is 0 m. 6 4 2 2 4 6 8 10 12 1 2 3 4 5 6 7 8 9 time (s) position(m) Find the average velocity in the time inter val 0 s to 2 s. Correct answer: 5 m / s. Explanation: Given : x f = 10 m and x i = 0 m . v av = x t = x f x i t v av = 10 m (0 m) 2 s 0 s = 5 m / s . 002 (part 2 of 5) 2 points Find the average velocity in the time interval 0 s to 4 s. Correct answer: 1 . 25 m / s. Explanation: Given : x f = 5 m and x i = 0 m . v av = 5 m (0 m) 4 s 0 s = 1 . 25 m / s . 003 (part 3 of 5) 2 points Find the average velocity in the time interval 2 s to 4 s. Correct answer: 2 . 5 m / s. Explanation: Given : x f = 5 m and x i = 10 m . v av = 5 m (10 m) 4 s 2 s = 2 . 5 m / s . 004 (part 4 of 5) 2 points Find the average velocity in the time interval 4 s to 7 s. Correct answer: 3 . 33333 m / s. Explanation: Given : x f = 5 m and x i = 5 m . v av = 5 m (5 m) 7 s 4 s = 3 . 33333 m / s . 005 (part 5 of 5) 2 points Find the average velocity in the time interval 0 s to 9 s. Correct answer: 0 m / s. Explanation: Given : x f = 0 m and x i = 0 m . v av = 0 m (0 m) 9 s 0 s = 0 m / s . Acceleration vs Time 01 02:04, trigonometry, numeric, > 1 min, wordingvariable. 006 (part 1 of 4) 3 points Consider the plot below describing the accel eration of a particle along a straight line with Version One Homework 2 Heinz 81204 Sep 02, 2004 2 an initial position of 30 m and an initial velocity of 4 m / s. 4 3 2 1 1 2 3 4 5 6 1 2 3 4 5 6 7 8 9 time (s) acceleration(m/s 2 ) What is the velocity at 3 s? Correct answer: 5 m / s. Explanation: In order to use the above graph, let x = x , 1 = 30 m , v = v , 1 = 4 m / s , ( t ,a ) = ( t , 1 ,a ) = (0 s , 0 m / s 2 ) , ( t 1 ,a 1 ) = ( t , 1 ,a 1 , 2 ) = (0 s , 3 m / s 2 ) , ( t 2 ,a 2 ) = ( t 2 , 3 ,a 1 , 2 ) = (4 s , 3 m / s 2 ) , ( t 3 ,a 3 ) = ( t 2 , 3 ,a 3 , 4 ) = (4 s , 5 m / s 2 ) , ( t 4 ,a 4 ) = ( t 4 , 5 ,a 3 , 4 ) = (9 s , 5 m / s 2 ) , and ( t 5 ,a 5 ) = ( t 4 , 5 ,a 5 ) = (9 s , 0 m / s 2 ) . Basic Concepts: The plot shows a curve of acceleration versus time . The change in velocity is the area ( a 1 , 2 t ) between the acceleration curve and the time axis v = v , 1 + a 1 , 2 t, where the acceleration is constant. Solution: With constant acceleration ( a 1 , 2 = 3 m / s 2 ), v = v , 1 + a 1 , 2 t (1) = (4 m / s) + ( 3 m / s 2 )(3 s) = 5 m / s ....
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This note was uploaded on 07/02/2011 for the course PHYS 201 taught by Professor Woodahl during the Fall '09 term at Indiana.
 Fall '09
 WOODAHL
 Physics, Work

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