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p201f04-homework02-sol

p201f04-homework02-sol - Version One Homework 2 Heinz 81204...

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Version One – Homework 2 – Heinz – 81204 – Sep 02, 2004 1 This print-out should have 22 questions. Multiple-choice 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 x -axis is shown. The objects initial position is 0 m. - 6 - 4 - 2 0 2 4 6 8 10 12 0 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 = Δ x Δ t = x f - x i Δ 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, wording-variable. 006 (part 1 of 4) 3 points Consider the plot below describing the accel- eration of a particle along a straight line with
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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 0 1 2 3 4 5 6 0 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 0 = x 0 , 1 = - 30 m , v 0 = v 0 , 1 = 4 m / s , ( t 0 , a 0 ) = ( t 0 , 1 , a 0 ) = (0 s , 0 m / s 2 ) , ( t 1 , a 1 ) = ( t 0 , 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 0 , 1 + a 1 , 2 t , where the acceleration is constant. Solution: With constant acceleration ( a 1 , 2 = - 3 m / s 2 ), v = v 0 , 1 + a 1 , 2 t (1) = (4 m / s) + ( - 3 m / s 2 ) (3 s) = - 5 m / s . Equations 1 and 3 are plotted below. - 1 0 1 2 0 1 2 3 4 5 6 7 8 9 time (s) velocity (m/s) [ × 0 . 1] 007 (part 2 of 4) 2 points What is the position at 3 s? Correct answer: - 31 . 5 m. Explanation: Basic Concepts: The change in position is the area ( v 0 , 1 t + 1 2 a t 2 ) between the veloc- ity curve and the time axis x = x 0 , 1 + v 0 , 1 t + 1 2 a t 2 . Solution: With constant acceleration ( a 1 , 2 = - 3 m / s 2 ), x = x 0 , 1 + v 0 , 1 t + 1 2 a 1 , 2 t 2 (2) = ( - 30 m) + (4 m / s) (3 s) + 1 2 ( - 3 m / s 2 ) (3 s) 2 = - 31 . 5 m .
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