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# hw4 - Overton Mays Homework 4 Due 4:00 am Inst Turner This...

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Overton, Mays – Homework 4 – Due: Jan 31 2005, 4:00 am – Inst: Turner 1 This print-out should have 11 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 4) 10 points Consider the plot below describing the ac- celeration of a particle along a straight line with an initial position of 0 m and an initial velocity of 0 m / s. - 7 - 6 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 0 1 2 3 4 5 6 7 8 9 time (s) acceleration (m/s 2 ) What is the velocity at 4 s? Correct answer: - 14 m / s. Explanation: In order to use the above graph, let x 0 = 0 m , v 0 = 0 m / s , ( t 0 , a 0 ) = (0 s , 0 m / s 2 ) , ( t 1 , a 0 ) = (1 s , 0 m / s 2 ) , ( t 1 , a 1 ) = (1 s , - 7 m / s 2 ) , ( t 3 , a 1 ) = (3 s , - 7 m / s 2 ) , ( t 3 , a 0 ) = (3 s , 0 m / s 2 ) , ( t 4 , a 0 ) = (4 s , 0 m / s 2 ) , ( t 4 , a 2 ) = (4 s , 5 m / s 2 ) , ( t 7 , a 2 ) = (7 s , 5 m / s 2 ) , ( t 7 , a 0 ) = (7 s , 0 m / s 2 ) , and ( t 9 , a 0 ) = (9 s , 0 m / s 2 ) . Basic Concepts: The plot shows a curve of acceleration versus time . The change in velocity is the area between the acceleration curve and the time axis v = v i + a Δ t , where the acceleration is constant. Solution: The velocity after 1 s, 3 s, 4 s, 7 s, and 9 s is v 1 = v 0 + a 0 [ t 1 - t 0 ] = (0 m / s) + (0 m / s 2 ) [(1 s) - (0 s)] = 0 m / s , v 3 = v 1 + a 1 [ t 3 - t 1 ] = (0 m / s) + ( - 7 m / s 2 ) [(3 s) - (1 s)] = - 14 m / s , v 4 = v 3 + a 0 [ t 4 - t 3 ] = ( - 14 m / s) + (0 m / s 2 ) [(4 s) - (3 s)] = - 14 m / s , v 7 = v 4 + a 2 [ t 7 - t 4 ] = (0 m / s) + (5 m / s 2 ) [(7 s) - (4 s)] = 1 m / s , v 9 = v 7 + a 0 [ t 9 - t 7 ] = (1 m / s) + (0 m / s 2 ) [(9 s) - (7 s)] = 1 m / s . The velocity graph below shows the area under the curve in the given acceleration vs time graph. - 14 - 12 - 10 - 8 - 6 - 4 - 2 0 2 0 1 2 3 4 5 6 7 8 9 time (s) velocity (m/s)

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Overton, Mays – Homework 4 – Due: Jan 31 2005, 4:00 am – Inst: Turner 2 002 (part 2 of 4) 0 points Calculate the position displacement after the car traveled the first 7 s. Correct answer: 47 . 5 m. Explanation: The change in position is the area between the velocity curve and the time axis x = x i + v i Δ t + 1 2 a Δ t , where the acceleration is constant.
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