Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 2 - Kinematics in One Dimension - Exercises and Problems: 28

Answer

At t = 0: $v = 0$ At t = 2 s: $v = 5~m/s$ At t = 4 s: $v = 20~m/s$ At t = 6 s: $v = 30~m/s$ At t = 8 s: $v = 30~m/s$

Work Step by Step

The change in velocity is equal to the area under the acceleration versus time graph. Note that the initial velocity is zero. At t = 0: $v = 0$ From t = 0 to t = 2 s: $\Delta v = \frac{1}{2}(5~m/s^2)(2~s)$ $\Delta v = 5~m/s$ At t = 2 s: $v = 0+\Delta v = 5~m/s$ From t = 2 s to t = 4 s: $\Delta v = 15~m/s$ At t = 4 s: $v = 5~m/s+\Delta v$ $v = 5~m/s+15~m/s$ $v = 20~m/s$ From t = 4 s to t = 6 s: $\Delta v = \frac{1}{2}(10~m/s^2)(2~s)$ $\Delta v= 10~m/s$ At t = 6 s: $v = 20~m/s+\Delta v$ $v = 20~m/s+10~m/s$ $v = 30~m/s$ From t = 6 s to t = 8 s: $\Delta v = 0$ At t = 8 s: $v = 30~m/s + \Delta v$ $v = 30~m/s + 0$ $v = 30~m/s$
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