Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 2 - Kinematics in One Dimension - Exercises and Problems: 83

Answer

The speed at the finish line is 12.5 m/s

Work Step by Step

We can find the speed at the end of the acceleration period; $v = (4.0~s)~a$ We can find the distance covered during the acceleration period; $d = \frac{1}{2}a(4.0~s)^2$ $d = (8.0~s^2)~a$ The remaining distance must be covered in 6.0 seconds. We can find the acceleration: $6.0~s = \frac{100.0~m-(8.0~s^2)~a}{(4.0~s)~a}$ $(32.0~s^2)~a = 100.0~m$ $a = 3.125~m/s^2$ We can find the speed at the finish line; $v = (4.0~s)~a$ $v = (4.0~s)(3.125~m/s^2)$ $v = 12.5~m/s$ The speed at the finish line is 12.5 m/s.
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