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: 81

Answer

The speed at the finish line is 12.5 m/s

Work Step by Step

Let $v_f$ be the speed at the end of the acceleration period. We can find an expression for the acceleration. $v_f = v_0+(4.0~s)~a$ $v_f = 0+(4.0~s)~a$ $a = \frac{v_f}{4.0~s}$ We can find an expression for the distance $d_1$ covered during the acceleration period. $d_1 = \frac{1}{2}a(4.0~s)^2$ $d_1 = \frac{1}{2}(\frac{v_f}{4.0~s})(4.0~s)^2$ $d_1 = (2.0~s)~v_f$ The remaining distance $d_2 $ must be covered in 6.0 seconds. We can find the final speed $v_f$. (Note that $d_2 = 100~m - d_1$) $d_2 = (6.0~s)~v_f$ $100~m - d_1= (6.0~s)~v_f$ $100~m - (2.0~s)~v_f= (6.0~s)~v_f$ $v_f = \frac{100~m}{8.0~s}$ $v_f = 12.5~m/s$ The speed at the end of the acceleration period is 12.5 m/s. Since the speed at the finish line is the same, the speed at the finish line is 12.5 m/s
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