Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 2 - Describing Motion: Kinematics in One Dimension - Problems - Page 44: 37


0.70 $m/s^{2}$

Work Step by Step

In order for the two runners to finish side by side they have to reach the finish line at the same time from their current position. You start off by taking Mary's location as the origin and using that to solve for Sally. Sally: $22 = 5.0+ 5.0t+\frac{1}{2}(-0.40)t^{2}$ --> $t^{2}-25t+85=0$ $t= 25 +/- $$\frac{\sqrt {25^2-4(85)}}{2}$ There is two answers for t, t = 4.059s and 20.94s The first value of t is when Sally first crosses the finish line. We then use this value of t which is 4.059s to find Mary's acceleration. Mary: $22= 0+4t+\frac{1}{2}at^{2}$ $a = $$\frac{\sqrt {22-4t}}{\frac{1}{2}t^{2}}$ $a = $$\frac{\sqrt {22-4(4.059)}}{\frac{1}{2}(4.059)^{2}}$ Solve for a and get $a= 0.70 m/s^{2}$
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