Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 2 - Kinematics in One Dimension - Problems - Page 49: 35


52.8 m

Work Step by Step

$x=ut+\frac{1}{2}at^{2}$ For both knights, u=0 $x=\frac{1}{2}at^{2}$ $t=\sqrt \frac{2x}{a}$ since $t\gt0$ Distance traveled by Sir George = $x_{G}$ Distance traveled by Sir Alfred = $88.0-x_{G}$ Sir George: $t=\sqrt \frac{2(x_{G})}{0.300}=\sqrt \frac{20x_{G}}{3}$ Sir Alfred: $t=\sqrt \frac{2(88.0-x_{G})}{0.200}=\sqrt (10(88.0-x_{G}))$ We now use a system of linear equations: $\sqrt \frac{20x_{G}}{3}=\sqrt (10(88.0-x_{G}))$ $ \frac{20x_{G}}{3}=10(88.0-x_{G})$ $20x_{G}=30(88.0-x_{G})$ $20x_{G}=2640-30x_{G}$ $50x_{G}=2640$ $x_{G}=52.8\; m$
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