Physics (10th Edition)

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

Chapter 2 - Kinematics in One Dimension - Problems: 36

Answer

a) 11 s b) 30 m

Work Step by Step

$x=ut+\frac{1}{2}at^{2}$ For both players, u=0 $x=\frac{1}{2}at^{2}$ Distance traveled by first player = $x_{1}$ Distance traveled by second player = $48-x_{1}$ First player: $x_{1}=\frac{1}{2}(0.50)t^{2}=0.25t^{2}$ Second player: $48-x_{1}=\frac{1}{2}(0.30)t^{2}$ $x_{1}=48-0.15t^{2}$ We now use a system of linear equations: a) $0.25t^{2}=48-0.15t^{2}$ $0.40t^{2}=48$ $t^{2}=120$ $t=\sqrt (120)$ since $t\gt0$ $t\approx 11\; s$ b) $x_{1}=0.25t^{2}$ $x_{1}=0.25(\sqrt 120)^{2}=30\; m$
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