Physics (10th Edition)

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

Chapter 2 - Kinematics in One Dimension - Problems: 10

Answer

80.5 m travelled east

Work Step by Step

Let due west be the negative direction Average velocity = $\frac{displacement}{time}$ = $\frac{x}{t}$ Time spent hiking west = $\frac{-6440}{-2.68}$ $\approx$ 2.40 $\times$ $10^{3}$ s Let distance hiked east = $x_{e}$ Time spent hiking east = $\frac{x_{e}}{0.447}$ Displacement = $x_{e}$ - 6440 Time = 2.40 $\times$ $10^{3}$ + $\frac{x_{e}}{0.447}$ Average velocity = displacement/time -1.34 = ($x_{e}$ - 6440) / (2.40 $\times$ $10^{3}$ + $\frac{x_{e}}{0.447}$) -3.22 $\times$ $10^{3}$ - 2.998$x_{e}$ = $x_{e}$ - 6440 3.998$x_{e}$ = 3.22$\times$ $10^{3}$ $x_{e}$ $\approx$ 80.5 m
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