Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 2 - Motion Along a Straight Line - Problems - Page 35: 42b

Answer

$$26.14\frac{m}{s}$$

Work Step by Step

Before 2.4s: $$ v_{me} = 110 \frac{km}{hr} · 1000 \frac{m}{km} ·\frac{1}{3600} \frac{hr}{sec} $$ $$= \frac{275}{9} \frac{m}{s}$$ $$ v_{police} = \frac{275}{9} - 2.5t $$ $$ d_{police} = \frac{275}{9}t - \frac{1}{2} · 5 · t^{2}$$ $$ d_{me} = \frac{275}{9} · t = \frac{275}{9} · 2.4 = \frac{220}{3} m$$ After 2.4s : $$ d_{me} = \frac{275}{9} (t - \frac{12}{5}) - \frac{5}{2} (t - \frac{12}{5})^{2} +\frac{220}{3} $$ $$ d_{me} =- \frac{5}{2}t^{2} + \frac{383}{9}t - \frac{72}{5}$$ $$ v_{me} =d_{me} \frac{d}{dt} = -5t + \frac{383}{9}$$ Equating $d_{me}$ and $d_{police}$: $$- \frac{5}{2}t^{2} + \frac{383}{9}t - \frac{72}{5} = \frac{275}{9}t - \frac{5}{2} t^{2}$$ $$ t =\frac{197}{60} \space s$$ Solving for velocity: $$ v_{me}(\frac{197}{60}) = -5(\frac{197}{60})+\frac{383}{9} = \frac{941}{36} = 26.14 \space \frac{m}{s}$$
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