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 37: 75a

Answer

The reaction time is $~~0.74~s$

Work Step by Step

We can convert $80.5~km/h$ to units of $m/s$: $(80.5~km/h)\times (\frac{1000~m}{1~km})\times (\frac{1~h}{3600~s}) = 22.36~m/s$ Let $T$ be the reaction time. The distance traveled before braking is $x_1 = 22.36~T$ Let $a$ be the deceleration during braking. We can find the distance $x_2$ the car travels during braking: $v^2 = v_0^2+2ax_2$ $0 = v_0^2+2ax_2$ $x_2 = -\frac{v_0^2}{2a}$ $x_2 = -\frac{(22.36)^2}{2a}$ $x_2 = -\frac{250}{a}$ We can write an equation for the total distance during the two phases: $x_1+x_2 = 56.7$ $22.36~T-\frac{250}{a} = 56.7$ We can convert $48.3~km/h$ to units of $m/s$: $(48.3~km/h)\times (\frac{1000~m}{1~km})\times (\frac{1~h}{3600~s}) = 13.42~m/s$ Let $T$ be the reaction time. The distance traveled before braking is $x_1 = 13.42~T$ Let $a$ be the deceleration during braking. We can find the distance $x_2$ the car travels during braking: $v^2 = v_0^2+2ax_2$ $0 = v_0^2+2ax_2$ $x_2 = -\frac{v_0^2}{2a}$ $x_2 = -\frac{(13.42)^2}{2a}$ $x_2 = -\frac{90}{a}$ We can write an equation for the total distance during the two phases: $x_1+x_2 = 56.7$ $13.42~T-\frac{90}{a} = 24.4$ $(-\frac{250}{90})\times(13.42~T-\frac{90}{a}) = (-\frac{250}{90})(24.4)$ $-37.28~T+\frac{250}{a} = -67.78$ We can add the two equations for the total distance in each case: $-14.92~T = -11.08$ $T = \frac{11.08}{14.92}$ $T = 0.74~s$ The reaction time is $~~0.74~s$
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