Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 2 - Kinematics in One Dimension - Exercises and Problems: 85

Answer

The magnitude of acceleration is $4.5~m/s^2$

Work Step by Step

To barely avoid a collision, the Enterprise needs to decelerate to a speed of 20 km/s while gaining 100 km on the other ship. The average speed during the deceleration period is 35 km/s, so the relative speed between the two ships is 15 km/s on average. We can find the time $t$ it takes to gain 100 km on the other ship; $t = \frac{100~km}{15~km/s} = 6.67~s$ We can then find the rate of the deceleration: $a = \frac{v-v_0}{t} = \frac{20~km/s-50~km/s}{6.67~s}$ $a = -4.5~km/s^2$ The magnitude of acceleration is thus $4.5~km/s^2=4500 m/s^2$.
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