## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 2 - Kinematics in One Dimension - Exercises and Problems: 83

#### 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 find the rate of the deceleration. $a = \frac{v-v_0}{t} = \frac{20~km/s-50~km/s}{6.67~s}$ $a = -4.5~m/s^2$ The magnitude of acceleration is $4.5~m/s^2$.

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