# Chapter 2 - Kinematics in One Dimension - Exercises and Problems - Page 66: 59

Bob is in jail.

#### Work Step by Step

We can find the distance the car moves in 0.60 seconds. $d = v~t = (50~m/s)(0.60~s)$ $d = 30~m$ We can find the distance $x$ the car travels during the braking period. $x = \frac{v^2-v_0^2}{2a}$ $x = \frac{0-(50~m/s)^2}{(2)(-10~m/s^2)}$ $x = 125~m$ The car travels a total distance of 125 m + 30 m which is 155 meters. Since the car needed to stop within 150 meters, Bob is in jail.

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