# Chapter 2 - Kinematics in One Dimension - Problems: 9

52 m

#### Work Step by Step

Let t = time elapsed since the bear was 26m behind $distance_{bear}$ = average speed $\times$ time = 6.0t d =$distance_{tourist}$ = average speed $\times$ time = 4.0t 6.0t $\leq$ 4.0t + 26 2t $\leq$ 26 t $\leq$ 13 $d_{min}$ = 4.0$t_{min}$ = 4.0 (13) = 52 m

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