# Chapter 2 - Kinematics in One Dimension - Problems - Page 49: 33

39.2 m

#### Work Step by Step

Distance travelled before car starts decelerating: $x=ut=(20.0)(0.530)=10.6\; m$ Distance travelled while decelerating: $v^{2}=u^{2}+2ax$ $x=\frac{v^{2}-u^{2}}{2a}=\frac{0^{2}-20.0^{2}}{2(-7.00)}\approx28.57\; m$ Stopping distance = 10.6 + 28.57 $\approx$ 39.2 m

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