## Physics (10th Edition)

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