### Driver

(a) The distance traveled during the reaction time happens at a constant speed: $t_{reaction}=x/v=(15m)/(20m/s)=0.75s.$
(b) The braking distance is proportional to the speed squared (Note that $100/5=20,400/20=20,900/45=20).$
In this case the distance is $x_{brake}=v^{2}/(20\frac{m}{s^{2}}).$ So $x_{brake}=(25\frac{m}{s})^{2}/(20\frac{m}{s^{2}})=31.25m.$
The reaction time distance is $x_{reac}=(25\frac{m}{s})(0.75s)=18.75m.$ The stopping distance is then 50m.