Question #201281

A driver of an ambulance going 70km/hr suddenly sees another car 35.0 m ahead. It takes the driver 0.62 s before he applies the brakes. Once he begins to brake, he reduces speed at a rate of 13.0 m/second squares. Determine if the driver will hit the car or not


Expert's answer

a=v−v0t=13a=\frac{v-v_0}{t}=13 m/s2

Time of breaking:

t=v−v0a=70/(3.6⋅13)=1.5 st=\frac{v-v_0}{a}=70/(3.6\cdot13)=1.5\ s

Breaking distance:

d=70⋅0.62/3.6+v0t−at2/2d=70\cdot0.62/3.6+v_0t-at^2/2

d=70⋅0.62/3.6+70⋅1.5/3.6−13⋅1.52/2=26.6 md=70\cdot0.62/3.6+70\cdot1.5/3.6-13\cdot1.5^2/2=26.6\ m

Since distance d < 35 m, the driver will not hit the car.


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