Question

The coefficient of kinetic friction between a rubber tire and a dry concrete road is 0.700. What is the distance in which a car will skid to a stop on such a road if its brakes are locked when it is moving at 85 kph?

Answer 1

Answer

Initial Velocity of car(u) =23.6m/s

Final velocity (v) =0 m/s

Acceleration (a) =-$\mu g=-0.700\times 9.81=-6.86m/s^2$

So equation of motion

$v^2=u^2-2as\\$

By putting values

$s=\frac{(23.6)^2}{2\times6.86}=40.6m$