Answer to Question #130936 in Physics for Inioluwa

Question

A motor car of mass 800kg travelling at 20ms(on top of it there is -1)is brought to rest by brakes in 100m.
Calculate the average braking force required

Answer 1

Let us write equations of motion of the car, assuming it moves with constant retardation (and hence constant braking force):

$S(t) = v_0 t - \frac{a t^2}{2}$, $v(t) = v_0 - a t$.

At the moment of time, when the car stops completely $v(t') = v_0 - a t' = 0$, from where the time it takes to completely stop is $t' = \frac{v_0}{a}$.

The horizontal distance covered in time $t'$ is given ($L = 800m)$, therefore: $L = S(t') = \frac{v_0^2}{a} - \frac{v_0^2}{2 a} = \frac{v_0^2}{2 a}$, from where the acceleration is $a = \frac{v_0^2}{2 L}$, and from second Newton's law $a = \frac{F}{m}$, so $\frac{v_0^2}{2 L} = \frac{F}{m}$, from where the force is $F = \frac{m v_0^2}{2 L} = 1600 N$.