Question

A car moving at 40 km/h is to be stopped by applying brakes in the next 4.0 m. If the car weighs 2000 kg, what average force must be applied on it ?

Answer 1

Given, u = 40 km/hr

$=\frac{40000}{3600}=11.11$ m/s

m = 2000 kg, v = 0, S = 4 m

Accelearation, a = $\frac{v^2-u^2}{2S}$

= $\frac{0^2-(11.11{)}^2}{2\times 4}=\frac{123.43}{8}$

= $-15.42m/s^2$ (deceleration)

$\therefore$ Braking force , F = ma

=$2000\times 15.42$

=30840 = 3.08 $\times {10}^{4}N$