Question

A 900-kg car is going 20 m/s along a level road. How large a constant retarding (friction)

force is required to stop it at a distance of 30m? (hint: find the negative acceleration).

Answer 1

$s=\frac{v^2-v_0^2}{2a}\to a=\frac{v^2-v_0^2}{2s}=\frac{0^2-20^2}{2\cdot30}=-6.67\ (m/s^2)$

$F=ma\to F=900\cdot(-6.67)=-6000\ (N)$ . Answer