# Solve the following :

Question:

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

Solution:

$u=40 \mathrm{~km} / \mathrm{hr}=\frac{40 \times 1000}{60 \times 60} \mathrm{~m} / \mathrm{s}=\frac{100}{a} \mathrm{~m} / \mathrm{s}$

By law of kinematics,

$v^{2}=u^{2}+2 a s$

$0=\left(\frac{100}{a}\right)^{2}+2 a \cdot 4$

$(v=0)$

$a=15.432 \mathrm{~m} / \mathrm{s}^{2}$

$F=m a=2000 \times 15.432=3.086 \times 10^{4} \mathrm{~N}$