Question

The shortest distance (in metres) in which an automobile moving at a speed of $90kmh^{- 1}$ can be stopped while travelling on a road whose coefficient of kinetic friction is $0.2$ is $P$ metres. Then, the value of $P/10$ is: (Take $g = 10 m / s^{2}$

Answer 1

By Newton's third law, $F=f_{s}=\mu R=\mu mg$
Also, $F=ma$
$\therefore μmg=ma$
$\therefore a=μg$
$\therefore a=0.2g$
we have,
$v^{2}=u^{2}-2as$
$\therefore 0^{2}=\left(\right.25\left(\left.\right)^{2}-2\times 0.2\times 10\times s$
$\therefore 0=625-4s$
$\therefore 4s=625$
$\therefore s=156.25m$