Question

A car moving with a speed of $50km/hr$, can be stopped by brakes after at least $6m$. If the same car is moving at a speed of $100km/hr$, the minimum stopping distance is

• A. $12m$
• B. $18m$
• C. $24m$
• D. $6m$

Answer 1

Answer: (C)

Given, $u=50km/hr=50\times \frac{5}{18}=\frac{250}{18}m/s$
$v=0$
$s=6m$ (at least)
By third equation by motion,
$v^2=u^2+2as$
$0=\left(\frac{250}{18}\times \frac{250}{18}\right)+2(-a)(s)$
$0=192.90-12a$
$a=\frac{192.90}{12}=16.07$
Given, $u=50km/hr=50\times \frac{5}{18}=\frac{250}{18}m/s$
$v=0$
$s=6m$ (at least)
By third equation by motion,
$v^2=u^2+2as$
$0=\left(\frac{250}{18}\times \frac{250}{18}\right)+2(-a)(s)$
$0=192.90-12a$
$a=\frac{192.90}{12}=16.07$