Question

An automobile travelling at $50km{h}^{-1}$, can be stopped at a distance of 40 m by applying brakes. If the same automobile is travelling at 90 $km{h}^{-1}$, all other conditions remaining same and assuming no skidding, the minimum stopping distance in metre is

• A. 72
• B. 92.5
• C. 102.6
• D. 129.6

Answer 1

Answer: (D)

By equation of motion $v^2-u^2$ = 2as where u is initial velocity, a is acceleration and s is displacement.
Given $u=50km{h}^{-1},v=0,=40m$
hence $a=\frac{u^2}{2s}=\frac{{\left(50\times \frac{5}{18}\right)}^2}{2\times 40}$
when u' = $90km{h}^{-1}$
so,s=$\frac{u^{\prime 2}}{2a}$
$\Rightarrow$ $\frac{{\left(90\times \frac{5}{18}\right)}^2\times 2\times 40}{2\times {\left(50\times \frac{5}{18}\right)}^2}$ or $s=129.6m$