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

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

Motion in a Straight Line

Solution:

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