Question

An automobile travelling with a speed of $72km{h}^{-1}$ , can be stopped within a distance of 30 m, by applying brakes. What will be the stopping distance, if the automobile speed is increased to $\sqrt{3}$ times and the same braking force is applied?

• A. 30 m
• B. 90 m
• C. 60 m
• D. 120 m

Answer 1

The correct option is B 90 m

Given Initial Velocity $u=72km{h}^{-1}$= 20$m{s}^{-1}$, Distance $s=30m$, Final Velocity $v=0$,

Let acceleration is $a$.

Using $v^2=u^2-2as$ $\Rightarrow 0={20}^2+2\times a\times 30$ $\Rightarrow a=-\frac{20}{3}$,

Now initial velocity is increases by $\sqrt{3}$ times and the same braking force is applied.

Let distance covered is $s$.

Using $v^2=u^2-2as$ $\Rightarrow 0=(20\sqrt{3}{)}^2+2\times (-\frac{20}{3})\times s$ $\Rightarrow s=90m$

Distance covered is 90 m.