Question

A car is moving along a straight horizontal road with speed $v_0$. If the coefficient of friction between the tyres and the road is $\mu$, the shortest distance in which the car can be stopped is

• A. $\frac{v_0^2}{2\mu g}$
• B. $\frac{v_0}{\mu g}$
• C. $(\frac{v_0}{\mu g}{)}^2$
• D. $\frac{v_0}{\mu }$

Answer 1

The correct option is A $\frac{v_0^2}{2\mu g}$

To stop the car in the shortest distance, we will have to switch off the engine, after which friction will stop the car in shortest distance.

Let the net friction on car is $f=\mu mg$ then the retarding acceleration will be

$a=\frac{f}{m}=\mu g$

Further, let it stop after travelling a distance of $H$ when the engine was switched off.

Using equation of motion,

$v^2-u^2=2as$

Here, $v=0;u=v_0;a=-\mu g;s=H$

$\Rightarrow 0-v_0^2=-2\times \mu g\times H$

$\therefore H=\frac{v_0^2}{2\mu g}$

Therefore, option $\left(a\right)$ is correct.

Alternate method:

Using Work-energy theorem,

$W_{net}=\Delta KE$

$W_f=K{E}_f-K{E}_i$

$-\mu mgH=0-\frac{1}{2}m{v_0}^2$

$\Rightarrow H=\frac{{v_0}^2}{2\mu g}$

Why this question? Concept- If engine is on and break is applied then net opposing force will be less because engine will apply force in forward direction. To maximize the net opposing force, the engine should be switched off.