Question

For a car moving on a straight line with initial speed $u$, the minimum distance over which it can be stopped is $s$. If it's initial speed becomes $nu$, what will be the minimum distance over which it can be stopped? Assume constant braking force.

• A. $\frac{s}{n}$
• B. $\frac{s}{n^2}$
• C. $ns$
• D. $n^{2}s$

Answer 1

The correct option is D $n^{2}s$

It's evident that car is in state of deceleration, and for constant braking force, acceleration $\left(a\right)=$ constant
Since,
$v^2=u^2+2as.....i$
When car stops, final velocity $v=0$
$\Rightarrow 0=u^2+2as$
$a=\frac{-u^2}{2s}....ii$
For the second case,
Initial velocity $=nu$
Final velocity $v=0$
From Eq. $i$:
$0=(nu{)}^2+2a{s}^{\prime }$
Substituting $a$ from Eq. $ii$
$\Rightarrow n^2{u}^2=-2\times \left(\frac{-u^2}{2s}\right)\times s^{\prime }$
$\therefore s^{\prime }=n^{2}s$
Hence $s^{\prime }$ is the required minimum distance over which car will stop in second case.