Question

When the speed of a car is $v$, the minimum distance over which it can be stopped is $s$. If the speed becomes $nv$, what will be the minimum distance over which it can be stopped during same retardation

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

Answer 1

The correct option is D

$n^{2}s$

Step 1:- Given data

Let the speed of the car be $v$
Minimum distance cover$=s$
Final speed $=0$

Step 2:- Calculate the minimum distance,

Now by the third equation of motion,

$v^2=u^2+2as$ or $v^2-u^2=2as$
Maximum retardation, $a=-v^2/2s$
When the initial velocity is $nv$, the acceleration is taken as negative as the velocity is decreasing, then the distance over which it can be stopped is given by
$s_n=\frac{u_0^2}{2\left(-a\right)}=\frac{(nv{)}^2}{2\left(v^2/2s\right)}=n^{2}s$

Thus, the minimum distance over which a car can be stopped during the same retardation is $n^{2}s$.

Hence, option D is the correct answer.