Question

A car is moving on a straight horizontal road with a speed $v$. When brakes are applied to give a constant retardation $a$, the car covers distance $S$ before coming to rest. If the car were moving with a speed $3v$ and the same retardation $a$ is applied, the distance that the car travels will be

• A. $3S$
• B. $6S$
• C. $9S$
• D. $27S$

Answer 1

The correct option is C $9S$

Acceleration $=-a$ (i.e retardation)
The shortest stopping distance $S$ is given by
$0-v^2=-2aS$ (From third equation of motion)
or $S=\frac{v^2}{2a}$.
Thus, for a given value of $a,S\propto v^2$.
If $v$ is increased by a factor of $3,$ $S$ will increase by a factor $(3{)}^2=9$.
Hence the correct choice is (c).