Question

A car moving with a velocity of $20{ms}^{-1}$ is stopped in a distance of $40m$. If the same car is travelling at double the velocity, the distance travelled by it for same retardation is

• A. $640m$
• B. $320m$
• C. $1280m$
• D. $160m$

Answer 1

The correct option is D $160m$

Initial speed of the car is $u$.

Final speed of the car $v=0$

Let the retardation of the car be $a$.

Using $v^2-u^2=2aS$

Or $0-u^2=2aS$

$⟹$ $S\propto u^2$ (for same $a$)

Given : $S=40m$ $u^{\prime }=2u$

$\therefore$ $\frac{S^{\prime }}{S}=\frac{(u^{\prime }{)}^2}{u^2}=2^2$

Or $\frac{S^{\prime }}{40}=4$

$⟹$ $S^{\prime }=160m$