Question

A car moving with a velocity of $10m/s$ can be stopped by the application of a constant force F in a distance of $20m$. If the velocity of the car is $30m/s$. It can be stopped by this force in:-

• A. $\frac{20}{3}m$
• B. $20m$
• C. $60m$
• D. $180m$

Answer 1

The correct option is D $180m$

Final velocity is zero.and acceleration is $-ve$
$0=u^2-2\times a\times ss=\frac{u^2}{2a}s\propto u^2$
if car moving with a velocity of $10m/s$ can be stopped by the application of a constant force F in a distance of $20m$, since the velocity is tripled stopping distance will become $3^2=9$times.

New stopping distance $=9\ast 20=180m$