Question

A car moving with a velocity of $10\text{m/s}$ can be stopped by the application of a constant force $F$ in a distance of $20\text{m}$. If the velocity of the car is $30\text{m/s}$, it can be stopped by this force in

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

Answer 1

The correct option is D $180\text{m}$

Since, the applied force is constant, the retardation produce by it in order to stop the car will be constant.
We have,
$v^2-u^2=2as\Rightarrow 0^2-u^2=-2as$
$\Rightarrow s\propto u^2$.
$\Rightarrow \frac{s_2}{s_1}=\frac{u_2^2}{u_1^2}\Rightarrow \frac{s_2}{20}=\frac{{30}^2}{{10}^2}$
$\Rightarrow s_2=180\text{m}$
Hence, in second condition the car will stop in a distance of
$s_2=180\text{m}$