Question

If a car travelling at $40\text{m/s}$ is to be brought to rest in a distance of $100\text{m}$, then its retardation should be

• A. $6{\text{m/s}}^2$
• B. $8{\text{m/s}}^2$
• C. $10{\text{m/s}}^2$
• D. $9{\text{m/s}}^2$

Answer 1

The correct option is B $8{\text{m/s}}^2$

As per the question, final velocity of the car will be $v=0$
Given, initial velocity $u=40\text{m/s}$, distance $s=100\text{m}$
As per $3^{rd}$ equation we have
$v^2-u^2=2as$
$\Rightarrow 0-{40}^2=2a\times 100\Rightarrow a=\frac{-1600}{200}=-8{\text{m/s}}^2$
Thus, the retardation of the car is $8{\text{m/s}}^2$