Question

If a train traveling $72kmph$ is to be brought to rest at a distance of $200metres,$ then its retardation should be

• A. $20m{s}^{-2}$
• B. $10m{s}^{-2}$
• C. $2m{s}^{-2}$
• D. $-1m{s}^{-2}$

Answer 1

The correct option is D

$-1m{s}^{-2}$

Step 1: Given data:

The train traveling speed $u=72kmph$

The distance,$s=200m$

Final velocity, $v=0$

Step 2: Formula used:

The equation of motion,

$v^2=u^2+2as$

Now substitute the values in the motion law,

$$\begin{gathered} 0={\left(72\times \left(\frac{5}{18}\right)\right)}^2+2a\left(200\right) \\ 0={\left(20\right)}^2+2a\left(200\right) \\ a=-1m/s^2 \end{gathered}$$

Therefore retardation should be$-1m{s}^{-2}$.

Hence, the correct option is (D).