Question

If a train travelling at $72\text{kmph}$ is to be brought to rest in a distance of $200\text{metres}$, 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

Speed of the train = 72 km/h = 20 m/s

Distance traveled = 200 m

Final velocity = 0

Step 2, Finding retardation

According to the question

$u=72\text{kmph}=20\text{m/s},v=0$

Let retardation of the train be $a$

By using third law of equation we have
$v^2=u^2-2as$

Or,
$\Rightarrow a=\frac{u^2}{2s}=\frac{(20{)}^2}{2\times 200}=1m/s^2$

Hence the retardation is 1 m/$s^2$