Question

An express train going towards Delhi is travelling at a speed of $90\text{kmph}$. Brakes are applied so as to produce a uniform acceleration of $–0.5{\text{m/s}}^2$. Find how far the train will go before it is brought to rest.

• A. $725\text{m}$
• B. $600\text{m}$
• C. $625\text{m}$
• D. $500\text{m}$

Answer 1

The correct option is C $625\text{m}$

Given:
Initial velocity, $u=90\text{kmph}$

$=\frac{90\times 1000}{3600}\text{m/s}$

$=25\text{m/s}$

Acceleration, $a=–0.5{\text{m/s}}^2$
Final velocity, $v=0$

Using $v^2=u^2+2as$,
$0={25}^2+(2\times -0.5\times s)$
$\Rightarrow s=625\text{m}$
Hence, the distance covered is $625\text{m}$ before the train comes to rest.