Question

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

• A. 600 m
• B. 725 m
• C. 625 m
• D. 500 m

Answer 1

The correct option is C

625 m

Here,

$u=90km/h$
$=\frac{90\times 1000}{3600}m/s$
$=25m/s$

$a=–0.5m/s^2$

$v=0$

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

$0={25}^2+(2\times -0.5\times S)$

$s=625m$

Hence, the distance covered is 625m.