Question

A train is travelling at a speed of $90kmph$. 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.

Answer 1

Given:
Initial speed of the train, $u=90kmph=90\times \frac{5}{18}=25m{s}^{-1}$
Final speed of the train, $v=0$ (as the train finally comes to rest)
Acceleration of the train, $a$ $=-0.5m{s}^{-2}$

Solving for the distance covered by the train:
Let $s$ be the distance covered.
According to the third equation of motion, $v^2=u^2+2as$
$\Rightarrow$$(0{)}^2=(25{)}^2+(2\times -0.5\times s)$
$\Rightarrow$$s=\frac{-{25}^2}{2\times -0.5}=625m$