Question

A train is moving with a velocity of 400 m/s with the application of brakes, retardation of $10m{s}^{-2}$ is produced. Calculate the following:

1) after how much time will it stop.

2) how much distance will it cover before it stops.

Answer 1

Given:

Initial velocity u = 400 m/s, $a=-10m{s}^{-2}\left(Negativesignsinceitisaretardation\right)$

Final velocity v = 0 m/s

1) Using the first equation of motion:

$v=u+at$

Or,

$\therefore t=\frac{v-u}{a}$

$\therefore t=\frac{0-400}{-10}$

$\therefore t=40s$

Hence the time taken is 40 sec

2) Using the third equation of motion

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

$\therefore s=\frac{v^2-u^2}{2a}$

$\therefore s=\frac{0-{400}^2}{2(-10)}$

$\therefore s=\frac{-400\times 400}{-20}$

$\therefore s=400\times 20$

$\therefore s=8000m$

Hence the distance traveled is 8000 m