Question

A train running at 108km/h is brought to a halt in 2 mins. Calculate the retardation produced by the brakes & the distance it travels before stopping.

Answer 1

Initial speed of train = u = 108 km/h
Final speed of train = v = 0 km/h
Time taken for the change = t = 2 min = 2/60 hr

Retardation produced by breaks =
$a = \frac{v - u}{t} = \frac{0 - 108}{ \frac{2}{60} } = - 5400 \: \frac{km}{ {hr}^{2} }$
Distance travelled before stopping =
$s = ut + \frac{1}{2} a {t}^{2} = 108 \times \frac{2}{60} + \frac{1}{2} \times ( - 5400) \times {( \frac{2}{60} )}^{2} = 3 \: km$

Hope this helps.

Answer 2

Explanation :

Initial velocity of the train, u = 108 km/s = 30 m/s

It comes to rest. So, its final velocity is 0

Time taken, t = 2 min = 120 s

Retardation produced is given by :

$a=\dfrac{v-u}{t}$

$a=\dfrac{0-30\ m/s}{120\ s}$

$a=-0.25\ m/s^2$

The retardation produced by the brakes is $-0.25\ m/s^2$.

Now, using third equation of motion :

$v^2-u^2=2as$

s is the distance travelled

$0^2-(30\ m/s)^2=2(-0.25\ m/s^2)s$

$s=1800\ m$

Hence, this is the required solution.