Question

A train travelling at $90km/hr$ is brought to rest by the application of brakes in a distance of $75m$.find the time in which the train is brought to rest Also calculate the distance covered by the train in the first half and the second half of this time interval.

Answer 1

Using equation of motion for calculating the time in which the train is brought to rest:

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

$\Rightarrow a=-54000km/h^2$

$0^2={90}^2+2a(75\times {10}^{-3})$

$v=u+at$

$0=90+(-54000)t$

$\Rightarrow t=0.00166h=0.00166\times 60min=0.1min=0.1\times 60s=6s$

So, the train comes to rest in 6 s.

Distance traveled in the first half, i.e., distance traveled in 3s:

$v=90+(-54000)\left(\frac{3}{3600}\right)=45km/h$

$(45{)}^2={90}^2+2(-54000)s$

$s=56.25m$

Thus, distance travelled in second half = (75-56.25)m = 18.75 m