Consider a train which can accelerate with an acceleration of `20cm//s^(2)` and slow down with deceleration of `100cm//s^(2)`. Find the minimum time for the train to travel between the stations 2.7 km apart. (180 s)

Explanation

Let the Acceleration of the train `alpha=20cm//s^(2)`
Decleration of the train `beta=100cm//s^(2)`
Distance between the two stations `s=2.7km=27xx10^(4)cm`
Let the minimum time for the train to travel between the two stations is t sec.
We know that `s=(alphabetat^(2))/(2(alpha+beta))`
`27xx10^(4)=(20xx100xxt^(2))/(2(20+100))`
`27xx10^(4)=(cancel(2)xx10^(cancel(3)^(2))xxt^(2))/(cancel(24_(12))xxcancel(10))`
`27xx12xx10^(4)=t^(2)xx10^(2)`
`t^(2)=(27xx12xxcancel(10^(4))^(10^(2)))/(cancel(10)^(2))`
`t^(2)=324xx10^(2)`
`t = sqrt(324xx10^(2))=180s`
`therefore` Time taken t = 180 s.