The speed of a train is reduced from `60km//h`, to `15km//h`, while it travels a distance of `450m`. If the retardation is uniform, find how much further it will travel before coming to rest?

Explanation

Here `u=60xx(5)/(18)=(50)/(3)m//s`
`v=15xx(5)/(18)=(25)/(6)m//s`
Using `v^(2)=u^(2)+2as`, we get
`((50)/(3))^(2)=((25)/(6))^(2)+2xxaxx450or`
`a=-(125)/(36xx12)m//s^(2)`
If `s^(1)` is the further distance travelled before coming to rest, then `s^(1)=(v^(2))/(2a)=(25)/(6)xx(25xx36xx12)/(6xx2xx125)=30m`