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

Explanation

Here, `u=60 xx (5)/(18)=(50)/(3)n`
`v=15 xx (5)/(18)=(25)/(6)m//s`
Using `v^(2)=u^(2)+2as`, we get
`((50)/(3))^(2)=((25)/(6))^(2)+2 xx a xx 450 or a=-(125)/(36 xx12)m//s^(2)`
If `s^(1)` is the further distance travelled before coming to rest, then `s^(1)=(v^(2))/(2a)=(25)/(6) xx (25 xx 36 xx 12)/(6 xx 2 xx 125)=30m`