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=60xx5/18=50/3` m/s
`v=15xx5/18=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/(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)`=30 m