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

Explanation

Here `u=60xx(5)/(18)=(50)/(3)n`
`v=15xx(5)/(18)=(25)/(6)m//s`
Using `v^(2)=u^(2)+2as,`we get
`((50)/(3))^(2)=((25)/(6))^(2)+2xxaxx450` 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)=30m`