A velocity of 60 km.`h^(-1)` of a train is reduced by the application of brakes. A retardation of 40 cm .`s^(-1)` is produced . After how much time will the train stop? What will be the velocity of the train after 20 s?

Explanation

Given, u= 60 km.`"h"^(-1)= (60 xx1000)/(60xx60) = (50)/(3) "m.s"^(-1)`
a = 40 `"cm.s"^(-2) = 0.4 "m.s"^(-2)`
and v = 0
Hence, from the relation v = u-at, we get
`0 = (50)/(3) - 0.4xxt " ""or",t = (50)/(3xx0.4) = 41.7s`
`:.` The train will stop after 41.7 s.
The velocity after 20s,
`v = (50)/(3) - 0.4 xx20`
`=(26)/(3) "m.s"^(-1) = 8.7"m.s"^(-1)`
`(8.7xx60xx60)/(1000) "km.h"^(-1)`
= 31.3 `"km.h"^(-1)`.