The brakes of a train which is travelling at `30 m//s` are applied as the train passes point A. The brakes produce a constant retardation of magnitude `3lambda m//s^(2)` until the speed of the train is reduced to `10 m//s`. The train travels at this speed for a distance and is then uniformly accelerated at `lambda m//s^(2)` until it again reaches a speed of `30 m//s` as it passes point B. the time taken by the train in travelling from A to B, a distance of 4 km, is 4 min. Sketch the speed time graph for this motion and calculate:-
(i) The value of `lambda` (ii) Distance travelled at `10 m//s`.

Answer

(i)` lambda=1/6`, (ii) `800 m`

Explanation

Area `=1/2 [10+30]xx20/(3lambda)+10xx(240-80/(3lambda))+1/2 [10+30]xx20/lambda=4000`

`400/(3lambda)+2400-800/(3lambda)+400/lambda=4000`
`(400-800+1200)/(3lambda)=1600`
`3lambda=800/1600=1/2 , lambda=1/6`
(ii) Dist. travelled`=10(240-80/(3xx1//6))=800 m`