A passenger and a good train are headed in the same derection on parallel tracks. The passenger train is `240 m` long and has a constant velocity `72 km h^(-1)` Beginning from the time the engine of the passenger train apprwaches the last wagon of the goods train it takes `25 s` to be in level with the engine of the goods train. It Took `30 s` more to complerly overtake the goods train. Determine the length and speed of the goods train.

Explanation

Velocity of passenger train in `ms^(-1)`,
`v_(P)=72xx(5)/(18) =ms^(-1)`
Let velocity goods train be `(V_(G))` and its length `x` merer, Velocity of passenger train to goods train.
`vec v_(PG) =vec v_(G) =vec v_(P) +(-vec v_(G))`
`|vec v_(PG)|=|vec v_(P)|-|vec v_(G)|=(20-v_(G))`
In `25 s`, the distance travelled by travelled by passenger train train relative to goods reain, `S_(PG)=x`. So,
`(20-V_(G)=20 x` (i)
In next `30 s`, the passener train completely overtakes the good train, which implies `S_(PG)=240 m`
`(20-V_(G))30=240`
On solving Egs. (i) and (ii) we get
`V_(B)=12 ms^(1)` and `x=200 m`,