A train, travelling at `20 kmh^(-1)` is approaching a platform. A bird is sitting on a pole on the platform. When the train is at a distance of 2km from pole, brakes are applied which produce a uniform deceleration in it. At that instant the bird flies towards the train at `60 kmh^(-1)` and after touching the nearest point on the train flies back to the pole and then flies towards the train continues repeating itself. Calculate how much distance the bird covers before the train stops ?

Answer

12 km

Explanation

Deaceleration of train,
`a=|(v^(2)-u^(2))/(2s)|=(20xx20)/(2xx2)=100 km//hr^(2)`
Time to reach platform `=20/100=1/5` hr
`:.` Total distance travelled by the bird`=vt=60xx1/5=12 km`