A bomber plane moves due east at `100 km h^-1` over a town `T` at a certain instant of time. Six minutes later, an interceptor plane sets off flying due north - east from the station `S` which is `40 km` south of `T`. If both maintain their courses, find the velocity with which the interceptor plane must fly in order to just overtake the bomber.

Answer

`(400 sqrt(2))/(3) km h^-1`

Explanation

Both should reach simultaneously at `A`.
`TS = 40 km rArr TA = 40 km`
Time taken by bomber to reach `A` :
`t = (40)/(100) = (2)/(5) h = 24 min`
So time taken by interceptor plane to reach `A` :
`24 - 6 = 18 min, v = (40 sqrt(2))/((80//60)) = (400sqrt(2))/(3) km h^-1`.
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