A swimmer crosses a flowing stream of width `omega` to and fro in time `t_(1)`. The time taken to cover the same distance up and down the stream is `t_(2)`. It `t_(3)` is the time the swimmer would take to swim a distance `2omega` in still water, then

Explanation

Let v be the river velocity and u the velocity of swimmer in still water. Then
`t_(1)=2((d)/(sqrt(u^(2)-v^(2))))…….(i)`
`t_(2)=(d)/(u+v)+(d)/(u-v)=(2ud)/(u^(2)-v^(2))…..(ii)`
and `t_(3)=(2d)/(u)…….(iii)`
from equation (i), (ii) and (iii)
`t_(1)^(2)=t_(2)t_(3)=sqrt(t_(2)t_(3))`