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 equations, (i), (ii), and (iii)
`t_(1)^(2) =t_(2)t_(3)