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)`. If `t_(3)` is the time the swimmer would take to swim a distance `2omega` in still water, then

Explanation

Let (u) and (v) be the velocity of a swimmer in still warer and river water respectively. Then ltbRgt ` t_1 = 2 (d/(sqrt (u^2 -v^2))`
` t_2 = d/(u-v) + d/(u-v) = ( 2 ud)/ ( u^2- v^2)`
and ` t_3 = (2 d)/u`
:. T_2 t_3 = ( 2 ud)/(u^2 -v^2) xx ( 1d)/u = ( 4 d^2) /(u^2-v^2) = t_1^1`.