Two swimmers start at the same time from point A one bank of a river to reach point B on the other bank, lying directly oppostie to point A. one of them crosses the river along the straight line AB, while the other swims at right angles to the stream and then walks the distance, which he has been carried awayby the stream to get to point B. Both swimmers reach point B at the same time. what was the velocity (assumed uniform) of his walking if velocity of both the swimmers in still water is 2.5 km `h^(-1)` and the stream velocity is 2 km `h^(-1)`?

Explanation

One swimmer moving along AB must swim at such an angle that his velocity along the stream is nu llified by stream velocity. If v is the velocity of the swimmer and u is the velocity of the stream, then u+`sin theta=0`. His velocity along the direction AB=`v cos theta`. Time taken by the swimmer to reac `B=(1)/(v cos theta)` where 1 is the width of the river.
Second swimmer swims in a direction perpendicu lar to the stream.
Time taken to reach the oppostie bank `-=(1)/(v)`.
Duirng this time the swimmer drifts by the stream and he reaches the point C.
Distance CB=velocitty xx time `=("ul")/(v)`.
If the swimmer walks this distance with uniform velocity u', taken `=("ul")/(vu')`
`therefore` Total time taken by second swimmer `=(1)/(v)+("ul")/(vu)`
Since both the swimmers reach point B simu ltaneously, we have
`(l)/(v cos theta)=(l)/(v)=("ul")/(vu') therefore (1)/(cos theta)1+(u)/(u')`
We have `u=2km h^(-1),v=2.5kmh^(-1)`
`cos theta=(v_(y))/(v)=(sqrt(v^(2)-u^(2)))/(v)=(sqrt(2.5^(2)-2^(2)))/(2.5)=(1.5)/(2.5)`
`therefore (2.5)/(1.5)=1+(2)/(u')rArru'=3kmh^(-1)`
i.e., velocity of walking of the second swimmer =`3kmh^(-1)` towards B.