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

In order to reach the point C, the swimmer should start in such a way that the vector sum of ` vecu and vecv` is along the line AC. For this, let the swimmer start at an angle `alpha` with line AB.
Velocity of swimmer w.r.t ground along x-axis, ` v_(x) = u+v sin alpha`
Velcoity of swimmer w.r.t ground along y -axis , ` v_(y) = v cos alpha`
Time taken to cross the river , ` t = (AB)/v_(y) = d/( v cos alpha)`
Displacement along x -axis.
` BC = v_(x)t`
` Rightarrow d tan theta = ( u + v sin alpha) t `
Form (i) , ` (v cos alpha) =d`
From (ii) , ` ( v sin alpha) t =d tan theta - ut`
` Rightarrow (vt cos alpha)^(2) + (vt sin alpha)^(2) = d^(2) + ( d tan theta - ut)^(2)`
` Rightarrow V^(2)t^(2) = d^(2) + d^(2) tan^(2) theta + u^(2) t^(2) - 2 udt tan theta`
` Rightarrow (v^(2) - u^(2)) t^(2) + ( 2 udt. tan theta) - d^(2) ( 1+ tan ^(2) theta) = 0`
On sloving for roots,
` t = ( dsqrt(v^(2) sec^(2) theta -u^(2))- ud tan theta)/((v^(2) -u^(2))) `