IIT JEE , AIEEE Forum - Physics , Mechanics

SuperCool

Can't draw figures here

Let one swimmer cross the river along the shortest path AB

Time taken = t1

${t}_{1}=\frac{d}{\sqrt{{v'}^{2}-{{v}_{0}}^{2}}}$

AB=d i.e. River Width

Second swimmer follows quickest path

${t}_{2}=\frac{d}{{v}^{'}}$

In time t2 drifting becomes (for 2nd swimmer)

$x={v}_{0}{t}_{2}=\frac{{v}_{0}d}{v'}$

Let t3 be the time for swimmer 2 to walk the distance x to come from C to B then

${t}_{3}=\frac{x}{u}=\frac{{v}_{0}d}{{v'u}}$

ATQ t1=t2+t3

I'm avoiding few simple steps

Finally you will get

$u=\frac{{v}_{0}}{(\frac{1-{v}_{0}^{2}}{v'^{2}})-1}=3Km/hr\\LaTeX Sucks$