IIT JEE , AIEEE Forum - Physics , Mechanics

shub

two swimmers leave point A on one bank of the river to reach point B lying right across on the other bank . one of them crosses the river along the straight line A B while the other swims at right angles to the stream and then walks the distance that he has been carried away by the stream to get to point B . what was the velocity u of his walking if both swimmers reached the destination simultaneously ? the stream velocity v= 2.0km\hour and the velocity v' of each swimmer with respect to water equals 2.5 kmper hour .

SuperCool

Is answer 3 Km/Hr?

budokai_tenkaichi_returns

nice question ...

achcha mix hai 2 questions ka...i like it

shub

hey supercool...ur answer is right...
please post the complete soln.

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

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

budokai_tenkaichi_returns

will u all give a minute here

and yeah

.http://www.goiit.com/posts/list/counselling-seriously-i-aint-studying-since-few-days-60243.htm..

shub

thanks a lot supercool....

SuperCool

No Problem,But Answering here sucks big time

Ask & Discuss Questions with Community & Experts