A-swimmer-crosses-a-flowing-river-of-width-d-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-dis

Question Number 14920 by Tinkutara last updated on 05/Jun/17

A swimmer crosses a flowing river of

width d 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 2 d in still water, then prove

that t_{1}^{2} = t_{2} t_{3} .

Answered by ajfour last updated on 05/Jun/17

Commented by ajfour last updated on 05/Jun/17

t o g o t o t h e o t h e r b a n k a n d r e t u r n

b a c k t o s a m e p o i n t (t o a n d f r o)

s w i m m e r m u s t s w i m a t a n a n g l e

θ (s e e f i g .) s u c h t h a t

v \sin θ = u \dots (i)

t_{1} = \frac{2 d}{v \cos θ} \dots . (i i)

t o g o u p s t r e a m a d i s t a n c e d a n d

r e t u r n b a c k t i m e t a k e n i s

t_{2} = \frac{d}{v - u} + \frac{d}{v + u} \dots . (i i i)

t o s w i m a d i s t a n c e 2 d i n s t i l l

w a t e r t i m e t a k e n i s

t_{3} = \frac{2 d}{v} .

t_{2} t_{3} = \frac{2 d^{2}}{v} (\frac{2 v}{v^{2} - u^{2}}) = \frac{4 d^{2}}{v^{2} - v^{2} \sin^{2} θ}

= {(\frac{2 d}{v \cos θ})}^{2} = t_{1}^{2}

[s e e (i i), (i i i), a n d (i)] .

Commented by Tinkutara last updated on 05/Jun/17

Thanks Sir!