A man wishes to swim across a river 0.5 m wide. If he can swim at the rate of 2 Km/ h in still water and the river flows at the rate of 1 km/hr. The angle (wrt flow of water) along which he should swim so as to reach a point exactly opposite his starting point, should be?

- Abhisar

- jamiebookeater

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- Abhisar

- Abhisar

- Vincent-Lyon.Fr

Hint: Draw a picture in order to add up the velocities.

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## More answers

- Michele_Laino

the velocity of the swimmer with respect to the water is 2 Km/hour, whereas the velocity of the current is 1 Km/hour, so the situation of your problem is like in my drawing:
|dw:1437510950238:dw|

- Michele_Laino

Now, I stay on the bank of the river and I use my stopwatch to measure the time needed to the swimmer to cross the river. within that time the swimmer travels the distance AB:
|dw:1437511307871:dw|

- Michele_Laino

and the water has traveled the distance CB, so we can write:
\[\begin{gathered}
A{C^2} + C{B^2} = A{B^2} \hfill \\
\hfill \\
{d^2} + {u^2}{t^2} = {v^2}{t^2} \hfill \\
\end{gathered} \]
where t is the time measured by my stopwatch, u is the speed of the water, and v is the speed of the swimmer with respect to the water

- Michele_Laino

so we have:
\[t = \frac{d}{{\sqrt {{v^2} - {u^2}} }} = \frac{{0.5}}{{\sqrt {5 - 1} }} = 0.25hours\]

- Michele_Laino

the space traveled by the water is therefore:
\[s = ut = 1 \times 0.25 = 0.25Km\]

- Michele_Laino

oops.. we have:
\[\begin{gathered}
t = \frac{d}{{\sqrt {{v^2} - {u^2}} }} = \frac{{0.5 \times {{10}^{ - 3}}}}{{\sqrt {5 - 1} }} = 2.5 \times {10^{ - 4}}hours \hfill \\
\hfill \\
s = ut = 1 \times 2.5 \times {10^{ - 4}} = 2.5 \times {10^{ - 4}}Km = 0.25meters \hfill \\
\end{gathered} \]

- Michele_Laino

|dw:1437511991612:dw|

- Michele_Laino

so we have:
\[0.25 = 0.5\tan \theta \]
then:
\[\tan \theta = \frac{1}{2}\]

- radar

At only .5 meter wide, why not just jump across it. lol

- radar

@Abhisar Is it copied correctly or should the width be .5 km ?

- Abhisar

@radar , lol sorry it's 0.5 Km

- Abhisar

@michele_laino , I tried it the same way but the answer given is 120°

- IrishBoy123

|dw:1437545407226:dw|

- radar

I was on a different method, determining time, then distance and still got the same answer (120 degrees) same as IrishBoy123.|dw:1437569372484:dw| So in effect the width of the river was irrelevant. Or was it?

- radar

@Abhisar Good luck with your studies.

- IrishBoy123

yes @Radar, width irrelevant. [unless it is 0.5m wide as per OP's question; because, as you say, he should just jump :-)) ]

- Abhisar

Thank you all :)

- Vincent-Lyon.Fr

Radar's sketch is ok, but can be simplified if you use velocities instead of displacements.
Since \(\sin \theta =\dfrac{v_{river}}{v_{swimmer}}=\dfrac 1 2\), then \(\theta=30°\)
Breadth of river is only needed if you have to work out the time it mill take to cross.
I think the river is 0.5 km wide, (not 0.5 m; a mere jump would do the trick)
The velocity wrt ground is \(2\cos 30°=2\times 0.87=1.74 \) km/h
So it will take him \(\dfrac {0.5}{1.74}=0.29\) h = 17,5 min to get across.

- radar

Yes, I've reached that conclusion now, but at first I thought the time it took to cross would have to be determined, but now I know that the velocities were the determining criteria, Thanks to all who worked this.

- Abhisar

WoW !
Thank you for all the work, all these discussions gave me lots of insights. thank you all :)

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