Abhisar
  • Abhisar
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?
Physics
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SOLVED
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jamiebookeater
  • jamiebookeater
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Abhisar
  • Abhisar
@Michele_Laino
Abhisar
  • Abhisar
@JoannaBlackwelder
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Hint: Draw a picture in order to add up the velocities.

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Michele_Laino
  • 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
  • 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
  • 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
  • Michele_Laino
so we have: \[t = \frac{d}{{\sqrt {{v^2} - {u^2}} }} = \frac{{0.5}}{{\sqrt {5 - 1} }} = 0.25hours\]
Michele_Laino
  • Michele_Laino
the space traveled by the water is therefore: \[s = ut = 1 \times 0.25 = 0.25Km\]
Michele_Laino
  • 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
  • Michele_Laino
|dw:1437511991612:dw|
Michele_Laino
  • Michele_Laino
so we have: \[0.25 = 0.5\tan \theta \] then: \[\tan \theta = \frac{1}{2}\]
radar
  • radar
At only .5 meter wide, why not just jump across it. lol
radar
  • radar
@Abhisar Is it copied correctly or should the width be .5 km ?
Abhisar
  • Abhisar
@radar , lol sorry it's 0.5 Km
Abhisar
  • Abhisar
@michele_laino , I tried it the same way but the answer given is 120°
IrishBoy123
  • IrishBoy123
|dw:1437545407226:dw|
radar
  • 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
  • radar
@Abhisar Good luck with your studies.
IrishBoy123
  • 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
  • Abhisar
Thank you all :)
Vincent-Lyon.Fr
  • 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
  • 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
  • Abhisar
WoW ! Thank you for all the work, all these discussions gave me lots of insights. thank you all :)

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