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Abhisar

  • one year ago

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?

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  1. Abhisar
    • one year ago
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    @Michele_Laino

  2. Abhisar
    • one year ago
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    @JoannaBlackwelder

  3. Vincent-Lyon.Fr
    • one year ago
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    Hint: Draw a picture in order to add up the velocities.

  4. Michele_Laino
    • one year ago
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    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|

  5. Michele_Laino
    • one year ago
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    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|

  6. Michele_Laino
    • one year ago
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    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

  7. Michele_Laino
    • one year ago
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    so we have: \[t = \frac{d}{{\sqrt {{v^2} - {u^2}} }} = \frac{{0.5}}{{\sqrt {5 - 1} }} = 0.25hours\]

  8. Michele_Laino
    • one year ago
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    the space traveled by the water is therefore: \[s = ut = 1 \times 0.25 = 0.25Km\]

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

  10. Michele_Laino
    • one year ago
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    |dw:1437511991612:dw|

  11. Michele_Laino
    • one year ago
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    so we have: \[0.25 = 0.5\tan \theta \] then: \[\tan \theta = \frac{1}{2}\]

  12. radar
    • one year ago
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    At only .5 meter wide, why not just jump across it. lol

  13. radar
    • one year ago
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    @Abhisar Is it copied correctly or should the width be .5 km ?

  14. Abhisar
    • one year ago
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    @radar , lol sorry it's 0.5 Km

  15. Abhisar
    • one year ago
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    @michele_laino , I tried it the same way but the answer given is 120°

  16. IrishBoy123
    • one year ago
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    |dw:1437545407226:dw|

  17. radar
    • one year ago
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    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?

  18. radar
    • one year ago
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    @Abhisar Good luck with your studies.

  19. IrishBoy123
    • one year ago
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    yes @Radar, width irrelevant. [unless it is 0.5m wide as per OP's question; because, as you say, he should just jump :-)) ]

  20. Abhisar
    • one year ago
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    Thank you all :)

  21. Vincent-Lyon.Fr
    • one year ago
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    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.

  22. radar
    • one year ago
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    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.

  23. Abhisar
    • one year ago
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    WoW ! Thank you for all the work, all these discussions gave me lots of insights. thank you all :)

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