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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. imqwerty
    • one year ago
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    really 0.5m wide river!!!

  2. ChillOut
    • one year ago
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    All right, let's draw that first.

  3. Abhisar
    • one year ago
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    Hahaha..nopes it should be 0.5 Km

  4. imqwerty
    • one year ago
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    lol ok :)

  5. ChillOut
    • one year ago
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    |dw:1437497187136:dw|

  6. ChillOut
    • one year ago
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    |dw:1437497254385:dw|

  7. imqwerty
    • one year ago
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    60 degrees

  8. Abhisar
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @imqwerty 30 degrees :) \(\color{blue}{\text{End of Quote}}\) No direct answers and btw it's incorrect ...

  9. ChillOut
    • one year ago
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    The net horizontal velocity must be equal 0, so \(v*cos( θ )-1=0\)

  10. ChillOut
    • one year ago
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    Can you finish it, @imqwerty? :)

  11. imqwerty
    • one year ago
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    let the man make angle theta with the direction opposite to the river flow his horizontal component of velocity = Velocity of man x cos(theta) = Vcos(theta)=2cos(theta) this should equal the velocity of river flow so as to reach the directly opposite point so 2cos(theta)=1 cos(theta)=1/2 theta=60 degrees

  12. Abhisar
    • one year ago
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    It's incorrect...

  13. Abhisar
    • one year ago
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    |dw:1437497568545:dw|

  14. Abhisar
    • one year ago
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    The man should swim across the line AO and you have to find angle AOD

  15. Abhisar
    • one year ago
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    Ring a bell?

  16. ChillOut
    • one year ago
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    Oh, right. I didn't read the part where it states it's the angle between the flow and the man xD

  17. Abhisar
    • one year ago
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    Yes, so ?

  18. ChillOut
    • one year ago
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    So it means that the angle must be in the second quadrant, 120 degrees.

  19. Abhisar
    • one year ago
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    That's correct!

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