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Abhisar

  • one year ago

A man wishes to swim across a river 0.5 Km 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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    @ganeshie8 @nincompoop @dan815

  2. Abhisar
    • one year ago
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    |dw:1437535101502:dw|

  3. anonymous
    • one year ago
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    Your sketch is accurate. You are after <AOB. You have two sides of a right triangle, OA=2 and AB=1. Trigonometry will give you the answer. You OK with the trig ratios?

  4. Abhisar
    • one year ago
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    Yes, I know this way to solve it but I was seeking a different way to do it. Would you mind checking it?

  5. anonymous
    • one year ago
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    Sure. Go ahead

  6. Abhisar
    • one year ago
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    Ok, since the river is 0.5 Km wide and the swimmer can swim with a speed of 2 km/h it will take 0.5/2 = 0.25 hours for him to reach the other side. The swimmer will actually aim along AO but due to the river flow he will reach at B. While reaching B he has travelled a horizontal distance of 1 * 0.25 = 0.25 Km due to river flow. |dw:1437535731457:dw|

  7. Abhisar
    • one year ago
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    Now AB/OB = tan <AOB which gives a different answer...

  8. Abhisar
    • one year ago
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    @ospreytriple

  9. anonymous
    • one year ago
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    He swims at 2 km/hr, but if he is travelling along segment OA, he is not making progress across the river at 2 km/hr, but rather at 2cos<AOB km/hr.

  10. Abhisar
    • one year ago
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    Yes but isn't he still making 2km/h across the vertical direction?

  11. anonymous
    • one year ago
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    Nope. He is travelling at 2 km/hr at 30 degrees CCW to the vertical

  12. anonymous
    • one year ago
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    So he is making progress across the river at 2cos 30 = 1.73 km/hr.

  13. Abhisar
    • one year ago
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    Yes, that makes sense now.

  14. anonymous
    • one year ago
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    Super. Good question.

  15. Abhisar
    • one year ago
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    Thank you.

  16. nincompoop
    • one year ago
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    what do you understand about the question

  17. nincompoop
    • one year ago
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    not the given, but the question it wants you to solve for.

  18. anonymous
    • one year ago
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    You're welcome

  19. Abhisar
    • one year ago
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    I already tried it according to my understanding above, turns out i was taking 2 km/h as vertical velocity of swimmer but it is actually his velocity across line Ao

  20. nincompoop
    • one year ago
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    |dw:1437536977448:dw|

  21. Abhisar
    • one year ago
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    You are doing the same mistake I guess. I thought this way too but actually the swimmer will have 2 km/h velocity along AO and a velocity of 2 cos 30 along the vertical direction.

  22. nincompoop
    • one year ago
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    I am showing you what happens if he just swam. I am not telling you that is the answer

  23. anonymous
    • one year ago
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    He is not swimming vertically. He is swimming along the 30 degree line.

  24. Abhisar
    • one year ago
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    |dw:1437537274293:dw|

  25. Abhisar
    • one year ago
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    It should be like this

  26. anonymous
    • one year ago
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    |dw:1437537285585:dw|

  27. nincompoop
    • one year ago
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    |dw:1437537906031:dw||dw:1437539262021:dw|

  28. nincompoop
    • one year ago
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    I treat them as forces and use the component method. I am not sure if 30 deg is an estimation or the actual angle.

  29. nincompoop
    • one year ago
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    |dw:1437540117642:dw|

  30. dan815
    • one year ago
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    |dw:1437542867378:dw|

  31. dan815
    • one year ago
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    |dw:1437542896575:dw|

  32. Abhisar
    • one year ago
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    Thanks all for the help.

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