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Mechanics Revision Thread

  1. Dec 21, 2007 #1
    Can you help me with some of these questions? I cant do the second part of Q1!!!
     

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  2. jcsd
  3. Dec 21, 2007 #2

    Hootenanny

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    magicuniverse,

    This is the second thread you've started where you haven't shown any work. Can I remind you that you are required to show the you have attempted any homework before posting as well as detailing any thoughts you have on your problem.

    "I have no idea!" doesn't count as detailing your thoughts.
     
  4. Dec 21, 2007 #3
    Well its not homework for a start. This is revision for an exam so by explaining/hinting at a method you arnt going to be put me at an unfair advantge.
     
  5. Dec 21, 2007 #4

    Hootenanny

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    Okay, so what have you done so far in question (1)?
     
  6. Dec 21, 2007 #5
    I have worked out the time to be 120s with the use of a diagram and orking out resultsnt vectors. Tried doing the second part but got it wrong, my answer was massively out so I though that I must be doing it worng and wanted to know how.
     
  7. Dec 21, 2007 #6

    Hootenanny

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    May i see your attempt?
     
  8. Dec 21, 2007 #7
    Yer but you cant see my diagram as I cant get it on here.

    Basically know that the currrent will drag the bo down, but the boy can swim upwards at 1m/s so if he can swim to the west at 3 m/s his velocity will be root(10) (from pythagorus) we then divide the distance 200m by root10. This comes out with 63.25s rather than the required 89.4s.
     
  9. Dec 21, 2007 #8

    Hootenanny

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    But surely if he swims at a speed of [itex]\sqrt{10}[/itex] m/s, that would mean that he is swimming at an angle of [itex]\tan^{-1}(1/3) \approx 18^o[/itex] above due west and would therefore miss the buoy?

    In addition, the the boy's maximum speed relative to the water is 3 m/s.
     
    Last edited: Dec 21, 2007
  10. Dec 21, 2007 #9
    I figured that the current was pulling him down? Il have alook at it, thanks.
     
  11. Dec 21, 2007 #10

    Hootenanny

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    Note my second point above, the boy's total speed cannot exceed 3 m/s
     
  12. Dec 21, 2007 #11
    Thanks I have to got the answer to 200/root(5), your time is appretiated. Any help with question 2 would be appretiated now thanks.
     
  13. Dec 22, 2007 #12

    Hootenanny

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    Good, and no problem :smile:
    What have you attempted thus far?
     
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