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Relative Motion

  1. May 11, 2004 #1
    Hi! I have an exercise here that I can't solve. Help please.

    "One of two cylists cycles in the north-west direction at a speed of 8 m/s making an angle of 53 with the west. The other cyclist cycles in north-east direction at a speed of 6m/s making an angle of 37 with the east. How does the second cyclist see the motion of first one?" (magnitude of velocity vector and direction)

    Thanks........ :
     
  2. jcsd
  3. May 11, 2004 #2

    jcsd

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    As usual show us what you've done.

    The obvious starting point is: What is the angle between the two cyclists?
     
  4. May 11, 2004 #3
    It must be 90
     
  5. May 11, 2004 #4

    jcsd

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    Yes, so what will the coponents of the relative velocity of the first cyclist (to the second cyclist)

    a) along his (the second cyclist's) diretcion of travel

    b) 90 degrees to his direction of travel
     
  6. May 11, 2004 #5
    I've found all Vx and Vy 's and Vrelative is 10m/s to North -West. But the answer in book is 10m/s to North-east
     
  7. May 11, 2004 #6

    jcsd

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    It sounds to me is what you've done is forgotten that the component of the first cyclist's velocity (relative to the second cyclist) along the the second cyclist's axis will be opposite and equal to the second cyclist's actual velocity.
     
  8. May 11, 2004 #7

    Doc Al

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    Your answer makes sense to me. Since they are separating along the east-west axis, the relative velocity (of the first with respect to the second) must have a westward component.
     
  9. May 12, 2004 #8
    I use this formula: Vrelative=Vc1-Vc2
     
  10. May 12, 2004 #9

    Doc Al

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    That is correct.
     
  11. May 12, 2004 #10
    Thanks. Now I am sure about my answer.
     
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