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Particle Position

  1. Sep 15, 2009 #1
    1. The problem statement, all variables and given/known data
    A watermelon seed has the following coordinates: x = -8.1 m, y = 8.8 m, and z = 0 m. Find its position vector as (a) a magnitude and (b) an angle relative to the positive direction of the x axis. If the seed is moved to the xyz coordinates (3.3 m, 0 m, 0 m), what is its displacement as (c) a magnitude and (d) an angle relative to the positive direction of the x axis?


    2. Relevant equations



    3. The attempt at a solution
    a= 11.9 Meters
    b= 133 degrees
    c= 15.2 Meters
    d = ????

    How do I find this final angle??
     
  2. jcsd
  3. Sep 15, 2009 #2

    kuruman

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    Find the x and y components of the displacement vector and get the angle the usual way.
     
  4. Sep 15, 2009 #3
    All I have is the magnitude of the displacement vector. I know I need to find the x and y components of it, but how do I do that without at least having one of the angles? Or do i already have it??
     
  5. Sep 15, 2009 #4

    kuruman

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    You have the coordinates of two points. The displacement is a vector whose components is the difference finish - start, (x2-x1, y2-y1, z2-z1)
     
  6. Sep 16, 2009 #5
    so my displacement vector's components are (11.4, -8.8, 0)? So the angle would be inverse tan of (-8.8/11.4)?
     
  7. Sep 16, 2009 #6

    kuruman

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    Correct.
     
  8. Sep 16, 2009 #7
    the angle I get from that calculation is -37.67..Now it says it wants an angle relative to the positive direction of the x-axis. Would that be it, or do I have to subtract that number from something else?
     
  9. Sep 16, 2009 #8

    kuruman

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    Draw the point on a sheet of paper. Connect it with a straight line to the origin. Then look at the angle that starts on the positive x-axis and goes counterclockwise to the line that you have drawn. That's the one you want. What is its relation to 37.67 degrees?
     
  10. Sep 16, 2009 #9
    ok, is it 360-37.67? which is going to be 322.33
     
  11. Sep 16, 2009 #10
    I put in 322.33, and it came out wrong! The correct angle was -37.67..Bummer
     
  12. Sep 16, 2009 #11

    kuruman

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    Bummer indeed. If this is one of those evil Web-based, calculated type questions like Webassign or Mastering Physics, I think you should explain your answer to your instructor. Chances are he/she will override the machine. I know I would. :wink:
     
  13. Sep 16, 2009 #12
    yea wiley plus..Horrible system!
     
  14. Sep 16, 2009 #13
    thanks anyways for your help!
     
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