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Average Velocity of a particle moving in a circle over a given interval

  1. Feb 17, 2011 #1
    1. The problem statement, all variables and given/known data

    [​IMG]


    2. Relevant equations

    d = 2.5
    c = pi*d = 7.854
    velocity/s ?= c * 2 = 15.7079


    3. The attempt at a solution
    Since PR is 1/4 of the circle and the particle moves around the circle 2 times per second, I thought the average velocity would be 1/8th of the velocity that it's traveling. I'm really confused on this one.

    EDIT: I only need to solve for (a), that's why I only have data pertaining to that. Not worried about B and C, but any intuition on those is welcome.
     
    Last edited: Feb 17, 2011
  2. jcsd
  3. Feb 17, 2011 #2
    What is the definition of average velocity?
     
  4. Feb 17, 2011 #3
    The change in x over time. The change in X is -1.25 right? and the time is 1/8 of a second... .125

    dx/dt = -1.25/.125 = 10?

    I'm not following
     
  5. Feb 17, 2011 #4

    gneill

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    Staff: Mentor

    If someone were to show you two snapshots, the first showing the particle at P at time zero, and the second at point R at a time 1/8 second later, and you had no idea about the circular course, or what path the particle took to get from P to R, how would you go about finding the average velocity?
     
  6. Feb 17, 2011 #5

    gneill

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    Staff: Mentor

    Why just x?
     
  7. Feb 17, 2011 #6
    What do you mean why just x?

    The change in x is -1.25 and the change in time is 0.125 seconds, right?
     
  8. Feb 17, 2011 #7

    gneill

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    Staff: Mentor

    So the particle didn't move along the circular path, but instead traveled a straight line along the x-axis and ended up at the origin? Is point R at the origin?
     
  9. Feb 17, 2011 #8
    It's at Y=1.25 and X=0. If I were to draw a straight line from P to R, the angles would be 45, 90 and 45 right? and the equation for the line would be:

    1.25^2 + 1.25^2 = C^2
    1.5625 + 1.5625 = C^2
    3.125 = C^2
    C = 1.767767

    So .... The vector is (1.767767, 135 degrees)?
     
    Last edited: Feb 17, 2011
  10. Feb 17, 2011 #9

    gneill

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    Okay! So you've got a distance vector. That gives you the displacement (magnitude of the vector) and the direction. Apparently the displacement happened in 1/8 second. So what's the average velocity vector?
     
  11. Feb 17, 2011 #10
    So if the displacement is 1.767767 and happened in 1/8th. I assume I multiply the displacement by 8 to get the velocity per second...

    So 14.1421 m/s?
     
  12. Feb 17, 2011 #11

    gneill

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    Yup!

    Remember though, that velocity is a vector. So you want to retain the angle information in your answer.
     
  13. Feb 17, 2011 #12
    Awesome. Your socratic method of teaching pissed me off to no end but it worked, and I'm very thankful for your help :tongue:
     
  14. Feb 17, 2011 #13

    gneill

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    Staff: Mentor

    Heh. Bet you'll remember how to find the average velocity though!

    Build a man a fire and he'll be warm for a day.
    Set a man on fire and he'll be warm for the rest of his life!
     
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