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Averaging of velocities!

  1. Oct 29, 2008 #1
    1. The problem statement, all variables and given/known data
    A particle starts from rest with constant acceleration. Find the ratio of space average velocity with time average velocity.


    Explain to me the underlined words in order that I may proceed to solve this problem.
     
  2. jcsd
  3. Oct 29, 2008 #2

    LowlyPion

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    Try writing out the equations for V as a function of t and V as a function of x. Then perhaps you can arrive at what taking the average over each function might yield?
     
  4. Oct 29, 2008 #3
    v=kt

    and v^2=2kx

    where t is time
    x is displacement
    and k is the constant acceleration
    now what shall I do?
     
  5. Oct 29, 2008 #4

    LowlyPion

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  6. Oct 29, 2008 #5
  7. Oct 29, 2008 #6

    LowlyPion

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    You have V(x) over the distance X in the same way that you have V(t) over T for how you might figure the time average velocity. You don't think you can come up with an expression for the distance averaged V?
     
  8. Oct 29, 2008 #7
    Time Average velocity is total displacement/ total time

    Is distance average velocity? total distance / total displacement ??? I really have no idea. Please help!!
     
  9. Oct 29, 2008 #8

    robphy

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    Can you derive the statement that
    "Time Average velocity is total displacement/ total time"?
    You may wish to first consider a simpler situation in which your total trip has just two velocities:
    v1 for time-interval t1, and then v2 for time-interval t2.
     
  10. Oct 29, 2008 #9
    Yup, I have done that. All I need to know is what is distance averaged velocity???
     
  11. Oct 29, 2008 #10
    Example: For 2km I go at a velocity 30m/s, then for 1km, I go at a velocity 15m/s. The distance-averaged velocity would be (2x30m/s+1x15m/s)/3=25m/s.

    The time-averaged velocity will be smaller than 25m/s, as I go an equal amount of time at the two different speeds: the time-averaged velocity is (30+15)/2 m/s=22.5 m/s.
     
  12. Oct 29, 2008 #11

    LowlyPion

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    OK. If Time averaged velocity is given by:

    [tex] V_{Tavg} = \frac{1}{T}\int_{0}^{T} V_{(t)} dt [/tex]

    Then won't the distance averaged velocity be given by:

    [tex] V_{Xavg} = \frac{1}{X}\int_{0}^{X} V_{(x)} dx [/tex]

    Now you have already supplied V as a function of x and V as a function of t.

    So ...
     
  13. Oct 29, 2008 #12
    [tex]<v>=\frac{1}{X}\int_{0}^{X} V(x) dx[/tex]
    right?
    But the thing is that the X is not eliminated!!!
     
  14. Oct 29, 2008 #13

    LowlyPion

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    T wasn't eliminated either.
     
  15. Oct 30, 2008 #14

    robphy

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    If I understood the problem correctly, it appears that the ratio desired is a dimensionless factor... a simple fraction.
     
  16. Oct 30, 2008 #15
    solved
     
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