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Homework Help: Weighted average

  1. Oct 30, 2007 #1
    Consider a velocity which is a function of position r, which does not vary linearly with time.

    Consider a body moving with this varying velocity between distance r1 to r2.

    Let us define the average velocity between r1 and r2 as (r2-r1)/time taken to travel between r2 and r1.

    I assumed the average would be found by:

    Integral[ v(r) dr {r2, r1}] / Integral [dr {r2, r1}]

    But this formula does not seem to work. Are there any special cases where this formula is not sufficient?
    Last edited: Oct 30, 2007
  2. jcsd
  3. Oct 30, 2007 #2

    Meir Achuz

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    You need the integral over time to get the average velocity.
  4. Oct 30, 2007 #3
    But time is a function of position, so there does not seem any reason why one could not use the position as an equivalent weight?

    So you suggest:

    Integral[v dt] / Integral[dt]

    r(t) => dr/dt = 1/f'
    => dt = f' dr

    Hence one could write Integral[v f' dr] / Integral[f' dr]

    But dr/dt = v
    hence f' = 1/v
    => vbar = Integral[dr] / Integral[dr /v]
    Last edited: Oct 30, 2007
  5. Oct 30, 2007 #4

    Shooting Star

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    The two integrals (1) and (2) are identical. Sorry, I don't get your point now...?
  6. Oct 31, 2007 #5
    Well Meir felt you couldn't take the weighted average by integrating over position, but I am trying to prove that you can by virtue of the fact that his suggestion, of integrating over time, can be written as an integral over position.
  7. Oct 31, 2007 #6


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    Ok. But Meir was responding to this in your original post:

    "Integral[ v(r) dr {r2, r1}] / Integral [dr {r2, r1}]"

    which is not going to give average velocity.

    But "vbar = Integral[dr] / Integral[dr /v]" in your 3rd post will give average velocity.
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