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Average and instantaneous velocity

  1. Feb 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Given the following formula for distance, find the average velocity on the interval [1,3]
    S(t) = t/(1+t^2)

    2. Relevant equations
    Vavg = (S(t0) - S(t1)) / (t0 - t1)

    or

    Vavg = (V0 + V1)/2


    3. The attempt at a solution

    I get two different answers and I need help understand why.

    Vavg = (S(1) - S(3)) / (1-3)

    Vavg = (1/2 - 3/10) / -2

    Vavg = -2/20 = -1/10

    Now using a different method, getting the instantaneous velocity at t = 1 and t = 3 by taking derivative of S(t) -> S'(t) = (1-t^2)/(1+t^2)^2

    Vavg = (S'(1) + S'(3))/2

    Vavg = (-8/100)/2 = (-2/25)/2 = -1/25

    Why am I getting different averages here?? I don't understand what I'm doing wrong.
     
    Last edited: Feb 26, 2013
  2. jcsd
  3. Feb 26, 2013 #2

    Dick

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    Judging by your work, I'm guessing you meant S(t)=t/(1+t^2), not S(t)=t/(1+t)^2. And you can't average instantaneous velocities to get average velocities. The first one is right, the second one isn't.
     
  4. Feb 26, 2013 #3
    Yes I'm sorry that's what I meant. But I don't understand, why can't you do what I did? If (v0 + v) / 2 = average velocity, and v0 and v are velocities at an instant in time, I don't understand how the second method doesn't work?
     
  5. Feb 26, 2013 #4

    Dick

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    Because they are two totally different notions of average velocity. If I start at rest in city A and drive to city B and park, then my beginning and ending velocities are zero. My average velocity, in the sense you want, isn't zero.
     
  6. Feb 26, 2013 #5
    Got it, thank you!
     
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