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Homework Help: Velocity question

  1. Feb 7, 2006 #1
    Does anyone happen to know why the instantaneous velocity at the midpoint of a time interval is equal to the average velocity over the same time interval?? I can't seem to prove this reasoning.

    Thanks!! :smile:
     
  2. jcsd
  3. Feb 7, 2006 #2

    russ_watters

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    Staff: Mentor

    You're talking about when acceleration is constant.... It's just from the definition of "average" - the 2 in the denominator is where you get the halfway point in time.
     
  4. Feb 8, 2006 #3

    andrevdh

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    Homework Helper

    First note that the average velocity is midway between the two velocities values, [itex]v_1,\ v_2[/itex], at the endpoints of the interval. Since
    [tex]v_1+\frac{1}{2}(v_2-v_1)=v_1+\frac{v_2}{2}-\frac{v_1}{2}[/tex]

    [tex]=\frac{v_1+v_2}{2}[/tex]
    We therefore need to show that the time at which the average velocity is reached is in the middle of the time interval. In the drawing time is on the horizontal x-axis and speed on the vertical y-axis. What needs to be proved then in the drawing is that [itex]AD=BC[/itex], It is clear that both these length are given by
    [tex]\frac{\Delta v}{2\tan(\theta)}[/tex]
     

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    Last edited: Feb 8, 2006
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