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Average Velocity vs. Instantaneous Velocity

  1. Feb 6, 2008 #1
    The graph below shows a plot of velocity vs. time for an object undergoing uniformly accelerated motion. The object has instantaneous velocity v1 at time t1 and instantaneous velocity v2 at time t2. Use the graph and the fact that, when the acceleration is constant, the average velocity can be written as vavg = (v1 + v2)/2 to explain how we know that the instantaneous velocity at the midpoint time is equal to vavg on the interval.




    V = Vo + at

    X - Xo = Vot + .5at2

    v2 = vo2 + 2a(X - Xo)

    X - Xo = .5(Vo + V)t

    Now I know the answer has to do with some how deriving one of these formulas to get it to equal the formula for average veolcity but I can't seem to remember how. I think you some how take the derivative of X - Xo = .5(Vo + V)t to get the answer maybe?
     
  2. jcsd
  3. Feb 9, 2008 #2
    What graph?
     
  4. Feb 12, 2009 #3
    pretty sure I know what the graph looks like cause I just had this problem in a homework lol.

    This answer doesn't require any deriving. Average velocity is simply the slope of the velocity over a certain interval, and instantaneous velocity is the slope at a certain point on the interval. We know that the acceleration graph is simply the graph of the slope of the velocity function.The slope of this graph is clearly linear. And because average velocity and instantaneous velocity are just values of the slope over their intervals or at certain points, if we take the average velocity from t1 to t2 it would be the same as the velocity at the midpoint between t1 and t2.

    Hopefully that makes some sense.
     
    Last edited: Feb 12, 2009
  5. Feb 12, 2009 #4
    you do realise that s/he posted this over a year ago, and probably doesnt need it anymore
     
  6. Feb 12, 2009 #5
    mmmm no unfortunately I did not realize that...this was a colossal waste of time.
     
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