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Average velocity is measured through distance

  1. Oct 18, 2011 #1
    Is average velocity equal to the average of two instantaneous velocities. I had a question on a calculus exam that gave an equation to a particle moving and was told to find both the average velocity on the time interval [0,1] and the instantaneous velocity at 1. I took the derivative of the equation to find the slope and found the instantaneous velocity at 0 and at 1 and divided by 2. I was told that was incorrect by my professor because "average velocity is measured through distance". He did not explain anymore. Can anyone elaborate on why my method was incorrect.
  2. jcsd
  3. Oct 18, 2011 #2


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    Gold Member

    If the particle travels 1m, and the velocity is 100m/s over 0.1m, and the velocity is 1m/s over 0.9m, the average isn't 50.5m/s. The particle is using (0.1/100)s + (0.9/1)s to travel 1m. That would be an average velocity of 1m / (0.001s + 0.9s) = 1.098779 m/s

  4. Oct 18, 2011 #3


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    Staff: Mentor

    In general, no; instead, it's the total distance traveled divided by the time interval.

    For the specific case of constant acceleration, it's possible to prove that this equals the average of the instantaneous velocities at the beginning and end of the time interval. This is a very commonly-used result, and many students mistakenly apply it to more general cases where it is invalid.
  5. Oct 18, 2011 #4
    Not really to my perspective. The average is only applied to uniform acceleration. Since you are in a calculus exam, you'd better use integration somehow. Perhaps, distance travelled can be calculated. And you can apply the definition, which is average velocity equals total distance over time elapsed.
  6. Dec 8, 2011 #5
    The average velocity is not only applied to uniform acceleration. It can be defined for any motion whatsoever!

    See my comments on the thread about the arbitrary amplitude pendulum and average velocity where i got confused between the time averaged velocity and the space averaged velocity. Here is the explicit argument which relates the two:

    Suppose we have some non uniform motion between point 0 and point X (in general this means that the velocity and the acceleration may be changing during the motion between 0 and X. The moving object is at x=0 at t=0 and at x=X at t=T).

    The time averaged velocity for this interval, <V(t)>, is given by <V(t)> = X/T

    This is equal to the integral <V(t)> = integral (V(t) dt) / integral dt limits 0 , T

    If you know the velocity at each point on the path from 0 to X ( call this V(x) at point x ) then you could calculate the

    space average velocity <V(x)> = integral ( V(x)dx) / integral(dx) integration limits 0, X

    Now in general the space average <V(x)> is NOT equal to the time average <V(t)>. In fact we can show that in general

    1/<V(t)> = < 1/V(x) > which is not the same as 1/<V(x)>

    (the average of the reciprocal of a function is not the same as the reciprocal of the average of the same function.)

    So if you are given the velocity as a function of space V(x) you can calculate the time it takes to go from x=0 to x=X by finding ther average value of 1/V(x) in the interval 0 to X and multiplying this by X the distance travelled during the motion. Or if you know the velocity as a function of the time you can calculate <V(t)> and get the time for the motion:

    T = X . <1/V(x)> = X /<V(t)>

    I will post the expilcit argument 1/<V(t)> = < 1/V(x) > if this is not clear.
    Last edited: Dec 8, 2011
  7. Dec 8, 2011 #6
    I agree with Post #3..it is EXACTLY correct...

    A trick in these forums is to figure out which answers are correct and which ones are not.
  8. Dec 8, 2011 #7


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    Staff: Mentor

    In general, the answer is a big NOOO! This is the oldest trick in the book. Try a few examples, and prove it to yourself. Then never forget the lesson learned. :smile:
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