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Homework Help: Average Velocity vs Average Velocity discrepencies?

  1. Feb 23, 2012 #1
    This question is about a discrepency between the method of finding average velocity via "change in distance over change in time" and the other method of "instantaneous velocity + instantaneous velocity all divided by 2"

    If I find average velocity of various points on a position vs position graph, then find instanenous velocity of each of those points, if I then recalculate average velocity based on the instantaneous velocities (as opposed to finding magnitude of average velocity based on the information provided in the position vs position graph), how come the average velocities of one method don't match up with the average velocities found using the other method?

    For instance, why do I get 18.9 when finding average velocity when I feel like I should be getting 17.2 as I did when I calculated average velocity in the beginning? What is accounting for the difference? Am I making an incorrect calculation, or is my calculator creating a discrepancy? Is the shorthand process of finding the derivative equations (as opposed to using the formal definition of derivative) for x(t) and y(t) not exact enough?

    So my question is, why am I getting 18.9m/s and not 17.2m/s?

    Here's an illustration of my question:

    EDIT: I removed the embeded illustration. Both question and answer are still clear without illustration.
    Last edited: Feb 23, 2012
  2. jcsd
  3. Feb 23, 2012 #2

    Doc Al

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    Staff: Mentor

    That should always work. (It's the definition of average velocity.)
    That will work over an interval of constant acceleration.

    Not sure what your diagram represents, but note that average velocity depends on the interval that you are studying.
  4. Feb 23, 2012 #3

    For instance, at time = 2, I get position (25, 23), and at time = 3, I get position (35, 37).

    If I find average velocity between those points by vector subtraction to get Δr and divide by a time of 1 second (which is the time interval I chose), I get the components of average velocity, then the magnitude of average velocity between those points is 17.2m/s.

    So I have established that average velocity between the 2s and 3s second time interval is 17.2m/s

    Then, if I take the derivative of the position functions and evaluate for the time intervals 2s and 3s, I get instanenous velocity for those two time intervals. I get an instaneous velocity of 17.2m/s at time = 2s and I get an instaneous velocity of 20.59m/s at time = 3s.

    So if I use those two instanenous velocity values to find average velocity between 2s and 3s, I would add V_1 + V_2 and divide by 2, correct? Well (17.2m/s + 20.59m/s)/2 equals 18.9m/s.

    Clearly, 17.20m/s ≠ 18.9m/s , but both values claim to be average velocity between 2s and 3s. Where is the discrepancy coming from?
  5. Feb 23, 2012 #4
    EDIT: I think I found my mistake. It was a calculation on my part. I am going to check now.

    EDIT 2: Yea, you are right, I calculated an incorrect value for instantaneous velocity (I accidently entered 2^3 power instead of 3^2 power on my calculator when evaluating t = 3 for the position function) I got 18.8 as the correct instantaneous velocity. And so the two methods of finding average velocities do in fact end up matching very closely along that interval.
    Last edited: Feb 23, 2012
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