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Average Velocity

  1. Aug 28, 2007 #1
    Question: Find the average velocity over the time interval from 1 to 5 seconds. You will need to interpolate to find the position at time t=5 s . Do not simply eyeball the position or you will likely not be able to obtain the solution to the desired accuracy.

    (Referring to attached graph: link http://i11.tinypic.com/6g4jl8i.jpg)

    How do you find the Average Velocity in a time interval of a position vs time graph when the position rises then decreases?

    some info from previous questions and work i did is as follows:
    velocity from t=1 to t=3 is 20 m/s
    position at t=5 is about 46.7 m
    velocity from t=4.5 to t=4.5 is (i think) about -26.7 m/s

    I tried a few times and got answers like 4.2 trying to average the time the velocity took in each part but it didnt work.

    The question says to interpolate, but apparently i dont know how to do that the right way since im gettin it wrong.

    any ideas?

    Attached Files:

    Last edited by a moderator: Aug 28, 2007
  2. jcsd
  3. Aug 28, 2007 #2
    46.7 sounds right for the position at t=5. Can you show us your work for trying to calculate the average velocity?
  4. Aug 28, 2007 #3
    Well, i tried to add each segment and multiply it by how much space it took, so like this:


    but that equals zero -.-, despite looking like itd give the average.
  5. Aug 28, 2007 #4
    The average velocity, which is a vector quantity, is the net displacement, divided by the time. Net being the important word.
  6. Aug 28, 2007 #5


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    When you calculate the average velocity, you might want to break the data (position vs. time) into samples .5s apart (since your smallest increment is .5) and then you can take the instantaneous velocity over each .5s increment, add them together, and divide by the total number of increments.

    an example:

    from 1 to 1.5 seconds, position increases by 10 so V = 10m/.5s = 20 m/s
    from 1.5 to 2 seconds.....
  7. Aug 28, 2007 #6


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    I don't think 4 is enough. You're times segments are differently spaced and you're averaging over time, so you may want to add those coefficients that represent seconds together (2 + 1.5 + 1.5 = 5)

    Why couldn't the average velocity be 0 anyway? I don't get exactly 0, by the way for that equation.
  8. Aug 28, 2007 #7
    ok, i did it with intervals of .5 and got like 3.33125 by dividing 26.65 by 8

    but thats not right either according to the stupid online questionnnnn

    thanks for the idea tho, maybe im missing a large picture.
    any ideas.
  9. Aug 28, 2007 #8
    Averaging the velocity I get 13.3 (1 second at 0, 2 seconds at 20, 1.5 seconds at 0 again, and 1.5 seconds at 80/3, over a total of 6 seconds). But if you calculate it considering the space you moved, then it is zero, as you finsih at the same point you started. Make sense?
  10. Aug 28, 2007 #9


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    you would have had to solved for 5.5s that way which would have been an extra step unless you kind of summed over several intervals like I'm going to below:

    what's the answer you're supposed to get?

    When I do it the intervals way I get:

    10 + 10 + 10 + 10 +0 + 0 + 0 - 1.5*40 = -20

    and there's 10 samples there (three of them are in the 1.5)

    so I get -20/10 = -2

    EDIT: my bad, I added 5.5s and 6s samples in, but problem only asks 1 to 5....
  11. Aug 28, 2007 #10
    it doesnt say, its this online question for my class, tells you when you're right but otherwise no, with 5 tries and i have one last one, after trying -2 and my own answer.

    btw, its the average from 1 to 5, not 0 to 6.
  12. Aug 28, 2007 #11


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    well, you got: 3.33125

    but I get:

    (10 + 10 + 10 + 10 + 0 + 0 + 0 - 13.445)/8 = 3.319375 (which I don't like either)

    and doing it your old way:

    2*20 + 1.5*0 - .5*26.67 (I think you did 1.5 here the first time)

    = 40 - 13.335 = 26.67

    and there's four seconds involved so:

    26.67/4 = 6.67

    I'd trust your way because I see an inconsistency in my way now. You just have to do your way right and not include the last second like you did in your first calculation.
  13. Aug 28, 2007 #12


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    You're making more complicated than it needs to be.

    The definition of average velocity is net displacement divided by time. What is the net displacement from position 1 to position 5? Divide that by 5-1 = 4 seconds. That's your answer.
  14. Aug 28, 2007 #13
    "You're making more complicated than it needs to be.


    so wait, itd be zero? cause it goes back to the same place, so 0/4 would be zero.
    that sounds too sumple tho; but i doubt its that since it says round to 2 sig figs.

    i jsut tried 6.7 (2 sig figs) as my last attempt and it worked, althought i dont get why you divide 26.67 by 4 and not the total number or intervals. however, 6.7 is the right answer.

    either way, thanks guys.
    if you can explain that little thing tho itd be helpful.
  15. Aug 28, 2007 #14


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    average velocity = net displacement/time elapsed

    The net displacement from 1s to 5s is 46.7m - 20m = 26.7m (just subtract the positions).

    time elapsed is 5-1 = 4s.

    26.7/4 = 6.7m/s
    Last edited: Aug 28, 2007
  16. Aug 28, 2007 #15
    or, draw a v-t graph,
    find area below the cruve(net distance), and divide it by 6
  17. Aug 28, 2007 #16
    nah wait i got it, i put 1.5 instead of .5 for the last velocity when i was interpolating, there we gooo.

    thanks guys.

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