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Finding Average Acceleration

  1. Sep 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Here's the graph:

    http://i.imgur.com/uUXPO.jpg

    What is the average acceleration between t=0s and t=2s?

    2. Relevant equations

    a avg = change in velocity / change in time ; (Vf-Vi)/(tf-ti)

    3. The attempt at a solution
    I know how to solve it if it was just a straight line where I could just plug in the numbers into the formula, but how do I calculate the average acceleration if the section above is non-straight? The part I don't get exactly is how to calculate the change in velocity itself since it's decreasing from 0s to 1s and then increasing from 1s to 2s.
    As I understand the acceleration in this problem is the slope?
    I tried calculating the slopes of those intervals separately and then adding them together and dividing by the change in time (2s), but I seem to be getting the wrong answer because my homework is online and it tells me if the answer plugged in is correct or not.

    0s to 1s interval contains points (0,4) and (1,2) so the slope is -2
    1s to 2s interval contains points (1,2) and (2,3) so the slope is 1
    then (-2)+1/2= -0.5 m/s^2 even without dividing and ending up with answer 1, the avg acceleration is still incorrect

    So since I am doing this completely wrong can somebody explain to me the whole concept, because for some reason I just don't seem to get it
     
  2. jcsd
  3. Sep 15, 2012 #2

    cepheid

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    Welcome to PF Daizin!

    That's the thing about an average: it doesn't take into account detailed information about what was going on over time intervals smaller than the interval you're averaging over (2 s). So really, all you do here is use the values of v and t at the two endpoints of the interval (t = 0 s and t = 2 s). These are the initial and final values. At the beginning of the interval, your velocity was +4 m/s, whereas at the end of the interval, it had decreased to +3 m/s, so on average, your acceleration during that time was some negative value (that you will compute) even though the details of what was happening during that interval are more complicated (you actually had positive instantaneous acceleration during parts of it).
     
  4. Sep 15, 2012 #3

    SammyS

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    It's much easier to work on this with the image displayed directly here.

    attachment.php?attachmentid=50856&stc=1&d=1347747644.jpg

    It looks like using (Vf-Vi)/(tf-ti) gives -0.5 m/s2.

    Averaging -2m/s2 and 1m/s2 also gives -0.5m/s2.

    So, what's the problem?
     

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    Last edited: Sep 15, 2012
  5. Sep 15, 2012 #4

    jedishrfu

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    calculate the area under the curve and then divide by the time interval.

    I get 12.5m as the area then dividing by 2 secs and the avg is ???
     
  6. Sep 15, 2012 #5

    SammyS

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    That gives the average velocity.

    The area under the velocity graph gives the displacement.

    The displacement divided by elapsed time gives the average velocity.
     
  7. Sep 15, 2012 #6
    so you're saying it should look like:
    (3-4)/(2-0) and yes, it does end up as -0.5 (or -1/2) and that's how I initially solved the problem at the very start but when I plugged this answer into the webassign homework, it listed it as an incorrect one, which is why I started seeking for help and maybe other formulas for this problem.

    Edit: nevermind I got it. Apparently I accidentally kept plugging in the answer into the wrong space which is why it kept telling me it was wrong (I know... I am so stupid) But hey, at least I am definitely sure about how to do all these now.
     
  8. Sep 15, 2012 #7

    SammyS

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    Glad that it worked out !

    By the Way: Welcome to PF !
     
  9. Sep 15, 2012 #8

    jedishrfu

    Staff: Mentor

    oops misread the problem and gave avg vel not avg accel
     
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