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Finding Instantaneous acceleration from a velocity-time graph

  1. Aug 31, 2010 #1
    Graph: http://s324.photobucket.com/albums/k327/ProtoGirlEXE/?action=view&current=Capture.jpg
    What is the instantaneous acceleration at t2 = 31 s?

    I know that the instantaneous acceleration is the slope of the tangent of that point and yes I do know I have to convert it to (m/s^2). But I am still not getting the right answers. Here are a list of answers I imputed and they are all wrong:
    -.1
    0
    -2
    -1
    -1.9
    1.9
    -.278
    -.52

    on the first few, I forgot to convert. I used 58/31 the last time.....

    The question had parts to it, but I don't know if they are necessary to solve this problem...here it is anyways:
    A) Compute the average acceleration during the time interval t = 0s to t = 10s. 1.7
    B) Compute the average acceleration during the time interval t = 30s to t = 40s. -1.7
    C) Compute the average acceleration during the time interval t = 10s to t = 30s. 0
    What is the instantaneous acceleration at t1 = 29 s. 0
    I already computed this and got them right.
     
    Last edited: Aug 31, 2010
  2. jcsd
  3. Aug 31, 2010 #2
    It looks like its either going to be 0 or its going to be at that corner, where the rate of change is not going to be defined.
     
  4. Aug 31, 2010 #3
    I put zero and it was wrong and I don't think undefined is an appropriate answer because 30 is exactly at that corner and it has an acceleration, which I used to solve part B with.
     
  5. Sep 2, 2010 #4
  6. Sep 2, 2010 #5
    nevermind I got it now.
     
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