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Acceleration Position vs. Time Graph

  1. Jun 5, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    Evaluate Ax, the x-component of the acceleration between M and S.

    2. Relevant equations
    a=x/t^2


    3. The attempt at a solution
    M= 48 m at 8.5 s
    S= 40 m at 11.5 s

    so change of x=-8 m, and change of t = 3s

    so a= -8 / (3^2), so -8/9 m/s^2 (or -0.889 m/s^2). Where did I go wrong?
     
  2. jcsd
  3. Jun 5, 2008 #2

    berkeman

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    Hint: figure out the velocities at M and at S, then the change in velocity between M and S, and then...
     
  4. Jun 5, 2008 #3
    Would the velocity at M be the velocity between R and M, and the velocity at S be the velocity between S and N, or do I have to calculate the instantaneous velocity at each point?
     
  5. Jun 5, 2008 #4

    Redbelly98

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    In this specific case, yes.

    In general, yes.
     
  6. Jun 5, 2008 #5
    They are the same thing since the velocity is uniform on those parts of the graph.
     
  7. Jun 6, 2008 #6
    So I ended up calculating 20/3 for the M velocity, and -20/1.5 for the S velocity, which gave me a change of velocity of -20. If the change in time was 3 seconds, then the x-component of the acceleration would be -20/3 m/s^2. But thats not the right answer given.

    Is the x component just the -20? I can't figure out what I did wrong.
     
  8. Jun 6, 2008 #7

    Redbelly98

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    Your values look correct, within any reasonable accuracy that can be read from this graph. You should be only slightly off from the "correct" answer. Just how far from the given answer is your value of -6.7 m/s^2?
     
  9. Jun 6, 2008 #8
    I gave this answer: -6.67 m/s^2, so I guess it was a mistake.
     
  10. Jun 6, 2008 #9

    Redbelly98

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    I'm not convinced you made a mistake. You got -6.67 m/s^2. Earlier you said this is not "the right answer given". So what is the right answer given, and how far off are you from it?
     
  11. Jun 6, 2008 #10
    I don't really know, because it is automated and answers aren't revealed till monday, but I can find out within a couple of days. I would like to know what the answer is myself. All I know is that the computer said my answer was wrong and I can't change it.
     
  12. Jun 6, 2008 #11

    Redbelly98

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    It's a little odd that they would require an "exact" answer, given the approximations inherent in reading a graph. You really have correctly solved the problem, but have probably read the numbers on the graph a little differently than the problem author.

    By the way, I get -12 m/s for the velocity at S.
     
  13. Jun 7, 2008 #12
    I found out that the correct answer was -6.33 m/s^2... seems kind of unfair to have an exact answer like that without having exact numbers.
     
  14. Jun 7, 2008 #13

    Redbelly98

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    Yes, I agree.
     
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