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Finding velocity at t=7

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Find each particle's velocity at t=7.0s. Work with the geometry of the graphs, not with kinematic equations.


    2. Relevant equations



    3. The attempt at a solution
    I'm not sure how to arrive at an answer. For the previous question (velocity from graph b) I found the value of -20 m/s by reading the graph and before that (velocity from graph a) I did a simple delta/delta[t] calculation and got -10 m/s. My issue is that I can't find an velocity at t=7.0s for graph C (the acceleration graph.) Please help.
    Thank you.
     

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    Last edited: Oct 3, 2012
  2. jcsd
  3. Oct 3, 2012 #2

    Ibix

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    Velocity is the integral of acceleration with respect to time. How could you read that off a graph?
     
  4. Oct 3, 2012 #3
    by integrating the area at that point to the x axis?
    1/2(2)(-20)
    ?
     
    Last edited: Oct 3, 2012
  5. Oct 3, 2012 #4

    Ibix

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    Integration is right, presuming you meant to the y axis. I don't think you've got to the right answer though, since there's more than one triangle to worry about in the 0-7s region of the graph. You can check your answer by counting squares for a graph this simple.

    Remember that the integral just gives you the velocity change over the time period, which isn't quite what you're asked for.
     
  6. Oct 3, 2012 #5
    how would i isolate
     
  7. Oct 3, 2012 #6
    Find a(t) and do antiderivative to find v(t).
     
    Last edited: Oct 3, 2012
  8. Oct 3, 2012 #7
    still can't get it.
    i tried calculating the area underneath both of those and got 10 m/s, but that wasn't the velocity of the acceleration graph... im not sure what to do
     
  9. Oct 3, 2012 #8
    as much as i hate to ask for a straight answer (with a minor explanation) from you guys, i think it would be beneficial because i simply can't get it and i've inputted about 12 answers into this program
     
  10. Oct 6, 2012 #9

    Ibix

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    Complete answers are forbidden by forum rules.

    The area under the graph can be obtained by counting squares between the graph and the x-axis, remembering that the area is negative when the graph is below the axis. Each square is 1s wide and 10ms-2 high. To start you off, the area between 5s and 7s is -20ms-1. You can work out the area between 0s and 5s.

    Do remember that the result is the velocity gain over the period, not the final velocity.
     
    Last edited: Oct 6, 2012
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