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Interpreting and converting an acceleration-time graph

  1. Mar 22, 2013 #1
    1. The problem statement, all variables and given/known data

    WL9HGAZ.png

    2. Relevant equations

    a = Δv/Δt
    v = Δx/Δt
    x = area under velocity graph

    3. The attempt at a solution

    According to my second attempt, the answer to "particle's speed at t = 20.0 s?" is not 15m/s either.
    Working out picture.
    I don't understand where I went wrong. I've always thought that the area under a velocity graph would give the distance.
    If t=20s isn't v=15m/s, then maybe that's why? But according to the a-t graph, 5x-3=-15. :S
     
    Last edited: Mar 22, 2013
  2. jcsd
  3. Mar 22, 2013 #2

    Ibix

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    Science Advisor

    Take another look at your graph for the deceleration phase. It slows down by 15m/s.
     
  4. Mar 22, 2013 #3
    Oh... I see now. It doesn't slow down to 15m/s but slows down 15m/s.
    So like this, right?
    lL42HNY.png
     
  5. Mar 23, 2013 #4

    SammyS

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    Staff Emeritus
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    Homework Helper
    Gold Member

    That looks good !
     
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