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Graph of elevator's velocity

  1. Feb 14, 2016 #1
    1. The problem statement, all variables and given/known data
    QQ9OQ0S.png

    2. Relevant equations


    3. The attempt at a solution
    This is more of a help with understanding problem. I understand a) and b), but for c), I don't understand:
    1. Why is the distance from t=12-16 -20 m and not +20 m? Isn't there a positive slope? I thought this question was asking for distance traveled, not displacement?
    2. My attempt at c) was to use the equation y(t)-y0 = 1/2(v0+v(t))t, but my answer was 325 m. y0=+75 m, v0=20 m/s, v(t)=5 m/s, and t=20s. What did I do wrong?
     
  2. jcsd
  3. Feb 14, 2016 #2

    TSny

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    Part (c) asks for distance traveled, but part (d) asks for the final position.
     
  4. Feb 14, 2016 #3
    So what did I do wrong in the equation I set up in d)?
     
  5. Feb 14, 2016 #4

    TSny

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    The average velocity is not 1/2(v0+v(t)) for part (d) because the acceleration is not constant over the entire time interval.
     
  6. Feb 14, 2016 #5
    Oh okay, I understand now. Thanks. But what about the distance from t=12-16 seconds? Why is it -20 m and not +20 m?
     
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