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Finding the displacement under a curve. Please Help

  1. Sep 2, 2011 #1
    GIV-Q-009_0010_-002-large.png

    1. What is the displacement between t= 0 s and t= 7 s?The "area' under the curve, between the two times is the displacement. The "area" is the area enclosed by the curve and the time axis (v=0 line). Those parts of the curve with negative velocity contribute negative area and those with positive velocity contribute positive area.
    Between t=0 s and 7 s,

    Δx =





    2. Relevant equations
    Area of Rectangle: (l*w)
    Area of Trapezoid: (1/2 b (h1+h2))
    Area of Triangle: (1/2bh)



    3. Rectangle: 1.5(-380)= -570
    Trapezoid:1/2*3*(-380+-280)=-825
    Triangle: 1/2*1.5*280=-420

    delta x = -(570+825+420)= -1815 ...this was my attempt and it was wrong. I don't know what I am doing wrong please help

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Sep 2, 2011
  2. jcsd
  3. Sep 2, 2011 #2
    Could you elaborate on how you got your answer for delta x?
     
  4. Sep 2, 2011 #3
    I used the points for t-0 and t-7 and came up with the three shapes on the graph..(the rectangle the trapezoid and the triangle) and then solved for their area under the curve and then combined.

    I don't have a scanner to post the graph that I drew the shapes on.
     
  5. Sep 3, 2011 #4

    SteamKing

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    You should be calculating the area between the curve and the horizontal axis where
    v = 0.
     
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