Finding the displacement under a curve.

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The displacement between t=0 s and t=7 s is determined by calculating the area under the velocity curve, which includes both positive and negative contributions. Negative velocity areas contribute negatively to the total displacement, while positive areas contribute positively. The user attempted to calculate the displacement using areas of rectangles, trapezoids, and triangles but arrived at an incorrect total of -1815. Clarification was requested on the calculation process, particularly regarding how to accurately determine the areas under the curve. Accurate calculation requires careful consideration of the areas relative to the horizontal axis where v=0.
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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 =


Homework 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

Homework Statement


Homework Equations


The Attempt at a Solution

 
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Could you elaborate on how you got your answer for delta x?
 
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.
 
You should be calculating the area between the curve and the horizontal axis where
v = 0.
 
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