1. Sep 2, 2011

### kimberley511

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. Sep 2, 2011

### davo789

Could you elaborate on how you got your answer for delta x?

3. Sep 2, 2011

### kimberley511

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.

4. Sep 3, 2011

### SteamKing

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