The function can be described as:
It is the line v(t)=6 from 0 to 3.
From t=3 to t=4, it is a line that passes through (3,6) and (4,0)
[tex]\begin{gathered} \frac{y-3}{x-6}=\frac{0-6}{4-3} \\ y-3=-6x+36 \\ y=-6x+39 \end{gathered}[/tex]
So v(t)=-6t+39 from 3 to 4
From 4 to 6 it passes through (0,4) and (6,-2)
[tex]\begin{gathered} \frac{y-4}{x-0}=\frac{-2-4}{6-0} \\ y-4=-x \\ y=-x+4 \end{gathered}[/tex]
v(t)=-t+4 from 4 to 6
From 6 to 7 the line passes through (6,-2) and (7,0) so the equation will be:
[tex]\begin{gathered} \frac{y-(-2)}{x-6}=\frac{0-(-2)}{7-6} \\ y+2=2x-12 \\ y=2x-14 \end{gathered}[/tex]
So v(t)=2t-14 from 6 to 7.
So the displacement from 0 to 7 is given by:
[tex]\int ^7_0v(t)\text{ dt=\lbrack}\int ^3_06\text{ +}\int ^4_3(-6t+39)+\int ^6_4(-t+4)\text{ }+\int ^7_6(2t-14)_{}\rbrack\text{ dt}[/tex]
Integrate to get:
[tex]6\lbrack3-0\rbrack-\frac{6}{2}\lbrack16-9\rbrack+39\lbrack4-3\rbrack-\frac{1}{2}\lbrack36-16\rbrack+4\lbrack6-4\rbrack+\frac{2}{2}\lbrack49-36\rbrack-14\lbrack7-6\rbrack=43[/tex]
So the displacement is 43 units.
