Question

What is the displacement over the first 7 seconds of this graph?

Answer 1

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)

`\begin{gathered} (y-3)/(x-6)=(0-6)/(4-3) \\ y-3=-6x+36 \\ y=-6x+39 \end{gathered}`

So v(t)=-6t+39 from 3 to 4

From 4 to 6 it passes through (0,4) and (6,-2)

`\begin{gathered} (y-4)/(x-0)=(-2-4)/(6-0) \\ y-4=-x \\ y=-x+4 \end{gathered}`

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:

`\begin{gathered} (y-(-2))/(x-6)=(0-(-2))/(7-6) \\ y+2=2x-12 \\ y=2x-14 \end{gathered}`

So v(t)=2t-14 from 6 to 7.

So the displacement from 0 to 7 is given by:

`\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}`

Integrate to get:

`6\lbrack3-0\rbrack-(6)/(2)\lbrack16-9\rbrack+39\lbrack4-3\rbrack-(1)/(2)\lbrack36-16\rbrack+4\lbrack6-4\rbrack+(2)/(2)\lbrack49-36\rbrack-14\lbrack7-6\rbrack=43`

So the displacement is 43 units.