Question

Okay I think so I can send it in the

Answer 1

`\begin{gathered} a)320ft^ \\ b)320ft \end{gathered}`

ANSWER

The area of the shaded rectangle is;

`\begin{gathered} (6-2)*80 \\ l=(6-2)=4 \\ b=80 \\ A=L* B \\ A=4*80 \\ =320ft \end{gathered}`

b) To estimate the change in the train's position on the time interval [2,6]. The constant velocity as measured at the midpoint of the interval equals;

`\begin{gathered} (2+6)/(2)=(8)/(2) \\ =4 \end{gathered}`

Hence, the midpoint of the interval is 4 sec.

From the graph, at t= 4 sec, the velocity is 80ft/sec.

During the interval [2,6] which is 4sec.

The change in position of the train is;

`\begin{gathered} \Delta s=velocity* time \\ =80*4 \\ =320ft \end{gathered}`