Question

Okay I think so I can send it in the

Answer 1

ANSWER

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

ANSWER

The area of the shaded rectangle is;

$\begin{gathered} (6-2)\times80 \\ l=(6-2)=4 \\ b=80 \\ A=L\times B \\ A=4\times80 \\ =320ft \end{gathered}$

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;

$\begin{gathered} \frac{2+6}{2}=\frac{8}{2} \\ =4 \end{gathered}$

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;

$\begin{gathered} \Delta s=velocity\times time \\ =80\times4 \\ =320ft \end{gathered}$