College Algebra, Chapter 3, 3.4, Section 3.4, Problem 26

Suppose that a man is a running...

$\begin{array}{|c|c|}
\hline\\
\text{Time } (s) & \text{Distance } (m) \\
\hline\\
32 & 200 \\
68 & 400 \\
108 & 600 \\
152 & 800 \\
203 & 1000 \\
263 & 1200 \\
335 & 1400 \\
412 & 1600 \\
\hline\\
\end{array} $

a.) What was the man's average speed (rate) between $68 \, s$ and $152 \, s$?


$
\begin{equation}
\begin{aligned}

\text{average speed } =& \frac{f(b) - f(a)}{b - a}
&& \text{Model}
\\
\\
\text{average speed } =& \frac{f(152 \, s) - f(68 \, s)}{152 - 68}
&& \text{Substitute } a = 68 \, s \text{ and } b = 152 \, s
\\
\\
\text{average speed } =& \frac{800 - 400}{84}
&& \text{Simplify}
\\
\\
\text{average speed } =& \frac{400}{84}
&& \text{Simplify}
\\
\\
\text{average speed } =& 4.76 \, m/s
&& \text{Answer}

\end{aligned}
\end{equation}
$


b.) What was the man's average speed (rate) between $263 \, s$ and $412 \, s$?


$
\begin{equation}
\begin{aligned}

\text{average speed } =& \frac{f(b) - f(a)}{b - a}
&& \text{Model}
\\
\\
\text{average speed } =& \frac{f(412 \, s ) - f(263 \, s) }{412 - 263}
&& \text{Substitute } a = 263 \, s \text{ and } b = 412 \, s
\\
\\
\text{average speed } =& \frac{1600 - 1200}{149}
&& \text{Simplify}
\\
\\
\text{average speed } =& 2.68 \, m/s
&& \text{Answer}

\end{aligned}
\end{equation}
$


c.) Calculate the man's speed for each lap. Is he slowing down, speeding up or neither?

$
\begin{array}{|c|c|c|}
\hline\\
\text{Time} \, (s) & \text{Distance} \, (m) & \displaystyle \text{Speed} \, \left( \frac{m}{s} \right) \\
\hline\\
32 & 200 & 6.25 \\
68 & 400 & 5.88 \\
108 & 600 & 5.56 \\
152 & 800 & 5.26 \\
203 & 1000 & 4.93 \\
263 & 1200 & 4.56 \\
335 & 1400 & 4.18 \\
412 & 1600 & 3.88 \\
\hline
\end{array}
$

The man's speed each lap is slowing down.