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.