Single Variable Calculus, Chapter 2, 2.1, Section 2.1, Problem 7

The table below shows the position of a cyclist at each time period.


$
\begin{equation}
\begin{aligned}

\begin{array}{|c|c|c|c|c|c|c|}
\hline\\
t( \text{seconds}) & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
s( \text{meters}) & 0 & 1.4 & 5.1 & 10.7 & 17.7 & 25.8\\
\hline
\end{array}

\end{aligned}
\end{equation}
$



a. Find the average velocity over the given time intervals:


$
\begin{equation}
\begin{aligned}

&\text{ (i) } [1, 3] & \text{ (ii) } [2, 3]\\
&\text{ (iii) } [3, 5] & \text{ (iv) } [3, 4]
\end{aligned}
\end{equation}
$





$
\begin{equation}
\begin{aligned}

\begin{array}{|c|c|c|c|c|c|}
\hline\\
& t_1 & t_2 & S_1 & S_2 & V_{ave} = \frac{S_2-S_1}{t_2-t_1} \\
\hline\\
\text{ (i) } & 1 & 3 & 1.4 & 10.7 & 4.65 \\
\hline\\
\text{ (ii) } & 2 & 3 & 5.1 & 10.7 & 5.6 \\
\hline\\
\text{ (iii) } & 3 & 5 & 10.7 & 25.8 & 7.55 \\
\hline\\
\text{ (iv) } & 3 & 4 & 10.7 & 17.7 & 7\\
\hline
\end{array}


\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\text{ (i) } V_{ave} =& \frac{10.7-1.4}{3-1} = 4.65\\
\text{ (ii) } V_{ave} =& \frac{10.7 - 5.1}{3 - 2} = 5.6\\
\text{ (iii) } V_{ave} =& \frac{25.8-10.7}{5-3} = 7.55\\
\text{ (iv) } V_{ave} =& \frac{17.7 - 10.7}{4-3} = 7
\end{aligned}
\end{equation}
$






b. Estimate the instantaneous velocity when $t=3$ using the graph of $s$ as a function of $t$.








Referring to the graph, the instantaneous velocity when $t=3$ is approximately equal to $\displaystyle 6.6 \frac{m}{s}$