Answer
See the full explanation below.
Work Step by Step
(a)
Consider the provided distance function:
$s\left( t \right)=10{{t}^{2}}$.
The value of function $s\left( t \right)=10{{t}^{2}}$ at ${{t}_{1}}=3$:
$\begin{align}
& s\left( 3 \right)=10\cdot {{\left( 3 \right)}^{2}} \\
& =90
\end{align}$
The value of function $s\left( t \right)=10{{t}^{2}}$ at ${{t}_{2}}=\text{4}$:
$\begin{align}
& s\left( 4 \right)=10\cdot {{\left( 4 \right)}^{2}} \\
& =160
\end{align}$
The rate of change is calculated by substitution of the value as ${{t}_{1}}=3\text{ seconds to }{{t}_{2}}=4\text{ seconds}$ and $s\left( {{t}_{1}} \right)=90\text{ and }s\left( {{t}_{2}} \right)=160$.
$\begin{align}
& \frac{\Delta s}{\Delta t}=\frac{s\left( 4 \right)-s\left( 3 \right)}{4-3} \\
& =\frac{160-90}{4-3} \\
& =70\text{ feet/sec}
\end{align}$
The ball’s average velocity for ${{t}_{1}}=3\text{ seconds to }{{t}_{2}}=4\text{ seconds}$ is $70\text{feet/sec}$.
(b)
Consider the provided distance function:
$s\left( t \right)=10{{t}^{2}}$.
The value of function $s\left( t \right)=10{{t}^{2}}$ at ${{t}_{1}}=3$:
$\begin{align}
& s\left( 3 \right)=10\cdot {{\left( 3 \right)}^{2}} \\
& =90
\end{align}$
The value of function $s\left( t \right)=10{{t}^{2}}$ at ${{t}_{2}}=\text{3}\text{.5}$:
$\begin{align}
& s\left( 3.5 \right)=10\cdot {{\left( 3.5 \right)}^{2}} \\
& =122.5
\end{align}$
The rate of change is calculated by substitution of the value as ${{t}_{1}}=3\text{ seconds to }{{t}_{2}}=3.5\text{ seconds}$ and $s\left( {{t}_{1}} \right)=90\text{ and }s\left( {{t}_{2}} \right)=122.5$.
$\begin{align}
& \frac{\Delta s}{\Delta t}=\frac{s\left( 3.5 \right)-s\left( 3 \right)}{3.5-3} \\
& =\frac{122.5-90}{3.5-3} \\
& =\frac{32.5}{0.5} \\
& =65\text{ feet/sec}
\end{align}$
The ball’s average velocity for ${{t}_{1}}=3\text{ seconds to }{{t}_{2}}=3.5\text{ seconds}$ is $65\text{feet/sec}$.
(c)
Consider the provided distance function:
$s\left( t \right)=10{{t}^{2}}$.
The value of function $s\left( t \right)=10{{t}^{2}}$ at ${{t}_{1}}=3$ ;
$\begin{align}
& s\left( 3 \right)=10\cdot {{\left( 3 \right)}^{2}} \\
& =90
\end{align}$
The value of function $s\left( t \right)=10{{t}^{2}}$ at ${{t}_{2}}=\text{3}\text{.01}$ ;
$\begin{align}
& s\left( 3.01 \right)=10\cdot {{\left( 3.01 \right)}^{2}} \\
& =90.601
\end{align}$
The rate of change is calculated by substitution of the value as ${{t}_{1}}=3\text{ seconds to }{{t}_{2}}=3.01\text{ seconds}$ and $s\left( {{t}_{1}} \right)=90\text{ and }s\left( {{t}_{2}} \right)=90.601$.
$\begin{align}
& \frac{\Delta s}{\Delta t}=\frac{s\left( 3.01 \right)-s\left( 3 \right)}{3.01-3} \\
& =\frac{90.601-90}{3.01-3} \\
& =\frac{0.601}{0.01} \\
& =60.1\text{ feet/sec}
\end{align}$
The ball’s average velocity for ${{t}_{1}}=3\text{ seconds to }{{t}_{2}}=3.01\text{ seconds}$ is $60.1\text{feet/sec}$.
(d)
Consider the provided distance function:
$s\left( t \right)=10{{t}^{2}}$.
The value of function $s\left( t \right)=10{{t}^{2}}$ at ${{t}_{1}}=3$:
$\begin{align}
& s\left( 3 \right)=10\cdot {{\left( 3 \right)}^{2}} \\
& =90
\end{align}$
The value of function $s\left( t \right)=10{{t}^{2}}$ at ${{t}_{2}}=\text{3}\text{.001}$ ;
$\begin{align}
& s\left( 3.001 \right)=10\cdot {{\left( 3.001 \right)}^{2}} \\
& =90.06001
\end{align}$
The rate of change is calculated by substitution of the value as ${{t}_{1}}=3\text{ seconds to }{{t}_{2}}=3.001\text{ seconds}$ and $s\left( {{t}_{1}} \right)=90\text{ and }s\left( {{t}_{2}} \right)=90.06001$.
$\begin{align}
& \frac{\Delta s}{\Delta t}=\frac{s\left( 3.001 \right)-s\left( 3 \right)}{3.001-3} \\
& =\frac{90.06001-90}{3.001-3} \\
& =\frac{0.06001}{0.001} \\
& =60.01\text{feet/sec}
\end{align}$
The ball’s average velocity for ${{t}_{1}}=3\text{ seconds to }{{t}_{2}}=3.001\text{ seconds}$ is $60.01\text{feet/sec}$.
