Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 16

Suppose that a particle moving in a straight line has an equation of motion $\displaystyle s = t^2 - 8t + 18$,
where $s$ and $t$ are measured in meters and seconds respectively.



a.) Find the average velocity over each time interval;



$
\begin{equation}
\begin{aligned}
& (i) [3,4] && (ii) [3.5,4]\\
& (iii)[4,5] && (iv) [4,4.5]
\end{aligned}
\end{equation}
$



b.) Determine the instantaneous velocity when $t=4$.



c.) Illustrate the graph of $s$ as a function of $t$ and draw the secant lines whose slopes are the average velocities.




$\displaystyle \text{a.) } \quad \text{average velocity } = \frac{\text{displacement}}{\text{time}} = \frac{f(t_2)-f(t_1)}{t_2-t_1}$



$f(t) = t^2 - 8t +18$



$
\quad
\begin{equation}
\begin{aligned}
\\
(i) [3,4]; \text{ average velocity } &= \frac{(4^2)-8(4)+18-[(3)^2-8(3)+18]}{4-3}\\
\\
&= 16-32+18-9+24-18\\
\\
&= -1\\
\\
(ii) [3.5,4]; \text{ average velocity }&= \frac{(4)^2-8(4)+18-[(3.5)^2-8(3.5)+18]}{4-3.5}\\
\\
&= \frac{16-32+18-12.25+28-18}{0.5}\\
\\
&= \frac{-1}{2}\\
\\
(iii) [4,5]; \text{ average velocity }&= \frac{(5)^2-8(5)+18-[(4)^2-8(4)+18]}{5-4}\\
\\
&= 25-40+18-16+32-18\\
\\
&= 1
\\
(iv) [4,4.5]; \text{ average velocity }&= \frac{(4.5)^2-8(4.5)+18-[(4)^2-8(4)+18]}{4.5-4}\\
\\
&= \frac{20.25-36+18-16-32+18}{0.5}\\
\\
&= \frac{1}{2}

\end{aligned}
\end{equation}
$



$\text{b.) }$ Based from the definition,


$
\quad
\begin{equation}
\begin{aligned}
f'(a) &= \lim \limits_{h \to 0} \frac{f(a+h)-f(a)}{h}\\
s &= f(t) = t^2 - 8t +18
\end{aligned}
\end{equation}
$



$
\quad
\begin{equation}
\begin{aligned}
f'(t) & = \lim \limits_{h \to 0} \frac{(t+h)^2 - 8 (t+h) + 18 - [(t)^2-8(t) + 18]}{h}\\
f'(t) & = \lim \limits_{h \to 0} \frac{\cancel{t^2}+2th+h^2-\cancel{8t}-8h + \cancel{18} - \cancel{t^2} + \cancel{8t} - \cancel{18}}{h}\\
f'(t) & = \lim \limits_{h \to 0} \frac{\cancel{h}(2t+h-8)}{\cancel{h}}\\
f'(t) & = \lim \limits_{h \to 0} (2t+h-8)\\
f'(t) &= 2t+0-8\\
f'(t) &= 2t-8\\
\end{aligned}
\end{equation}
$



$
\quad
\text{when } t=4\\
\displaystyle
\qquad f'(t) = 2(4) - 8 = 0 \frac{m}{s}
$


$\text{c.) }$