Exercise 1.2.18

Question

Jade and her roommate Jari commute to work each morning, traveling west on I-10. One morning Jade left for work at 6:50 am, but Jari left 10 minutes later. Both drove at a constant speed. The graphs show the distance (in miles) each of them has traveled on I-10, $t$ minutes after 7:00 am.\begin{enumerate}[label=(\alph*)]
        \item Use the graph to decide which driver is traveling faster.
        \item Find the speed (in mi/h) at which each of them is driving.
        \item Find linear functions $f$ and $t$ that model the distances traveled by Jade and Jari as functions of $t$ (in minutes).
    \end{enumerate}

Khagan Gulusoy · Accepted Answer

\begin{enumerate}[label=(\alph*)]
    \item Let $d_1$ be the function of distance traveled Jade's train, $d_2$ - the function of distance traveled of Jari's train. The graph of $d_2$ looks steeper than the graph of $d_1$. Therefore, Jari is traveling faster than Jade.

\item To find the speed, we need to know two coordinates that belong to the graph. We need to find the distance traveled $(y_2-y_1)$ for the time $(x_2-x_1)$ and then divide the distance by the time.
    \begin{itemize}
        \item 	extit{(Jade's train speed)} We can see that points $(0,10),\ (6,16)$ lie on the graph. Therefore, we can find the speed of the Jades train as follows:
        $$v_1=\frac{16-10}{6-0}=1\mathrm{~mi}/	ext{h,}$$
        where $v_1$ is the velocity of the Jades train.
        \item 	extit{(Jari's train speed)} We know that points $(0,0),\ (6,7)$ lies on the graph. Therefore, we can find the speed of the Jaris train as follows:
        $$v_2=\frac{7-0}{6-0}=1\frac{1}{6}\mathrm{~mi}/	ext{h,}$$
        where $v_2$ is the velocity of the Jades train.
    \end{itemize}

\item For any affine function $f$ and for two points $(x_1,y_1), \ (x_2,y_2)$ belonging to its graph, this formula is true:
    \begin{equation}\label{eee}
        \forall x\in \mathbb{R}:\quad \frac{x-x_1}{x_2-x_1}=\frac{f(x)-y_1}{y_2-y_1},
    \end{equation}
    \begin{itemize}
        \item 	extit{(Jade's train)} We can see that points $(0,10)$ and $(6,16)$ belong to the graph. Thus, the upper statement translates to: $$\frac{t-0}{6-0}=\frac{d_1(t)-10}{16-10}.$$ $$\frac{t}{6}=\frac{d_1(t)-10}{6}.$$
$$t=d_1(t)-10.$$ $$d_1(t)=t+10.$$
\item 	extit{(Jari's train speed)} We can see that points $(0,0)$ and $(6,7)$ belong to the graph. Thus, the statement ef{eee} translates to: $$\frac{t-0}{6-0}=\frac{d_2(t)-0}{7-0}.$$ $$\frac{t}{6}=\frac{d_2(t)}{7}.$$ $$d_2(t)=\frac{7}{6}t.$$

\end{itemize}
\end{enumerate}

Exercise 1.2.18

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Exercise 1.2.18

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