Exercise 1.3.61

Question

A ship is moving at a speed of $30 \mathrm{~km} / \mathrm{h}$ parallel to a straight shoreline. The ship is $6 \mathrm{~km}$ from shore and it passes a lighthouse at noon.\begin{enumerate}[label=(\alph*)]
        \item Express the distance $s$ between the lighthouse and the ship as a function of $d$, the distance the ship has traveled since noon; that is, find $f$ so that $s=f(d)$.
        \item Express $d$ as a function of $t$, the time elapsed since noon; that is, find $g$ so that $d=g(t)$.
        \item Find $f \circ g$. What does this function represent?
    \end{enumerate}

Khagan Gulusoy · Accepted Answer

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\end{figure}\begin{enumerate}[label=(\alph*)]
\item Let $A$ be the point the ship reaches at noon. Let C be the location point of the lighthouse. Then, at the point $A$, the ship is $6 \mathrm{~km}$ away from the point $C$. Therefore, at the distance $d$ from the point $A$ the ship is $\sqrt{d^{2}+36} \ \mathrm{~km}$ away from the lighthouse. Thus, the function $f$ should be defined as follows:
$$f(d)=\sqrt{d^{2}+36} 	ext {.}$$

\item Since with every hour the distance $d$ increases by $30 \mathrm{~km}$, the function $g$ should be defined as follows:
$$g(t)=30t.$$
\item We get:
$$f \circ g(t)=f(g(t))=f(30 t)=\sqrt{900 t^{2}+36} 	ext {.}$$

The function $f\circ g$ represents the value of the distance between the lighthouse and the ship at a particular time $t$.

\end{enumerate}

Exercise 1.3.61

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Exercise 1.3.61

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