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%\markboth{Dr. Matthew M. Conroy - University of Washington}{Dr. Matthew M. Conroy - University of Washington}


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{\LARGE{
\hfil Math 120 A Spring 2017 \hfil \pp
\hfil Mid-Term Exam Number One\hfil  \pp
\hfil April 20, 2017 \hfil \pp
\hfil Answers \hfil \pp
}}

There were two versions.

In version A, in problem 1, Tristan starts at the point $(10,-3)$.

\begin{enumerate}
\item (a) $x=10-\frac{3}{2}t, y=-3+\frac{11}{2}t$
(b) $x=5+2t, y=4+\frac{2}{3}t$ (c) $\frac{924}{641}\approx 1.441497$ hours after midnight
\item Let $D(t)$ be the distance from Anna to her starting point after $t$ hours.
Then \[
D(t) = \begin{cases}
2t & \text{ if } 0 \le t \le 3, \\
\sqrt{36+9(t-3)^2} & \text{ if } 3 \le t \le 4, \\
\sqrt{(6+5(t-4))^2+9} & \text{ if } 4 \le t \le 6.
\end{cases}
\]
\item $\frac{-4+\sqrt{51}+\sqrt{91}}{8} 
\approx 1.585102555$ hours.
\item (a) $-3$ (b) $10x+5h+1$
\end{enumerate}

In version B, in problem 1, Tristan starts at the point $(9,-1)$.
\begin{enumerate}
\item (a) $x-9-\frac{3}{2}t, y=-1+\frac{7}{2}t$
(b) $x=4+2t, y=4-t$ 
(c) $\frac{16}{13} \approx 1.23076$
\item Let $D(t)$ be the distance from Anna to her starting point after $t$ hours.
Then
\[
D(t) = \begin{cases}
4t & \text{ if } 0 \le t \le 3 \\
\sqrt{144+(6(t-3))^2} & \text{ if } 3 \le t \le 4 \\
\sqrt{(12+10(t-4))^2+36} & \text{ if } 4 \le t \le 6.
\end{cases}
\]
\item $\frac{-8+\sqrt{75}+\sqrt{91}}{12} \approx 0.8499705$ hours.
\item (a) $-2$ (b) $6x+3h+1$
\end{enumerate}
\end{document}


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