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\noindent
\vfil \noindent
\large
\hfil Math 120 B, C - Autumn 2008 \hfil \pp
\hfil Mid-Term Exam Number One\hfil  \pp
\hfil October 16, 2008 \hfil \pp
\hfil Answers \hfil \pp
\normalsize

There were two versions of the exam.

Version A - In problem 1, Jo initially runs SOUTH.

\begin{enumerate}

\item 
\[
D(t) = \left\{ \begin{array}{ll}
3t & \mbox{if $0 \le t \le 90$,} \\
\sqrt{270^2+(4(t-90))^2} & \mbox{if $90 \le t \le 121$,} \\
\sqrt{124^2+(270+2.5(t-121))^2} & \mbox{if $121 \le t \le 251$,}
\end{array}
\right.
\]

\item 115.698 meters

\item 1.75719 seconds

\item $8a-6$

\end{enumerate}

Version B - In problem 1, Jo initially runs WEST.

\begin{enumerate}

\item 
\[
D(t) = \left\{ \begin{array}{ll}
2t & \mbox{if $0 \le t \le 70$,} \\
\sqrt{140^2+(3(t-70))^2} & \mbox{if $70 \le t \le 103$,} \\
\sqrt{99^2+(-140-3.5(t-103))^2} & \mbox{if $103 \le t \le 188$,}
\end{array}
\right.
\]

\item 64.743 meters

\item 1.63775 seconds

\item $12a-8$


\end{enumerate}


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