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\noindent
\vfil \noindent
\large
\hfil Math 120 C - Autumn 2009 \hfil \pp
\hfil Mid-Term Exam Number One\hfil  \pp
\hfil October 22, 2009 \hfil \pp
\hfil Answers \hfil \pp
\normalsize


There were two versions of the exam.

\bigskip

{\textbf{Version A}} - In problem 1, Pedro begins by walking north.

\begin{enumerate}

\item The distance function is
\[
d(t) =  \left \{ \begin{array}{ll}
2.5t & \mbox{if $0 \le t \le 3$} \\
7.5-4(t-3) & \mbox{if $3 \le t \le 4$} \\
\sqrt{(-5(t-4))^2+3.5^2} & \mbox{if $4 \le t \le 6$} 
\end{array}
\right.
\]

\item (a) 4.787 hours; (b) With the origin at the center of the forest: x=2.9302, y=0.4876

\item 20.66667 years after 2000

\item $\dsps x = -2+0.9t, y = -3+0.6t$

\end{enumerate}

\bigskip 

{\textbf{Version B}} - In problem 1, Pedro begins by walking east.

\begin{enumerate}

\item The distance function is
\[
d(t) = \left \{ \begin{array}{ll}
3.5t & \mbox{if $0 \le t \le 3$} \\
10.5-4(t-3) & \mbox{if $3 \le t \le 5$} \\
\sqrt{2.5^2+(2(t-5))^2} & \mbox{if $5 \le t \le 12$} 
\end{array}
\right.
\]


\item (a) 1.823 hours; (b) With the origin at the center of the forest: x=1.91329, y=1.06358

\item 7.63636 years after 2000

\item $\dsps x=-1+\frac{3}{5}t, y = -2 + \frac{1}{2} t$

\end{enumerate}



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