%Aaron Titus
%miSolutions M&I practice problems and solutions
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\begin{document}
\noindent
A golf ball rolls across the green toward the hole as shown below. Due to the grass and the slope of the green, it breaks toward the hole.
\begin{figure}[htbp]
\begin{center}
\includegraphics[scale=0.6]{1170002.jpg}
\caption{A golf ball rolling towards the hole.}
\label{default}
\end{center}
\end{figure}
\noindent
\href{./physlets/1170002/1170002.html}{View Simulation} \\
\noindent
The positions of the golf ball at each second as it rolls toward the hole are given below.
\begin{table}[htdp]
\caption{Positions and clock readings for the golf ball}
\begin{center}
\begin{tabular}{|c|c|}
\hline
t (s) & $\vec{r}$ (m)\\
\hline
0.0 & $\triple{-8.00}{0}{0}$\\
\hline
1.00 & $\triple{-5.00}{1.00}{0}$ \\
\hline
2.00 & $\triple{-2.88}{1.36}{0}$ \\
\hline
3.00 & $\triple{-1.60}{1.14}{0}$ \\
\hline
4.00 & $\triple{-1.10}{0.80}{0}$ \\
\hline
\end{tabular}
\end{center}
\label{default}
\end{table}
\begin{enumerate}
\item What is the average velocity of the golf ball from $t=0$ to $t=1.0$ s?
\item What is the magnitude of its average velocity and what is its direction from $t=0$ to $t=1.0$ s?
\item If the golf ball continues with the same average velocity, what position would it be located at the clock reading $t=2.0$ s?
\item How does the predicted position at $t=2.0$ s compare to the actual position? Why is the predicted and actual position different?
\end{enumerate}
\section*{Solution}
\noindent
(a) The average velocity of the golf ball between t=0 s and t=1.0 s is
\begin{eqnarray*}
\vec{v}_{avg} & = & \frac{\Delta \vec{r}}{\Delta t} \\
& = & \frac{<-5,1,0> - <-8,0,0>\ \meter}{1.00 \ \second}\\
& = & <3.00,1.00,0>\ \meter \per \second
\end{eqnarray*}
\noindent
(b) The magnitude of the average velocity is
\begin{eqnarray*}
|\vec{v}_{avg}| & = & \sqrt{v_{avg,x}^2+v_{avg,y}^2+v_{avg,z}^2} \\
& = & \sqrt{(3)^2+(1)^2+(0)^2} \\
& = & 3.16\ \meter \per \second\\
\end{eqnarray*}
\noindent
The direction is given by the unit vector.
\begin{eqnarray*}
\hat{v}_{avg} & = & \frac{\vec{v}_{avg}}{|\vec{v}_{avg}|} \\
\\
& = & \frac{<3.00,1.00,0>\ \cancel{\meter \per \second}}{3.16\ \cancel{\meter \per \second}} \\
& = & <0.949,0.0.316,0>
\end{eqnarray*}
\noindent
(c) Use the position update equation, and assume that the ball's velocity is constant from $t=1.00\ \second$ to $t=2.00\ \second$ with the same value as from $t=0.0\ \second$ to $t=1.00\ \second$. The initial position is the position at $t=1.00\ \second$. The final position is the position at $t=2.00\ \second$.
\begin{eqnarray*}
\vec{r}_f & = & \vec{r}_i+\vec{v}_{avg}\Delta t \\
& = & <-5.00,1.00,0>\ \meter + (<3.00,1.00,0>\ \meter \per \second) (1.00\ \second) \\
& = & <-2.00,2.00,0>\ \meter
\end{eqnarray*}
\noindent
A picture showing a vector diagram for the calculation of $\vec{r}_f$ at $t=2.00\ \second$ is shown below. You will notice that the predicted position and actual position at $t=2.00\ \second$ are not the same.\\
\begin{figure}[htbp]
\begin{center}
\includegraphics[scale=0.6]{1170002c-solution.jpg}
\caption{Initial position, predicted final position, and predicted displacement for the time interval from t=1.0 s to t=2.0 s.}
\end{center}
\end{figure}
\noindent
(d) $\vec{r}_{predicted}=<-2.00,2.00,0>\ \meter$ and $\vec{r}_{actual}=<-2.88,1.36,0>\ \meter$ . Though it is within a meter in the x an y directions, the predicted position and the actual position at $t=2.00\ \second$ are not equal. In fact, one can argue that they aren't even very close. \\
\noindent
The predicted position at $t=2.00\ \second$ and the actual position at $t=2.00\ \second$ are off for two reasons. \\
\noindent
(1) We are not using small enough time intervals. The approximation that the velocity is constant will given a reasonably good prediction for the final position only if the time interval is small. \\
\noindent
(2) The ball is curving toward the hole due to interactions with gravity and with the grass. Because of these interactions, the ball's velocity is not constant. Yet, in using the position update equation we assumed that the velocity was constant. If the velocity of the golf ball was constant, then the predicted and measured values for its position at $t=2.00\ \second$ would be equal. However, because the ball's velocity is not constant during the time interval from $t=1.00\ \second$ to $t=2.00\ \second$, the predicted position of the ball at $t=2.00\ \second$ is not equal to the actual position of the ball at $t=2.00\ \second$.
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