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1. Make two graphs. In the first one, plot the following points
$
\begin{tabular}{c|c}
t & v \\
\hline
0 & 12 \\
15 & 12 \\
15 & 11 \\
30 & 11 \\
45 & 10 \\
60 & 10 \\
60 & 8 \\
75 & 8 \\
75 & 7 \\
90 & 7 \\
\end{tabular}
$
In the second, plot the following points.
$
\begin{tabular}{c|c}
t & v \\
\hline
0 & 11 \\
15 & 11 \\
15 & 10 \\
30 & 10 \\
45 & 10 \\
45 & 8 \\
60 & 8 \\
60 & 7 \\
75 & 7 \\
75 & 0 \\
90 & 0 \\
\end{tabular}
$
Connect the dots in the obvious way and you should get two stepped
graphs. $\frac{area\ of\ each\ graph}{60\ minutes}$ is the distance
travelled. The first graph represents the upper bound while the latter
is the lower bound. You will find that the distance is $\frac{1}{4}
\sum_{i=1}^6 v_i$ for the upper bound and $\frac{1}{4} \sum_{i=2}^7
v_i$ for the lower bound, where $v_i=\{12,11,10,10,8,7,0\}$.
2. If you did 1 correctly, you'll find the upper bound is 14.5 and the
lower bound is 11.5. A good estimate one might say is the average,
which assumes all the accelerations between data points were constant.
The average is merely $\frac{11.5+14.5}{2}=13$.
3. This is a really ambiguous question. If one does not make the same
assumption denoted in 1, then it is impossible. If one does make that
assumption, but does not insist that the given datapoints be part of
the speed log, then it is still impossible. So I will make the
assumption that we make the same assumption about never increasing
velocity and that the given datapoints are still valid. In this case,
we need to make a speed vs. time graph with a total drop of 12 mph.
Thus, we must have $12\left(\frac{x}{60}\right)=0.1$, where $x$ is the
number of minutes between logging of datapoints. In our case, $x=0.5$.
Since I'm not actually getting paid for this answer, I won't give any
more explanation. |