\section{Problem 1}
\subsection{Statement}
The most critical stage of a 100m race is as the athlete accelerates
out of the blocks. The athlete will be able to maintain the speed they
manage to reach in this phase. Below are the values of an athlete's
acceleration in the first 0.8 seconds of a practice run. What is the
athlete's speed 0.8 seconds into the run?
\begin{center}
\begin{tabular}{|c|c|}
\hline
\textbf{Time (sec)} & \textbf{Acceleration (m/s$^2$)}\\\hline
0 &20.4\\
0.1 & 20.3\\
0.2 & 19.4\\
0.3 & 17.3\\
0.4 & 14.0\\
0.5 & 9.8\\
0.6 & 5.3\\
0.7 & 1.6\\
0.8 & 0\\
\hline
\end{tabular}
\end{center}

\subsection{Estimation from the Data}
If we assume that the athlete maintains each of these accelerations for the full interval, then we could calculate their speed as follows:
\begin{center}
\begin{tabular}{|c|c|r|}\hline
Acceleration & Length of interval & Increase in speed\\\hline
20.4 & 0.1 & $20.4 \times 0.1 =2.04  $\\
20.3 & 0.1 & $20.3 \times 0.1 =2.03  $\\
19.4 & 0.1 & $19.4 \times 0.1 =1.94  $\\
17.3 & 0.1 & $17.3 \times 0.1 =1.73  $\\
14.0 & 0.1 & $14.0 \times 0.1 =1.40  $\\
9.8  & 0.1 & $9.8  \times 0.1 =0.98   $\\
5.3  & 0.1 & $5.3  \times 0.1 =0.53   $\\
1.6  & 0.1 & $1.6  \times 0.1 =0.16   $\\
\hline
&&\rule{0pt}{18pt}$\dps\sum_{i=0}^{8}a_i\times0.1=10.81$\\\hline
\end{tabular}
\end{center}
Each of these multiplications can be represented as an area of a rectangle and together can be draw as:
\begin{center}
\includegraphics[width=0.75\textwidth]{Problem1/M4_P1_1.eps}
\end{center}
Total area is 10.81.
\subsection{Improved approximations using a fitted model}
The athlete does not, however, maintain their acceleration over the
whole 0.1 second interval, their acceleration is continually
decreasing. If we had their acceleration every 0.05 seconds and
repeated the calculation above we would obtain a more accurate
approximation. To find the athlete's acceleration at these
intermediate times we first fit an interpolation function to the
data. If we find the best fitting model of the form $y=at^4+bt^3+c$
using the techniques of Module 3 we would find that this data is
approximated by with the function
\[
a(t)=150.51 t^4-160.28t^3+20.42
\]
So the speed would be $\dps\sum_{i=0}^{15}a(0.05\times i)\times 0.05=10.299$. This can be calculated in Mathematica with the commands:
\begin{verbatim}
acc[t_]=150.51 t^4-160.28t^3+20.42
Sum[acc[0.05*i]*0.05,{i,0,0.8/0.05-1}]
\end{verbatim}
As before, we can represent this a the sum of the areas of rectangles
(each rectangle has a height of $a(0.05\times
i)$ and a width 0.05. These are pictured below:
\begin{center}
\includegraphics[width=0.7\textwidth]{Problem1/M4_P1_2.eps}
\end{center}
It would be even more accurate if only assumed constant acceleration
for 0.01 second time intervals,that is, calculate
$\dps\sum_{i=0}^{79}a(0.01\times i)\times 0.01$. The Mathematica command is
\verb+Sum[acc[0.01*i]*0.01,{i,0,0.8/0.01-1}]+ and the result is
9.89. The illustration of this is
\begin{center}
\includegraphics[width=0.7\textwidth]{Problem1/M4_P1_3.eps}
\end{center}
Here the bars are so narrow that sum in approximately equal to the area under the curve.
\subsection{Best possible approximation}
The shorter the time interval over which we assume that the
acceleration is constant the more accurate the approximation is. So if
we take find the limit of these approximations as the interval width
tended to zero we would have best possible approximation of the
athlete's speed, and the \emph{exact} value of the area under the
curve. That is $\dps\lim_{dt\to 0}\sum_{i=0}^{0.8/dt-1}a(dt\times
i)\times dt$ in Mathematica this is calculated with
\verb+Limit[Sum[acc[dt*i]*dt,{i,0,0.8/dt-1}],dt->0]+, and the answer
we get is 9.78853. So the athletes speed, to 3 significant figures, at the end of this spurt of
acceleration is 9.79 m/s.

\subsection{Another method}
Acceleration is, by definition, the instantaneous rate of change of
velocity (speed). So if $a(t)$ represents the acceleration of a body
at time $t$ and $v(t)$ represents the speed of that body at time $t$
then
\[
\dif{v}{t}=a(t)
\]
In this problem the `body' is the athlete, we know their acceleration
($a(t)=149.54 t^4-159.51t^3+20.42$) and wish to know the athlete's
speed at $t=0.8$. In other words if we differentiated the function we
want to know about (speed) we would get $149.54 t^4-159.51t^3+20.42$
(acceleration), so we have to `undo' the differentiation that led to
$149.54 t^4-159.51t^3+20.42$. Another way to say this, is that to find
the speed we need to solve the differential equation
\[
\dif{v}{t}=150.51 t^4-160.28t^3+20.42
\]
This equation is also called the model of the problem.
Mathematica will solve this differential equation in response to the
command 
\begin{verbatim}
DSolve[v'[t]==150.51 t^4-160.28t^3+20.42,v[t],t]
\end{verbatim}
The solution it gives is $20.42t-40.07t^4+30.1t^5+C_1$, where $C_1$
denotes the arbitrary constant. We do not have to worry about the value
of this constant if we think of our problem as finding the increase in
speed from $t=0$ to $t=0.8$ then we are finding $v(0.8)-v(0)$ and this
is 
\[
(20.42\times 0.8-40.07\times 0.8^4+30.1\times 0.8^5+C_1)-(0+C_1)\]
and the value of this is 9.79, to 3 significant figures.