# [l2h] Another pstricks question

Mark marknewlyn at yahoo.co.uk
Wed Dec 1 08:10:33 CET 2004

Hi Ross

Thats a much quicker fix - thanks. I get some strange behaviour with it
though, see for example:

http://www-pdsf.nersc.gov/~mhorner/rectilinear_motion-selfcontained/node5.html

Do you know what causes the gray band on the right of the plots with,
what I think is, the word group under the plot?
I attached my latex file for the webpages above.

Thanks for all the help,

Mark

Ross Moore wrote:

>
> On 01/12/2004, at 12:38 PM, Mark Horner wrote:
>
>> Hi Ross
>>
>> Thanks very much for the explanation and solution. I have tried a
>> few  of them and things look fine.
>
>
> That's good.
>
> Try simply putting the following in the document preamble:
>
> \usepackage{html}
> \begin{imagesonly}
>   \let\PSTpspicture\pspicture
>   \let\PSTendpspicture\endpspicture
>   \def\HTMLpspicture{\begingroup\setbox0=\hbox\bgroup\PSTpspicture}
>
> \def\HTMLendpspicture{\PSTendpspicture\egroup\colorbox{white}{\box0}\end
> group}
>   \let\pspicture\HTMLpspicture
>   \let\endpspicture\HTMLendpspicture
> \end{imagesonly}
>
> This obviates the need to explicitly put in the \framebox (or
> \colorbox{white}{....} )
> around every {pspicture}.
>
> The code rebinds \pspicture to first set the picture's contents
> within  a temporary
> box-register  (in fact within \box0 ) before putting a white
> rectangle  under it,
> via the   \colorbox{white}{\box0} .
>
> It all happens inside a \begingroup .... \endgroup to preserve any
> previous
> contents of \box0 . So there may be problems with any \label, \index
> or  similar
> commands within the {pspicture} creating flags to be accessed from
> outside it.
> But that's a pretty rare thing, that probably doesn't work in
> LaTeX2HTML anyway.
>
> Also, it only happens when using LaTeX2HTML via the  {imagesonly}
> environment
> construction, from the  {html}  package.
>
>
>>
>> I really appreciate the help as you've saved me a tremendous amount
>> of  work :-D
>
>
> This little hack may save you some more. :-)
>
> Cheers
>
>     Ross
>
>
>>
>> Thanks again,
>>
>> Mark
>>
>
>

--
--
Mark Horner
Jabber/AIM/Yahoo:   marknewlyn

Co-author:
http://www.nongnu.org/fhsst
http://savannah.gnu.org/projects/fhsst

"Life is but a seg-fault away ...

Life received signal SIGSEGV, Segmentation fault.
0x42074d40 in calloc () from /lib/i686/liblife.so.6"

-------------- next part --------------
\documentclass[10pt, a4paper, titlepage]{report}
\usepackage{color}
\usepackage{ulem}
\usepackage{amsmath, amsthm, amsfonts, amssymb}
\usepackage{pst-all}
\usepackage{html}
\begin{imagesonly}
\let\PSTpspicture\pspicture
\let\PSTendpspicture\endpspicture
\def\HTMLpspicture{\begingroup\setbox0=\hbox\bgroup\PSTpspicture}
\def\HTMLendpspicture{\PSTendpspicture\egroup\colorbox{white}{\box0}\end group}
\let\pspicture\HTMLpspicture
\let\endpspicture\HTMLendpspicture
\end{imagesonly}
% Copyright (c) 2003 "Free High School Science Texts"
%===========================================================================%
%This is to remove the Underfull warnings when compiling

%===========================================================================%
% Here are the locally defined macros which are designed to make the FHSST
% authors' lives just that little bit more bearable ;-)

% Note To Self (comes up parenthesised and red)
\newcommand{\nts}[1]{(\textcolor{red}{NOTE TO SELF: #1})}
% Meters per second : \ms = ms^{-1}
\newcommand{\ms}{\mathrm{m.s^{-1}}}
% \momen = the standard symbol for momentum
\newcommand{\momen}{\overrightarrow{p}}
% \kener = the standard symbol for kinetic energy
\newcommand{\kener}{K}
%\newcommand{\kener}{{E_k}}

%===========================================================================%

% This is an environment for conveniently writing out the syllabus for a
% particular section. using \item for each relevant entry from the syllabus
\newenvironment{syllabus}{\ttfamily The syllabus requires:\begin{itemize}}{\end{itemize}}

\newcounter{pworkedexamplecounter} % physics worked example counter
\setcounter{pworkedexamplecounter}{1}
\newcounter{mworkedexamplecounter} % maths worked example counter
\setcounter{mworkedexamplecounter}{1}
\newcounter{cworkedexamplecounter} % chemistry worked example counter
\setcounter{cworkedexamplecounter}{1}
\newlength{\saveparindentlength}
\setlength{\saveparindentlength}{\parindent}
% Essay counter :)
\newcounter{essaycounter} % physics worked example counter
\setcounter{essaycounter}{1}
% My attempt to define an essay environment
% Will take 3 arguments
%   1 - Essay title
%   2 - Authors name and details block
%   3 - About the author block

\newenvironment{essay}[4]%
{\begin{center}\begin{pspicture}(-5,0)(5,0.01)
\psline[linecolor=red]{-}(5,0)(7,0)\psline[linecolor=red]{-}(7,0)(7,-1)
\psline[linecolor=red]{-}(-7,0)(-5,0)\psline[linecolor=red]{-}(-7,0)(-7,-1)
\end{pspicture}\end{center}
\vspace{.5cm}\setlength{\parindent}{5pt}
\textbf{\textit{ Essay \arabic{essaycounter}} \\ #1 \small{by #2}}%
\par\vspace{.5cm} #4 \\\par\vspace{0.25cm}\textit{\textbf{Author: #2}}\\\vspace{0.5cm}%
\small{#3}\\ %
}%
{%
\begin{center}\begin{pspicture}(0,0)(5,0.01)%
\psline[linecolor=red]{-}(0,0)(5,0)%
\end{pspicture}\end{center}%
\setlength{\parindent}{\saveparindentlength}}

% My attempt to define a worked example environment
\newcounter{stepcounter}% All this does now is write the text in bold
% and reset the step counter to 1
\newenvironment{pwex}[1]%
{\begin{center}\begin{pspicture}(0,0)(5,0.01)
\psline[linecolor=red]{-}(0,0)(5,0)
\end{pspicture}\end{center}
\begin{quotation}\noindent\vspace{.5cm}\setlength{\parindent}{0pt}
\textbf{\textit{Worked Example \arabic{pworkedexamplecounter}} \\ #1}%
\par\vspace{.5cm}}%
{\end{quotation}\begin{center}\begin{pspicture}(0,0)(5,0.01)
\psline[linecolor=red]{-}(0,0)(5,0)
\end{pspicture}\end{center}
\setlength{\parindent}{\saveparindentlength}}

\newenvironment{cwex}[1]%
{\begin{quotation}\vspace{.5cm}\setlength{\parindent}{0pt}
\textbf{\textit{Worked Example \arabic{cworkedexamplecounter}} : #1}%
\par\vspace{0.5cm}}%
{\end{quotation}\setlength{\parindent}{\saveparindentlength}}

\newenvironment{mwex}[1]%
{\begin{quotation}\vspace{.5cm}\setlength{\parindent}{0pt}
\textbf{\textit{Worked Example \arabic{mworkedexamplecounter}} : #1}%
\par\vspace{0.5cm}}%
{\end{quotation}\setlength{\parindent}{\saveparindentlength}}

% can we deprecate the use of this, and use \westep instead
\newcommand{\step}
{\par \pagebreak[1]
\textit{Step \arabic{stepcounter} : \nts{step is deprecated, use westep instead.}}\par
\newcommand{\westep}[1]
{\par \pagebreak[1]
\textit{Step \arabic{stepcounter} : #1}\par

\ifx\setlinejoinmode\undefined
\newcommand{\setlinejoinmode}[1]{}
\fi
\ifx\setlinecaps\undefined
\newcommand{\setlinecaps}[1]{}
\fi
% This way define your own fonts mapping (for example with ifthen)
\ifx\setfont\undefined
\newcommand{\setfont}[2]{}
\fi

\newcommand{\pscompass}%
{%
\pscircle[linewidth=2pt](0cm,0cm){.2}%
\pscircle*(0cm,0cm){.1}%
%North
\pspolygon(0cm,0.375cm)(0cm,0.75cm)(0.1875cm,0.1875cm)(0cm,0.375cm)
\pspolygon[fillcolor=black,fillstyle=solid](0cm,0.375cm)(0cm,0.75cm)(-0.1875cm,0.1875cm)(0cm,0.375cm)
%\pspolygon*(0,0.5)(0,1)(-.25,0.25)(0,0.5)
%South
\pspolygon[fillcolor=black,fillstyle=solid](0cm,-0.375cm)(0cm,-0.75cm)(0.1875cm,-0.1875cm)(0cm,-0.375cm)
\pspolygon(0cm,-0.375cm)(0cm,-0.75cm)(-0.1875cm,-0.1875cm)(0cm,-0.375cm)
%\pspolygon(0,-0.5)(0,-1)(-.25,-0.25)(0,-0.5)
%\pspolygon*(0,-0.5)(0,-1)(.25,-0.25)(0,-0.5)
%East
\pspolygon[fillcolor=black,fillstyle=solid](0.375cm,0cm)(0.5625cm,0cm)(0.1875cm,0.1875cm)(0.375cm,0cm)
\pspolygon(0.375cm,0cm)(0.5625cm,0cm)(0.1875cm,-0.1875cm)(0.375cm,0cm)
%\pspolygon(0.5,0)(0.75,0)(.25,-0.25)(0.5,0)
%West
\pspolygon(-0.375cm,0cm)(-0.5625cm,0cm)(-0.1875cm,0.1875cm)(-0.375cm,0cm)
\pspolygon[fillcolor=black,fillstyle=solid](-0.375cm,0cm)(-0.5625cm,0cm)(-0.1875cm,-0.1875cm)(-0.375cm,0cm)

%\pspolygon[fillcolor=black](-0.5,0)(-0.75,0)(-.25,0.25)(-0.5,0)
%\pspolygon*[fillcolor=black](-0.5,0)(-0.75,0)(-.25,-0.25)(-0.5,0)
\rput(0cm,1.3cm){\small{N}}
\rput(0cm,-1.3cm){\small{S}}
\rput(1.05cm,0cm){\small{E}}
\rput(-1.05cm,0cm){\small{W}}
}

\newcommand{\psrings}%
{%
\pscircle*[linewidth=0.5pt](0cm,0cm){.01}
\pscircle[linewidth=.5pt,linecolor=gray](0cm,0cm){.25}%
\pscircle[linewidth=.5pt,linecolor=gray](0cm,0cm){.5}%
\pscircle[linewidth=.5pt,linecolor=gray](0cm,0cm){0.75}%
\pscircle[linewidth=.5pt,linecolor=gray](0cm,0cm){1}%
}
\newcommand{\psintfact}%
{%
\rput(0,0){\begin{tabular}{c}\textbf{In$\ \$teresti$\$ing}\\ \textbf{Fact:}%
\end{tabular}}
}

\newcommand{\psfact}%
{%
\begin{pspicture}(-.5,-.5)(.5,.5)
\psintfact
\PstLens[LensSize=.85,LensRotation=-60,LensMagnification=1.3](-.04,-.1){\psintfact}
\end{pspicture}
}

%Defining our own aside environment!
\newenvironment{IFact}[1]%
{\vspace{3mm}\sffamily
\begin{center}\begin{pspicture}(0,0)(5,0.01)
\psline[linecolor=red]{-}(0,0)(5,0)
\end{pspicture}
\begin{minipage}[t]{0.85\textwidth}%
\rput(-1.5,0){\psfact} #1%
\end{minipage}%
}%}
{\begin{pspicture}(0,0)(5,0.01)
\psline[linecolor=red]{-}(0,-0.1)(5,-0.1)
\end{pspicture}\end{center}\vspace{3mm}\fontfamily{\familydefault}}%

%Defining our own aside environment!
\newenvironment{Aside}[1]%
\begin{minipage}[t]{0.85\textwidth}%
\textit{Aside:} #1%
\end{minipage}%
}} {\vspace{3mm}}%

%%Defining variables for random dot placement in pstricks (chem)
%\newdimen\XPos
%\newdimen\YPos
%\newdimen\ZPos

%===========================================================================%
% These next 3 lines are so that Emacs can view this file individually
% by using the C-c C-b' command of AUCTeX. C-c C-c' will act on the
% top level file (full book). See the file SETUP to see how to get
% this feature (and many others) working on your version of GNU
% Emacs. SETUP also contains install instructions for a lot of the
% extra adon LaTeX packages which FHSST makes use of.
%
%%% Local Variables:
%%% TeX-master: "main.tex"
%%% TeX-master: "main"
%%% End:
\begin{document}

% state: editable by all, maintained by Mark Horner <marknewlyn at yahoo.co.uk>

\chapter{Rectilinear Motion}
\label{pr}

\section{What is rectilinear motion?}

Rectilinear motion means motion along a straight line.  This is a useful topic to study for learning how to describe the movement of cars along a straight road or of trains along straight railway tracks.  In this section you have only 2 directions to worry about: (1) along the direction of motion, and (2) opposite to the direction of motion.\\

\noindent
To illustrate this imagine a train heading east.
\begin{center}
\begin{pspicture}(-3.0,-1.5)(3.0,1.5)
\psline[linewidth=2pt]{->}(0.6,0)(1.0,0)
\psline[linewidth=1pt]{-}(-2.0,-0.5)(-1.9,-0.4)
\psline[linewidth=1pt]{-}(-1.8,-0.5)(-1.7,-0.4)
\psline[linewidth=1pt]{-}(-1.6,-0.5)(-1.5,-0.4)
\psline[linewidth=1pt]{-}(-1.4,-0.5)(-1.3,-0.4)
\psline[linewidth=1pt]{-}(-1.2,-0.5)(-1.1,-0.4)
\psline[linewidth=1pt]{-}(-2.0,-0.4)(1.0,-0.4)
\psline[linewidth=1pt]{-}(-1.1,-0.5)(0.9,-0.5)
\psline[linewidth=1pt]{-}(-1.0,-0.5)(-0.9,-0.4)
\psline[linewidth=1pt]{-}(-0.8,-0.5)(-0.7,-0.4)
\psline[linewidth=1pt]{-}(-0.6,-0.5)(-0.5,-0.4)
\psline[linewidth=1pt]{-}(-0.4,-0.5)(-0.3,-0.4)
\psline[linewidth=1pt]{-}(-0.2,-0.5)(-0.1,-0.4)
\psline[linewidth=1pt]{-}(0,-0.5)(0.1,-0.4)
\psline[linewidth=1pt]{-}(0.2,-0.5)(0.3,-0.4)
\psline[linewidth=1pt]{-}(0.4,-0.5)(0.5,-0.4)
\psline[linewidth=1pt]{-}(0.6,-0.5)(0.7,-0.4)
\psline[linewidth=1pt]{-}(0.8,-0.5)(0.9,-0.4)
\psline[linewidth=1pt]{-}(-2.0,-0.5)(-2.0,-0.8)
\psline[linewidth=1pt]{-}(-0.9,-0.5)(-0.9,-0.8)
\psline[linewidth=1pt]{-}(-2,-0.5)(-0.9,-0.5)
\pspolygon[linewidth=1pt](-0.6,0.4)(0.6,0.4)(0.6,-0.4)(-0.6,-0.4)
\rput(0.0,0.0){Train} \rput(2.7,0){\pscompass} \rput(-1.5,-0.7){P}
\end{pspicture}
\end{center}

If it is accelerating away from the station platform (P), the direction of acceleration is the same as the direction of the train's velocity - east.  If it is braking the direction of acceleration is opposite to the direction of its motion, i.e. west.

\section{Speed and Velocity}

Let's take a moment to review our definitions of velocity and speed by looking at the worked example below:
\pagebreak
\begin{pwex}{Speed and Velocity}

\begin{center}
\begin{pspicture}(4.9,1)(9.1,6.1)
\psline[linestyle=dashed,arrowscale=2]{->}(5,2)(8,6)
\pcline[offset=12pt]{|-|}(5,2)(8,6)
%\lput*{:U}{50m}
\psline[arrowscale=2]{->}(5,2)(8,2)
\pcline[offset=12pt]{|-|}(5,1)(8,1)
\lput*{:U}{30m}
\psline[arrowscale=2]{->}(8,2)(8,6)
\pcline[offset=12pt]{|-|}(9,2)(9,6)
\rput(4.75,1.75){A}
\rput(8.25,1.75){B}
\rput(8.25,6.25){C}
\lput*{:U}{40m}
%Right angle in corner (8,2)
\psline[linestyle=dashed,arrowscale=2](7.7,2)(7.7,2.3)
\psline[linestyle=dashed,arrowscale=2](7.7,2.3)(8,2.3)
\end{pspicture}
\end{center}

\textbf{Question:} A cyclist moves from A through B to C in 10 seconds.  Calculate both his speed and his velocity.
\step  Analyse the question to determine what is given. The question
explicitly gives
\begin{itemize}
\item the distance between A and B
\item the distance between B and C
\item the total time for the cyclist to go from A through B to C
\end{itemize}
all in the correct units! \marginpar{Remember to check the units!}

\step What is being asked? We are asked to calculate the average speed and the average velocity of the cyclist.

His speed - a scalar -  will be
\begin{eqnarray*}
v&=&\frac{s}{t}\\
&=&\frac{30m+40m}{10s}\\
&=&7\frac{m}{s}
\end{eqnarray*}

Since velocity is a vector we will first need to find the resultant displacement of the cyclist.  His velocity will be
\begin{displaymath}
\overrightarrow{v}=\frac{\overrightarrow{s}}{t}
\end{displaymath}

The total displacement is the vector from A to C, and this is just the resultant of the two displacement vectors, ie.
\begin{displaymath}
\overrightarrow{s} = \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}
\end{displaymath}
\noindent Using the rule of Pythagoras:
\begin{eqnarray*}
\overrightarrow{s} &=& \sqrt{{(30m)}^2+{(40m)}^2}\\
&=& 50m \ in \ the \ direction \ from \ A \ to \ C\\
\therefore \overrightarrow{v} &=&\frac{50m}{10s}\\
&=& 5 \frac{m}{s} \ in\ the \ direction \ from \ A \ to \ C
\end{eqnarray*}

\end{pwex}

For this cyclist, his velocity is not the same as his speed because there has been a change in the direction of his motion.  If the cyclist traveled directly from A to C {\it without} passing through B his speed would be
\begin{eqnarray*}
v &=& \frac{50m}{10s}\\
&=& 5 \frac{m}{s}
\end{eqnarray*}

and his velocity would be
\begin{eqnarray*}
\overrightarrow{v} &=& \frac{50m}{10s}\\
&=& 5 \frac{m}{s} \ in \ the\  direction \ from \ A \ to \ C
\end{eqnarray*}

In this case where the cyclist is not undergoing any change of direction (ie. he is traveling in a straight line) the magnitudes of the speed and the velocity are the same.  This is the defining principle of rectilinear motion.

\begin{center}
\begin{tabular}{c}
\textbf{Important:}
For motion along a \emph {straight line} the magnitudes of speed and\\velocity are the same, and the magnitudes of the distance and displacement are the same.
\end{tabular}
}
\end{center}

\section{Graphs}
In physics we often use graphs as important tools for picturing certain concepts.  Below are some graphs that help us picture the concepts of displacement, velocity and acceleration.

\subsection{Displacement-Time Graphs}
Below is a graph showing the displacement of the cyclist from A to C:
\begin{center}
\begin{pspicture}(-0.5,-0.5)(2.5,5.5)
\psaxes[labels=none, ticks=none]{->}(2.5,5.2)
\psline[linewidth=1pt]{-}(0,0)(2,5)
\psline[linewidth=1pt,linestyle=dashed]{-}(1,2.5)(1.5,2.5)(1.5,3.75)
\rput(1.25,2.2){$\Delta t$}
\rput(1.9,3.1){$\Delta \overrightarrow{s}$}
\rput(-0.2,0){A} \rput(2.2,5){C} \rput(2,-0.2){10} \rput(-0.3,5){50}
\rput{90}(-0.3,2.5){displacement (\emph{m})}
\rput(1.2,-0.3){time (\emph{s})}
\end{pspicture}
\end{center}

This graphs shows us how, in 10 seconds time, the cyclist has moved from A to C.  We know the gradient (slope) of a graph is defined as the change in y divided by the change in x, i.e $\frac{\Delta y}{\Delta x}$.  In this graph the gradient of the graph is just $\frac{\Delta \overrightarrow{s}}{\Delta t}$ - and this is just the expression for velocity.

\begin{center}
\begin{tabular}{c}
\textbf{Important:}
The slope of a displacement-time graph gives the velocity.
\end{tabular}
}
\end{center}
The slope is the same all the way from A to C, so the cyclist's velocity is constant over the entire displacement he travels.

\par
In figure {\ref{fig:displacement_time_graphs} are examples of the displacement-time graphs you will encounter.

\begin{figure}[!h] %use a figure to group them all and give one caption and label
\begin{tabular}{ccc}
\begin{pspicture}(-0.5,-0.5)(3,3)
\psaxes[labels=none,ticks=none]{->}(3,3)
\psline(0,1)(3,1)
\rput{90}(-0.3,1.5){displacement}
\rput(1.5,-0.3){time}
\rput(-0.3,-0.3){a)}
\end{pspicture}
&
\begin{pspicture*}(-0.5,-0.5)(3,3)
\psaxes[labels=none,ticks=none]{->}(3,3)
\psline(0,0)(3,2.5)
\rput{90}(-0.3,1.5){displacement}
\rput(1.5,-0.3){time}
\rput(-0.3,-0.3){b)}
\end{pspicture*}
&
\begin{pspicture*}(-0.5,-0.5)(3,3)
\psaxes[labels=none,ticks=none]{->}(3,3)
\psplot[plotstyle=curve]{0}{3}{x x mul}
\rput{90}(-0.3,1.5){displacement}
\rput(1.5,-0.3){time}
\rput(-0.3,-0.3){c)}
\end{pspicture*}

\end{tabular}
\caption{Some common displacement-time graphs:}\label{fig:displacement_time_graphs}
\end{figure}

\begin{enumerate}
\item[a)] shows the graph for an object stationary over a period of time.  The gradient is zero, so the object has zero velocity.
\item[b)] shows the graph for an object moving at a constant velocity.  You can see that the displacement is increasing as time goes on.  The gradient, however, stays constant (remember: its the slope of a straight line), so the velocity is constant.  Here the gradient is positive, so the object is moving in the direction we have defined as positive.
\item[c)] shows the graph for an object moving at a constant acceleration.  You can see that both the displacement and the velocity (gradient of the graph) increase with time.  The gradient is increasing with time, thus the velocity is increasing with time and the object is accelerating.
\end{enumerate}

\subsection{Velocity-Time Graphs}

Look at the velocity-time graph below:

\begin{center}
\begin{pspicture}(-0.5,-0.5)(3.5,3.5)
\psaxes[labels=none, ticks=none]{->}(3.5,3.5)
\psline[linewidth=1pt]{-}(0,0)(3,3)
\psline[linewidth=1pt,linestyle=dashed]{-}(1,1)(2,1)(2,2)
\rput(1.5,0.7){$\Delta t$}
\rput(2.3,1.5){$\Delta \overrightarrow{v}$}
\rput(-0.2,0){A} \rput(3.2,3){B}
\rput{90}(-0.3,1.5){velocity}
\rput(1.5,-0.3){time}
\end{pspicture}
\end{center}

This is the velocity-time graph of a cyclist traveling from A to B at a constant acceleration, i.e. with steadily increasing velocity.  The gradient of this graph is just $\frac{\Delta \overrightarrow{v}}{\Delta t}$ - and this is just the expression for acceleration.  Because the slope is the same at all points on this graph, the acceleration of the cyclist is constant.

\begin{center}
\begin{tabular}{c}
\textbf{Important:}
The slope of a  velocity-time graph gives the acceleration.
\end{tabular}
}
\end{center}

Not only can we get the acceleration of an object from its velocity-time graph, but we can also get some idea of the displacement traveled.  Look at the graph below:

\begin{center}
\begin{pspicture}(-1.0,-1.0)(3.5,4)
\psset{yunit=.6cm}
\psaxes[labels=none, ticks=none]{->}(3.5,6)
\psline[linewidth=1pt]{-}(0,5)(3,5)
\psline[linewidth=1pt,linestyle=dashed]{-}(3,0)(3,5)
\rput(-0.4,5){10} \rput(3.2,-0.3){5}
\rput{90}(-0.3,2.5){velocity (\emph{m/s})}
\rput(1.5,-0.3){time (\emph{s})}
\end{pspicture}
\end{center}

This graph shows an object moving at a constant velocity of $10m/s$ for a duration of 5$s$.  The area between the graph and the time axis (the \nts{SHADED} area) of the above plot will give us the displacement of the object during this time.  In this case we just need to calculate the area of a rectangle with width 5\emph{s} and height 10\emph{m/s}

\begin{eqnarray*}
\mathrm{area \ of \ rectangle} &=& \mathrm{height} \times \mathrm{width}\\
&=& \overrightarrow{v} \times t\\
&=& 10 \frac{m}{s} \times 5 s\\
&=& 50 m\\
&=& \overrightarrow{s}= \mathrm{displacement}\\
\end{eqnarray*}

So, here we've shown that an object traveling at 10\emph{m/s} for 5\emph{s} has undergone a displacement of 50\emph{m}.

\begin{center}
\begin{tabular}{c}
\textbf{Important:}
The area between a velocity-time graph and the time'\\axis gives the displacement of the object.
\end{tabular}
}
\end{center}

Here are a couple more velocity-time graphs to get used to:
\begin{center}
\begin{figure}[!h] %use a figure to group them all and give one caption and label
\begin{tabular}{cc}
\begin{pspicture*}(-0.5,-0.5)(3,3)
\psaxes[labels=none,ticks=none]{->}(3,3)
\psline(0,2)(3,2)
\rput{90}(-0.3,1.5){velocity}
\rput(1.5,-0.3){time}
\rput(-0.3,-0.3){a)}
\end{pspicture*}
&
\begin{pspicture*}(-0.5,-0.5)(3,3)
\psaxes[labels=none,ticks=none]{->}(3,3)
\psline(0,2.8)(2.8,0)
\rput{90}(-0.3,1.5){velocity}
\rput(1.5,-0.3){time}
\rput(-0.3,-0.3){b)}
\end{pspicture*}

\end{tabular}
\caption{Some common velocity-time graphs:}\label{fig:velocity_time_graphs}
\end{figure}
\end{center}

\par
In figure {\ref{fig:velocity_time_graphs} are examples of the displacement-time graphs you may encounter.
\begin{enumerate}
\item[a)] shows the graph for an object moving at a constant velocity over a period of time.  The gradient is zero, so the object is not accelerating.
\item[b)] shows the graph for an object which is decelerating.  You can see that the velocity is decreasing with time.  The gradient, however, stays constant (remember: its the slope of a straight line), so the acceleration is constant.  Here the gradient is negative, so the object is accelerating in the opposite direction to its motion, hence it is decelerating.
\end{enumerate}

\subsection{Acceleration-Time Graphs}

In this chapter on rectilinear motion we will only deal with objects moving at a constant acceleration, thus all acceleration-time graphs will look like these two:

\begin{figure}[!h]
\begin{tabular}{cc}
\begin{pspicture*}(-0.5,-0.5)(3.5,3.5)
\psaxes[labels=none, ticks=none]{->}(3.5,3.5)
\psline[linewidth=2pt]{-}(0,0)(2.5,0)
%\rput(-0.2,0){A} \rput(3.2,3){B}
\rput{90}(-0.3,1.5){acceleration}
\rput(1.5,-0.3){time}
\rput(-0.3,-0.3){a)}
\end{pspicture*}

\begin{pspicture*}(-0.5,-0.5)(3.5,3.5)
\psaxes[labels=none, ticks=none]{->}(3.5,3.5)
\psline[linewidth=1pt]{-}(0,1)(2.8,1)
%\rput(-0.2,0){A} \rput(3.2,3){B}
\rput{90}(-0.3,1.5){acceleration}
\rput(1.5,-0.3){time}
\rput(-0.3,-0.3){b)}
\end{pspicture*}
\end{tabular}
\end{figure}

Here is a description of the graphs below:
\begin{enumerate}
\item[a)] shows the graph for an object which is either stationary or traveling at a constant velocity.  Either way, the acceleration is zero over time.
\item[b)] shows the graph for an object moving at a constant acceleration.  In this case the acceleration is positive - remember that it can also be negative.
\end{enumerate}

We can obtain the velocity of a particle at some given time from an acceleration time graph - it is just given by the area between the graph and the time-axis.  In the graph below, showing an object at a constant positive acceleration, the increase in velocity of the object after 2 seconds corresponds to the \nts{shaded} portion.
\begin{center}
\begin{pspicture*}(-1,-1)(3.5,3.5)
\psaxes[labels=none, ticks=none]{->}(3.5,3.5)
\psline[linewidth=1pt]{-}(0,1)(2.8,1)
\psline[linewidth=1pt,linestyle=dashed]{-}(2,0)(2,1)
\rput(-0.2,1){5} \rput(2,-0.2){2}
\rput{90}(-0.6,1.5){acceleration ($m/s^2$)}
\rput(1.5,-0.6){time (\emph{s})}
%\rput(-0.3,-0.3){b)}
\end{pspicture*}
\end{center}

\begin{eqnarray*}
\mathrm{area \ of \ rectangle} &=& \overrightarrow{a} \times t \\
&=& 5 \frac{m}{s^2} \times 2s\\
&=& 10 \frac{m}{s}\\
&=& \overrightarrow{v}\\
\end{eqnarray*}

Its useful to remember the set of graphs below when working on problems.  Figure \ref{fig:relation} shows how displacement, velocity and time relate to each other.  Given a displacement-time graph like the one on the left, we can plot the corresponding velocity-time graph by remembering that the slope of a displacement-time graph gives the velocity.  Similarly, we can plot an acceleration-time graph from the gradient of the velocity-time graph.
\pagebreak

\begin{figure} %use a figure to group them all and give one caption and label
\begin{tabular}{ccccc}
\begin{pspicture*}(-0.,-2.)(3,3) %asterisk means clipping is on!
\psaxes[labels=none](3,3)
\parabola(3,3)(0.0,0.0)
\rput(1.5,-1){Displacement}
\end{pspicture*}
&
\begin{pspicture}(-0.,-2.)(1,3)
\psline[linewidth=.1cm]{<->}(0,1.5)(1,1.5)
\psline[linewidth=.1cm]{<->}(0,-1)(1,-1)
\end{pspicture}
&
\begin{pspicture*}(-0.,-2.)(3,3)
\psaxes[labels=none](3,3)
\psline(0,0.0)(3,3)
\rput(1.5,-1){Velocity}
\end{pspicture*}
&
\begin{pspicture*}(-0.,-2.)(1,3)
\psline[linewidth=.1cm]{<->}(0,1.5)(1,1.5)
\psline[linewidth=.1cm]{<->}(0,-1)(1,-1)
\end{pspicture*}
&
\begin{pspicture*}(-0.,-2.)(3,3)
\psaxes[labels=none](3,3)
\psline(0,1.5)(3,1.5)
\rput(1.5,-1){Acceleration}
\end{pspicture*}
\end{tabular}
%\caption{The Relationship Between Common Graphs}\label{fig:relation}
\caption{A Relationship Between Displacement, Velocity and Acceleration}
\label{fig:relation}
\end{figure}

\subsection{Worked Examples}

\begin{pwex}{Relating displacement-, velocity-, and acceleration-time graphs}
\textbf{Question:} Given the displacement-time graph below, draw the corresponding velocity-time and acceleration-time graphs, and then describe the motion of the object.

\begin{pspicture}(-3,-2)(5,5)
\psset{yunit=0.5cm}
\psaxes[labels=none]{->}(-2,-1)(5,7)
\psline[linewidth=1pt]{-}(-2,0)(0,0)
\psplot[linewidth=1pt,plotstyle=curve]{0}{2}{x 2 exp}
\psline[linewidth=1pt]{-}(2,4)(4,5)
\psline[linewidth=1pt,linestyle=dashed]{-}(0,-1)(0,0)
\psline[linewidth=1pt,linestyle=dashed]{-}(4,-1)(4,5)
\psline[linewidth=1pt,linestyle=dashed]{-}(2,-1)(2,4)
\rput{90}(-2.5,3.5){displacement (\emph{m})}
\rput(1,-1.7){time (\emph{s})}
\rput(-2,-1.5){0}
\rput(0,-1.5){2}
\rput(2,-1.5){4}
\rput(4,-1.5){6}
\end{pspicture}

\step  Analyse the question to determine what is given. The question
explicitly gives a displacement-time graph.
3 things:
\begin{enumerate}
\item Draw a velocity-time graph
\item Draw an acceleration-time graph
\item Describe the behaviour of the object
\end{enumerate}

Lets start by working on the velocity-time graph.
For the first 2 seconds we can see that the displacement remains constant - so the object is not moving, thus it has zero velocity during this time.  We can reach this conclusion by another path too:  remember that the gradient of a displacement-time graph is the velocity.  For the first 2 seconds we can see that the displacement-time graph is a horizontal line, ie. it has a gradient of zero.  Thus the velocity during this time is zero and the object is stationary.

For the next 2 seconds, displacement is increasing with time so the object is moving.  Looking at the gradient of the displacement graph we can see that it is not constant.  In fact, the slope is getting steeper (the gradient is increasing) as time goes on.  Thus, remembering that the gradient of a displacement-time graph is the velocity, the velocity must be increasing with time during this phase.

For the final 2 seconds we see that displacement is still increasing with time, but this time the gradient is constant, so we know that the object is now travelling at a constant velocity, thus the velocity-time graph will be a horizontal line during this stage.

So our velocity-time graph looks like this one below.  Because we haven't been given any values on the vertical axis of the displacement-time graph, we cannot figure out what the exact gradients are and hence what the values of the velocity are.  In this type of question it is just important to show whether velocities are positive or negative, increasing, decreasing or constant.

\begin{pspicture}(-3,-2)(5,3.5)
\psset{yunit=0.5cm}
\psaxes[labels=none]{->}(-2,-1)(5,5)
\psline[linewidth=2pt]{-}(-2,-1)(0,-1)
\psline[linewidth=1pt]{-}(0,-1)(2,3)
\psline[linewidth=1pt]{-}(2,3)(4,3)
%\psline[linewidth=1pt,linestyle=dashed]{-}(0,-1)(0,0)
\psline[linewidth=1pt,linestyle=dashed]{-}(4,-1)(4,3)
\psline[linewidth=1pt,linestyle=dashed]{-}(2,-1)(2,3)
\rput{90}(-2.5,2.5){velocity (\emph{m/s})}
\rput(1,-1.7){time (\emph{s})}
\rput(-2,-1.5){0}
\rput(0,-1.5){2}
\rput(2,-1.5){4}
\rput(4,-1.5){6}
\end{pspicture}

Once we have the velocity-time graph its much easier to get the acceleration-time graph as we know that the gradient of a velocity-time graph is the just the acceleration.

For the first 2 seconds the velocity-time graph is horizontal at zero, thus it has a gradient of zero and there is no acceleration during this time.  (This makes sense because we know from the displacement time graph that the object is stationary during this time, so it can't be accelerating).

For the next 2 seconds the velocity-time graph has a positive gradient.  This gradient is not changing (i.e. its constant) throughout these 2 seconds so  there must be a constant positive acceleration.

For the final 2 seconds the object is traveling with a constant velocity.  During this time the gradient of the velocity-time graph is once again zero, and thus the object is not accelerating.

The acceleration-time graph looks like this:

\begin{pspicture}(-3,-2)(5,3.5)
\psset{yunit=0.5cm}
\psaxes[labels=none]{->}(-2,-1)(5,5)
\psline[linewidth=2pt]{-}(-2,-1)(0,-1)
\psline[linewidth=1pt]{-}(0,2)(2,2)
\psline[linewidth=2pt]{-}(2,-1)(4,-1)
%\psline[linewidth=1pt,linestyle=dashed]{-}(0,-1)(0,0)
\psline[linewidth=1pt,linestyle=dashed]{-}(0,-1)(0,2)
\psline[linewidth=1pt,linestyle=dashed]{-}(2,-1)(2,2)
\rput{90}(-2.5,2.5){acceleration (\emph{$m/s^2$})}
\rput(1,-1.7){time (\emph{s})}
\rput(-2,-1.5){0}
\rput(0,-1.5){2}
\rput(2,-1.5){4}
\rput(4,-1.5){6}
\end{pspicture}

A brief description of the motion of the object could read something like this: At $t=0s$ and object is stationary at some position and remains stationary until $t=2s$ when it begins accelerating.  It accelerates in a positive direction for 2 seconds until $t=4s$ and then travels at a constant velocity for a further 2 seconds.
\end{pwex}

\begin{pwex}{Calculating distance from a velocity-time graph}
\textbf{Question: }The velocity-time graph of a car is plotted below.  Calculate the displacement of the car has after 15 seconds.

\begin{pspicture}(-2,-3)(17,5)
\psset{xunit=0.5cm,yunit=0.8cm}
\psaxes[labels=none]{->}(0,0)(0,-2.5)(16,6)
\psline[linewidth=1pt]{-}(0,0)(5,4)
\psline[linewidth=1pt]{-}(5,4)(12,4)
\psline[linewidth=1pt]{-}(12,4)(15,-2)
\psline[linewidth=1pt,linestyle=dashed]{-}(5,0)(5,4)
\psline[linewidth=1pt,linestyle=dashed]{-}(12,0)(12,4)
\psline[linewidth=1pt,linestyle=dashed]{-}(15,0)(15,-2)
\rput{90}(-1,1){velocity (\emph{m/s})}
\rput(8,-0.5){time (\emph{s})}
\rput(-0.5,4){4}
\rput(-0.5,-2){-2}
\rput(5,-0.5){5}
\rput(12,-0.5){12}
\rput(15,0.5){15}
\end{pspicture}

We are asked to calculate the displacement of the car.  All we need to remember here is that the area between the velocity-time graph and the time axis gives us the displacement.

For $t=0s$ to $t=5s$ this is the triangle on the left:
\begin{eqnarray*}
Area \triangle &=& \frac{1}{2}b \times h\\
&=& \frac{1}{2}5s\times 4m/s\\
&=&10m
\end{eqnarray*}

For $t=5s$ to $t=12s$ the displacement is equal to the area of the rectangle
\begin{eqnarray*}
Area \Box &=& w \times h\\
&=&7s \times 4m/s\\
&=&28m
\end{eqnarray*}

For $t=12s$ to $t=14s$ the displacement is equal to the area of the triangle above the time axis on the right
\begin{eqnarray*}
Area \triangle &=& \frac{1}{2}b \times h\\
&=& \frac{1}{2}2s \times 4m/s\\
&=&4m
\end{eqnarray*}

For $t=14s$ to $t=15s$ the displacement is equal to the area of the triangle below the time axis
\begin{eqnarray*}
Area \triangle &=& \frac{1}{2}b \times h\\
&=& \frac{1}{2}1s \times 2m/s\\
&=&1m
\end{eqnarray*}

Now the total displacement of the car is just the sum of all of these areas.  HOWEVER, because in the last second (from $t=14s$ to $t=15s$) the velocity of the car is negative, it means that the car was going in the opposite direction, i.e. back where it came from!  So, to get the total displacement, we have to add the first 3 areas (those with positive displacements) and subtract the last one (because it signifies a displacement in the opposite direction).
\begin{eqnarray*}
\overrightarrow{s}&=&10 +28 +4 -1\\
&=&41m\ in\ the\ positive\ direction
\end{eqnarray*}
\end{pwex}

\begin{pwex}{Velocity from a displacement-time graph}
\textbf{Question: }Given the diplacement-time graph below,
\begin{enumerate}
\item what is the velocity of the object during the first 4 seconds?
\item what is the velocity of the object from $t=4s$ to $t=7s$?
\end{enumerate}

\begin{pspicture}(-2,-1)(8,3.5)
\psset{xunit=0.75cm,yunit=0.5cm}
\psaxes[labels=none,ticks=none]{->}(0,0)(8,5)
\psline[linewidth=1pt]{-}(0,0)(4,4)
\psline[linewidth=1pt]{-}(4,4)(7,4)
%\psline[linewidth=1pt]{-}(12,4)(15,-2)
\psline[linewidth=1pt,linestyle=dashed]{-}(4,0)(4,4)
%\psline[linewidth=1pt,linestyle=dashed]{-}(12,0)(12,4)
%\psline[linewidth=1pt,linestyle=dashed]{-}(15,0)(15,-2)
\rput{90}(-1,2.5){s (\emph{m})}
\rput(3,-0.7){t (\emph{s})}
\rput(-0.5,4){2}
\rput(4,-0.3){4}
\rput(7,-0.3){7}
%\rput(12,-0.5){12}
%\rput(15,0.5){15}
\end{pspicture}

\begin{enumerate}
\item The velocity is given by the slope of a displacement-time graph.  During the first 4 seconds, this is
\begin{eqnarray*}
\overrightarrow{v}&=&\frac{\Delta s}{\Delta t}\\
&=&\frac{2m}{4s}\\
&=&0.5m/s
\end{eqnarray*}
\item For the last 3 seconds we can see that the displacement stays constant, and that the gradient is zero.  Thus $\overrightarrow{v}=0m/s$
\end{enumerate}
\end{pwex}

\begin{pwex}{From an acceleration- to a velocity-time graph}
\textbf{Question:  }Given the acceleration-time graph below, assume that the object starts from rest and draw its velocity-time graph.
\begin{center}
\begin{pspicture}(-1,-3)(7,3)
%\psset{xunit=0.75cm,yunit=0.5cm}
\psaxes[labels=none,ticks=none]{->}(0,0)(0,-2.5)(7,2.5)
\psline[linewidth=1pt]{-}(0,2)(2,2)
\psline[linewidth=2pt]{-}(2,0)(4,0)
\psline[linewidth=1pt]{-}(4,-2)(6,-2)
\psline[linewidth=1pt,linestyle=dashed]{-}(2,0)(2,2)
\psline[linewidth=1pt,linestyle=dashed]{-}(4,0)(4,-2)
\psline[linewidth=1pt,linestyle=dashed]{-}(6,0)(6,-2)
\rput{90}(-0.7,0){a ($m/s^2$)}
\rput(6,0.3){t (\emph{s})}
\rput(-0.5,2){2}
\rput(-0.5,-2){-2}
\rput(2,-0.3){2}
\rput(4,0.3){4}
\rput(6.1,-0.3){6}
\end{pspicture}
\end{center}

\begin{center}
\begin{pspicture}(-1,-1)(7,4)
\psset{yunit=0.75cm}
\psaxes[labels=none,]{->}(0,0)(7,5)
\psline[linewidth=1pt]{-}(0,0)(2,4)
\psline[linewidth=1pt]{-}(2,4)(4,4)
\psline[linewidth=1pt]{-}(4,4)(6,0)
%\psline[linewidth=1pt,linestyle=dashed]{-}(2,0)(2,4)
%\psline[linewidth=1pt,linestyle=dashed]{-}(4,0)(4,-2)
%\psline[linewidth=1pt,linestyle=dashed]{-}(6,0)(6,-2)
\rput{90}(-0.5,2.5){v ($m/s$)}
\rput(3.5,-0.6){t (\emph{s})}
\rput(-0.4,4){4}
%\rput(-0.4,-2){-2}
\rput(2,-0.3){2}
\rput(4,-0.3){4}
\rput(6,-0.3){6}
\end{pspicture}
\end{center}

\end{pwex}

\pagebreak
\section{Equations of Motion}

This section is about solving problems relating to uniformly accelerated motion.  We'll first introduce the variables and the equations, then we'll show you how to derive them, and after that we'll do a couple of examples.

\begin{eqnarray*}
u &=& \mbox{starting velocity (m/s) at}\ t = 0 \\
v &=& \mbox{final velocity (m/s) at time}\ t \\
s &=& \mbox{displacement (m)} \\
t &=& \mbox{time (s)} \\
a &=& \mbox{acceleration (m/s$^2$)}\\
\end{eqnarray*}

\begin{eqnarray}
v &=& u + at \label{eq:eq1}\\
s &=& \frac{(u + v)}{2} t\label{eq:eq2}\\
s &=& ut + \frac{1}{2}at^2 \label{eq:eq3}\\
v^2 &=& u^2 + 2as\label{eq:eq4}
\end{eqnarray}

Make sure you can rhyme these off, they are \emph{very} important!
There are so many different types of questions for these equations. Basically when you are answering a question like this:
\begin{enumerate}
\item Find out what values you have and write them down.
\item Figure out which equation you need.
\item \emph{Write it down!!!}
\item Fill in all the values you have and get the answer.
\end{enumerate}

\medskip\par
\underline{Equation \ref{eq:eq1}}\\
By the definition of acceleration
\begin{displaymath}
a = \frac{\Delta v}{t}
\end{displaymath}
\noindent where $\Delta v$ is the change in velocity, i.e. $\Delta v = v - u$.
Thus we have
\begin{eqnarray*}
a &=& \frac{v-u}{t} \nonumber\\
v &=& u +at
\end{eqnarray*}

\medskip \par
\underline{Equation \ref{eq:eq2}}\\
In the previous section we saw that displacement can be calculated from the area between a velocity-time graph and the time-axis. For \emph{uniformly accelerated motion} the most complicated velocity-time graph we can have is a straight line.  Look at the graph below - it represents an object with a starting velocity of \emph{u}, accelerating to a final velocity \emph{v} over a total time \emph{t}.

\begin{center}
\begin{pspicture}(-1,-1)(3,3)
%\psset{yunit=0.75cm}
\psaxes[labels=none,ticks=none]{->}(0,0)(3,3)
\psline[linewidth=1pt]{-}(0,1)(2,2)
\psline[linewidth=1pt,linestyle=dashed]{-}(0,1)(2,1)
\psline[linewidth=1pt,linestyle=dashed]{-}(2,0)(2,2)
\psline[linewidth=1pt,linestyle=dashed]{-}(0,2)(2,2)
%\psline[linewidth=1pt,linestyle=dashed]{-}(4,0)(4,-2)
%\psline[linewidth=1pt,linestyle=dashed]{-}(6,0)(6,-2)
\rput{90}(-0.8,1.5){velocity ($m/s$)}
\rput(1,-0.8){time (\emph{s})}
\rput(-0.4,1){\emph{u}}
\rput(-0.4,2){\emph{v}}
\rput(2,-0.4){\emph{t}}
%\rput(4,-0.3){4}
%\rput(6,-0.3){6}
\end{pspicture}
\end{center}

To calculate the final displacement we must calculate the area under the graph - this is just the area of the rectangle added to the area of the triangle.

\begin{eqnarray*}
Area \triangle &=& \frac{1}{2}b \times h\\
&=& \frac{1}{2}t \times (v-u)\\
&=& \frac{1}{2}vt - \frac{1}{2}ut
\end{eqnarray*}

\begin{eqnarray*}
Area \Box &=& w \times h\\
&=& t \times u\\
&=& ut
\end{eqnarray*}

\begin{eqnarray*}
Displacement &=& Area \Box + Area \triangle\\
s &=& ut + \frac{1}{2}vt -\frac{1}{2}ut\\
&=& \frac{(u+v)}{2}t
\end{eqnarray*}

\medskip \par
\underline{Equation \ref{eq:eq3}}\\
This equation is simply derived by eliminating the final velocity $v$ in equation \ref{eq:eq2}.  Remembering from equation \ref{eq:eq1} that
\begin{displaymath}
v=u+at
\end{displaymath}
then equation \ref{eq:eq2} becomes
\begin{eqnarray*}
s &=& \frac{u+u+at}{2}t\\
&=&\frac{2ut+at^2}{2}\\
&=&ut+\frac{1}{2}at^2
\end{eqnarray*}

\medskip \par
\underline{Equation \ref{eq:eq4} }\\
This equation is just derived by eliminating the time variable in the above equation.  From Equation \ref{eq:eq1} we know
\begin{displaymath}
t=\frac{v-u}{a}
\end{displaymath}
\noindent Substituting this into Equation \ref{eq:eq3} gives
\begin{eqnarray}
s &=& u(\frac{v-u}{a})+\frac{1}{2}a(\frac{v-u}{a})^2 \nonumber\\
&=& \frac{uv}{a}-\frac{u^2}{a}+\frac{1}{2}a(\frac{v^2-2uv+u^2}{a^2})\nonumber \\
&=& \frac{uv}{a}-\frac{u^2}{a}+\frac{v^2}{2a}-\frac{uv}{a}+\frac{u^2}{2a}\nonumber\\
2as &=& -2u^2+v^2+u^2 \nonumber\\
v^2& =&u^2+2as
\end{eqnarray}

This gives us the final velocity in terms of the initial velocity, acceleration and displacement and is independent of the time variable.

\begin{pwex}{}
\textbf{Question:  }A racing car has an initial velocity of 100$m/s$ and it covers a displacement of 725$m$ in 10$s$.  Find its acceleration.
\step We are given the quantities $u$, $s$ and $t$ - all in the correct units.  We need to find $a$.
\step We can use equation \ref{eq:eq3}
\begin{displaymath}
s=ut +\frac{1}{2}at^2
\end{displaymath}
\step Rearranging equation \ref{eq:eq3} we have
\begin{displaymath}
a=\frac{2(s-ut)}{t^2}
\end{displaymath}
Substituting in the values of the known quantities this becomes
\begin{eqnarray*}
a &=& \frac{2(725m-100\frac{m}{s}\cdot10s)}{10^2s^2}\\
&=&\frac{2(-275m)}{100s^2}\\
&=&-5.5\frac{m}{s^2}
\end{eqnarray*}
The racing car is accelerating at -5.5$\frac{m}{s^2}$, or we could say it is decelerating at 5.5$\frac{m}{s^2}$.
\end{pwex}

\begin{pwex}{}
\textbf{Question:  }An object starts from rest, moves in a straight line with a constant acceleration and covers a distance of 64$m$ in 4$s$.  Calculate
\begin{itemize}
\item its acceleration
\item its final velocity
\item at what time the object had covered half the total distance
\item what distance the object had covered in half the total time.
\end{itemize}
\step We are given the quantities $u$, $s$ and $t$ in the correct units.
\step To calculate the acceleration we can use equation \ref{eq:eq3}.  Rearranging it we have:
\begin{displaymath}
a=\frac{2(s-ut)}{t^2}
\end{displaymath}
Substituting in the values of the known quantities this becomes
\begin{eqnarray*}
a &=& \frac{2(64m-0\frac{m}{s}4s}{4^2s^2}\\
&=&\frac{128m}{16s^2}\\
&=&8\frac{m}{s^2}
\end{eqnarray*}
\step To calculate its final velocity we can use equation \ref{eq:eq1} - remember we now also know the acceleration of the object.
\begin{eqnarray*}
v&=&u+at\\
&=&0\frac{m}{s}+(8\frac{m}{s^2})(4s)\\
&=&32\frac{m}{s}
\end{eqnarray*}
\step The time at which the object had covered half the total distance.  Half the distance is 32$m$.  Here we have the quantities $s$, $u$ and $a$ so we first use equation \ref{eq:eq4} to calculate the velocity at this distance:
\begin{eqnarray*}
v^2&=&u^2+2as\\
&=&(0m)^2+2(8m/s^2)(32m)\\
&=&512m^2/s^2\\
v&=&22.6m/s
\end{eqnarray*}
Now we can use equation \ref{eq:eq2} to calculate the time:
\begin{eqnarray*}
t&=&\frac{2s}{u+v}\\
&=&\frac{(2)(32m)}{0m/s+22.6m/s}\\
&=&2.8s
\end{eqnarray*}
\step The distance the object had covered in half the time.  Half the time is $2s$.  Thus we have $u$, $a$ and $t$ - all in the correct units.  We can use equation \ref{eq:eq3} to get the distance:
\begin{eqnarray*}
s&=&ut+\frac{1}{2}at^2\\
&=&(0m/s)(2s)+\frac{1}{2}(8\frac{m}{s^2})(2s)^2\\
&=&16m
\end{eqnarray*}
\end{pwex}

\begin{pwex}{}
\textbf{Question:  }A ball is thrown vertically upwards with a velocity of $10m/s$ from the balcony of a tall building.  The balcony is 15$m$ above the ground and gravitational accleration is $10m/s^2$. Find a) the time required for the ball to hit the ground, and b) the velocity with which it hits the ground.\\
\step In this case it often helps to make the problem easier to understand if we draw ourselves a picture like the one below:

\begin{center}
\begin{pspicture}(-2,-1.0)(4,4.5)
\psaxes[labels=none,ticks=none]{-}(-0,0)(-1,0)(4,4.5)
\psline[linewidth=1pt]{-}(0,3)(0.2,3)(0.2,3.2)(0,3.2)
\psline[linewidth=1pt,linestyle=dashed]{|-|}(0.4,3.2)(0.4,4.2)
\rput(0.6,3.7){$s_1$}
\psline[linewidth=1pt,]{->}(0.9,4.2)(0.9,3.8)
\rput(1.4,4){$v_1$}
\psline[linewidth=1pt,]{->}(0.9,3)(0.9,3.4)
\rput(1.4,3.1){$u_1$}
\rput(-1.4,3.2){balcony}
\rput(2,-0.6){ground}
\psline[linewidth=1pt,linestyle=dashed]{|-|}(2.0,0)(2.0,4.2)
\rput(2.3,2.2){$s_2$}
\psline[linewidth=1pt,]{->}(2.4,0.5)(2.4,0.1)
\rput(2.9,0.3){$v_2$}
\psline[linewidth=1pt,]{->}(2.4,4.2)(2.4,3.8)
\rput(2.9,4){$u_2$}
\psline[linewidth=1pt]{->}(3.7,3.4)(3.7,3.0)
\rput(4.0,3.5){$a_1,a_2$}
\end{pspicture}
\end{center}

First the ball goes upwards with gravitational acceleration slowing it until it reaches its highest point - here its speed is $0m/s$ - then it begins descending with gravitational acceleration causing it to increase its speed on the way down.  We can separate the motion into 2 stages:\\
Stage 1 - the upward motion of the ball\\
Stage 2 - the downward motion of the ball.\\
We'll choose the upward direction as positive - this means that gravitation acceleraton is negative - and
we'll begin by solving for all the variables of Stage 1.  So far we have these quantities:
\begin{eqnarray*}
u_1&=&10m/s\\
v_1&=&0m/s\\
a_1&=&-10m/s^2\\
t_1&=&?\\
s_1&=&?
\end{eqnarray*}
Using equation \ref{eq:eq1} to find $t_1$:
\begin{eqnarray*}
v_1&=&u_1+a_1t_1\\
t_1&=&\frac{v_1-u_1}{a_1}\\
&=&\frac{0m/s-10m/s}{-10m/s^2}\\
&=&1s
\end{eqnarray*}

We can find $s_1$ by using equation \ref{eq:eq4}
\begin{eqnarray*}
v_1^2&=&u_1^2+2a_1s_1\\
s_1&=&\frac{v_1^2-u_1^2}{2a}\\
&=&\frac{(0m/s)^2-(10m/s)^2}{2(-10m/s^2)}\\
&=&5m
\end{eqnarray*}

For Stage 2 we have the following quantities:
\begin{eqnarray*}
u_2&=&0m/s\\
v_2&=&?\\
a_2&=&-10m/s^2\\
t_2&=&?\\
s_2&=&-15m-5m=20m
\end{eqnarray*}

We can determine the final velocity $v_2$ using equation \ref{eq:eq4}:
\begin{eqnarray*}
v_2^2&=&u_2^2+2a_2s_2\\
&=&(0m/s)^2+2(-10m/s^2)(-20m)\\
&=&400(m/s)^2\\
v_2&=&20m/s \mbox{ downwards}
\end{eqnarray*}

Now we can determine the time for Stage 2, $t_2$, from equation \ref{eq:eq1}:
\begin{eqnarray*}
v_2&=&u_2+a_2t_2\\
t_2&=&\frac{v_2-u_2}{a_2}\\
&=&\frac{-20m/s-0m/s}{-10m/s^2}\\
&=&2s
\end{eqnarray*}

Finally,\\
a) the time required for the stone to hit the ground is $t=t_1+t_2=1s+2s=3s$\\
b) the velocity with which it hits the ground is just $v_2=-20m/s$

\end{pwex}

These questions do not have the working out in them, but they are all done in the manner described on the previous page.
\vspace{0.5em}\\
\textbf{Question:} A car starts off at 10 m/s and accelerates at 1 m/s$^2$ for 10 seconds. What is it's final velocity?\\
\vspace{0.5em}\\
\textbf{Question:} A car starts from rest, and accelerates at 1 m/s$^2$ for 10 seconds. How far does it move?\\
\vspace{0.5em}\\
\textbf{Question:} A car is going 30 m/s and stops in 2 seconds. What is it's stopping distance for this speed?\\
\vspace{0.5em}\\
\textbf{Question:} A car going at 20 m/s stops in a distance of 20 m/s.
\begin{enumerate}
\item What is it's deceleration?
\item If the car is 1 Tonne (1000 Kg, or 1 Mg) how much force do the brakes exert?
\end{enumerate}

\pagebreak
\section{Important Equations and Quantities}

\begin{table}[!hp]
\begin{center}
\begin{tabular}{|c|c|c|c|}\hline \hline
% after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
\multicolumn{4}{|c|}{\textbf{Units}}\\ \hline
Quantity & Symbol & Unit & Base S.I. Units   \\ \hline
Displacement & $\overrightarrow{s}$  & - & $m$ + direction   \\
Velocity & $\overrightarrow{u}$,$\overrightarrow{v}$ &-& $m.s^{-1}$ + direction \\
Distance & $s$ & - &  $m$ \\
Speed & $v$ & - &   $m.s^{-1}$\\
Acceleration & $\overrightarrow{a}$ &-&  $m.s^{-1}$ + direction \\\hline \hline
\end{tabular}
\end{center}
\caption{Units used in \textbf{Rectilinear Motion} } \label{table:rectilinear_motion::units}
\end{table}

\end{document}
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