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"
\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 file is licensed under the terms of the GNU Free Documentation
% License, please see http://www.gnu.org/licenses/fdl.html
%===========================================================================%
%This is to remove the Underfull warnings when compiling
\hbadness=10000
%===========================================================================%
% 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}}%
\addtocounter{essaycounter}{1}%
\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}%
\addtocounter{pworkedexamplecounter}{1}\setcounter{stepcounter}{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}%
\addtocounter{cworkedexamplecounter}{1}\setcounter{stepcounter}{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}%
\addtocounter{mworkedexamplecounter}{1}\setcounter{stepcounter}{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
\addtocounter{stepcounter}{1}}
\newcommand{\westep}[1]
{\par \pagebreak[1]
\textit{Step \arabic{stepcounter} : #1}\par
\addtocounter{stepcounter}{1}}
\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}
%\psshadowbox{
\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]%
{\vspace{3mm}\psshadowbox{
\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 <[EMAIL PROTECTED]>
\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.
\par\textbf{Answer:}
\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}
\psshadowbox{
\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}
\psshadowbox{
\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}
\psshadowbox{
\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}
\psshadowbox{
\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}
\par\textbf{Answer:}
\step Analyse the question to determine what is given. The question
explicitly gives a displacement-time graph.
\step What is asked? \\
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}
\textbf{Answer: }
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}
\textbf{Answer: }
\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}
\textbf{Answer: }
\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.
\nts {SHADING}
\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.
\par\textbf{Answer: }
\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}
\textbf{Answer: }
\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.\\
\textbf{Answer: }
\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?\\
\textbf{Answer:} 20 m/s\\
\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?\\
\textbf{Answer:} 50 m\\
\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?\\
\textbf{Answer:} 30 m\\
\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}
_______________________________________________ latex2html mailing list [EMAIL PROTECTED] http://tug.org/mailman/listinfo/latex2html
