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\begin{document}
\title{zen Module 2 Lecture Notes: Motion Along a Line}
\author[1]{Richard Piccioni}%
\affil[1]{College of Marin}%
\vspace{-1em}
\date{\today}
\begingroup
\let\center\flushleft
\let\endcenter\endflushleft
\maketitle
\endgroup
\selectlanguage{english}
\begin{abstract}
Abstract content goes here%
\end{abstract}%
\sloppy
% Be sure to replace at least the position-time graphs for accelerated motion, but ideally all graphs.
\setcounter{section}{2}
\subsection{Introduction}
The goal of \emph{kinematics} is to describe the motion of objects. There are many ways to do that in words:
\begin{center}
``The object is moving to the right and speeding up,''
\end{center}
... in a diagram:\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.42\columnwidth]{figures/pgl-strobe-vec-00/physGL.rgpStrobe.01}
\caption{{``Strobe'' depictions of a moving object
{\label{886970}}%
}}
\end{center}
\end{figure}
... in a mathematical formula:
\begin{center}
$x(t) = (12.0~m) + (6.12~\frac{m}{s})t + (1.71~\frac{m}{s^2})t^2$
\end{center}
... as a computer program:
\begin{verbatim}
# Here is a little Python code to predict position and velocity
t, x, v, a = 0, 12.0, 6.12, 3.42 # s, m, m/s, m/s/s
t_f, Dt = 10, 0.01 # s
while t <= t_f:
v += a*Dt
x += v*Dt
t += Dt
print(t,x,v)
\end{verbatim}
... or an animation.\href{http://www.physgl.org/index.php/welcome/share/29c91ae49167a631fd0bad720cec85c9}{(Click here and drag up the Controls panel to run.} Use your browser's Back button to return).\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.84\columnwidth]{figures/physGL-rgp-Strobe-Animation/physGL-rgp-Strobe-Animation}
\caption{{Animated strobe diagram
{\label{986324}}%
}}
\end{center}
\end{figure}
In this lab activity you will observe some motions and get some practice describing those motions in \textit{graphs}. The object you will be observing will be a small cart that rolls along a metal track with very little friction.
\subsubsection{Position vs. time graphs for constant velocity}
Imgine an object on a line. The \textit{signed distance} between the object and a particular, fixed point on that line (the \textit{origin}) is called the object's \textit{position} ($x$). We'll say that if the object is to the right of the origin, $x$ is positive and if the object is to the left, $x$ is negative. If the object happens to be located at the origin, $x = 0$.
Imagine also you have a clock. At any particular moment, or, as physicists say, any particular \textit{instant in time}, the clock has a reading $t$. Finally, imagine you have a way of knowing $x$, the position of the object, at any clock reading (any time) $t$.
Physics is mostly about change. We measure changes in position by calculating \emph{displacement} ($\Delta x$) using the formula:
\begin{equation}\label{eqn:disp}
\Delta x = x_f - x_0
\end{equation}
and changes in time ($\Delta t$) by calculating the \emph{time interval}:
\begin{equation}\label{eqn:interval}
\Delta t = t_f - t_0.
\end{equation}
Displacement can be positive ($\Delta x > 0$) or negative ($\Delta x < 0$), depending upon which direction the object has moved. Notice that while $\Delta x$ can have either sign, $\Delta t$ can only be positive: moving backwards in time is not an option, unfortunately.
We know for sure an object has moved if its position $x_f$ at time $t_f$ differs from its position $x_0$ at an earlier time $t_0$. But it doesn't work the other way around: even though $x_f = x_0$, the object \textit{could} have moved during the time interval $\Delta t$. It's just that at the end of the time interval, the object was back to where it started. The only thing we can say is that displacement was zero, that is, $\Delta x = 0$.
To get away from all that confusion, we first focus on a special kind of motion, that does have that ambiguity, \textit{uniform motion}. Uniform motion is motion with unchanging speed and direction. As far as we know, it was Galileo Galilei (1564-1642) who first pointed out that if (and only if) the motion of an object is uniform, the displacements during any two equal time intervals will themselves be equal. That means, \textit{if the motion is uniform}, the ratio $\frac{\Delta x}{\Delta t}$ will have the same value, no matter what $\Delta t$ you are looking at. And if $\Delta x = 0$, you know the object never moved.
Because it does not change, the quantity $\frac{\Delta x}{\Delta t}$ has a special name. It's called the \textit{velocity along the x-axis}. Let's save some ink and just call it the \textit{velocity}. To calculate the velocity of an object whose motion is uniform, we just use this equation:
\begin{equation}\label{eqn:vx_diffs}
v = \frac{x_f-x_0}{t_f-t_0}
\end{equation}
Soon we will learn about the very special conditions needed for the motion of an object to be uniform (Spolier Alert: the vector sum of the forces acting on the object must be zero). We will also learn how to think about velocity when motion is \textit{not} uniform. In the meantime, we can make the following statement:
\begin{center}
\emph{If (and only if) an object's motion is uniform, the object's velocity is constant.}
\end{center}
During any time interval $\Delta t$ of uniform motion, velocity has the same \textit{sign} as the displacement. If the object happens to be \emph{stationary}, then $x_f = x_0$, so $x_f - x_0$ = 0, so its velocity is zero.
Table~\ref{tab:x_of_t} lists twenty clock readings (times) and the corresponding positions of an object. As far as we can tell, the motion is uniform (the velocity is constant). You should know why.\selectlanguage{english}
\begin{table}
\caption{{\label{tab:x_of_t} Position and time data for an object in uniform motion}}
\centering
\begin{tabular}{cccc}
Time (s) & Position (m)\\
\hline
0.0 & 10.0\\
1.0 & 12.0\\
2.0 & 14.0\\
3.0& 16.0\\
4.0& 18.0\\
5.0& 20.0\\
6.0& 22.0\\
7.0& 24.0\\
8.0& 26.0\\
9.0& 28.0\\
10.0& 30.0\\
11.0& 32.0\\
12.0& 34.0\\
13.0& 36.0\\
14.0& 38.0\\
15.0& 40.0\\
16.0& 42.0\\
17.0& 44.0\\
18.0& 46.0\\
19.0& 48.0\\
20.0& 50.0\\
\end{tabular}
\end{table}
Fig. \ref{470553}, is a graph used to plot the data in Table \ref{tab:x_of_t}. Notice that in this graph, the vertical axis is position ($x$) and the horizontal axis is time $t$. That can be confusing at first, but it works out better in the long run.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/pos-vs-t/pos-vs-t}
\caption{{Position vs. time graph for an object moving with uniform motion
(constant velocity)
{\label{470553}}%
}}
\end{center}
\end{figure}
Not too surprisingly, the points from Table~\ref{tab:x_of_t}, when plotted in Fig. \ref{470553}, lie on a line. The quantity we are calling velocity is equal to the \emph{slope} of that line. That should make sense because slope is \emph{rise over run}. In this graph, that's $\frac{\Delta x}{\Delta t}$.
If the vertical axis crosses the horizontal axis at $t_0$, then the ``y-intercept'' (where the line hits the vertical axis) is the object's \emph{initial position} $x_0$. That's true in Fig. \ref{470553} if $t_0 = 0~s$.
Recall that the general equation for a line is $y(x)=mx+b$, where $y$ is the quantity plotted on the vertical axis, $x$ is the quantity plotted on the horizontal axis, $m$ is the slope, and $b$ is the y-intercept. We write the \emph{equation} for motion with constant velocity (uniform motion) like this:
\begin{equation}\label{eqn:const_v}
x(t)=x_0+vt
\end{equation}
This doesn't look exactly like $y(x)=mx+b$, more like $y(x) = b + mx$, but it amounts to the same thing.
Examples of position vs. time graphs for constant velocity (uniform motion) are shown in Figure \ref{255519} below, where $t_0$ is set to equal zero.
% really unfortunate annotations ``= + const.'' and ``=-const.''. Change ASAP.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.92\columnwidth]{figures/xvst/xvst}
\caption{{Position vs. time graphs for zero, positive, and
negative~\emph{constant~}velocities. The expression ``= + const'' means
``equal to a positive constant'' and ``= - const'' means ``equal to a
negative constant''. ~
{\label{255519}}%
}}
\end{center}
\end{figure}
\subsubsection{Velocity vs. time graphs for constant acceleration}
Velocities aren't always constant, but they often change in a simple way. If the \emph{changes in velocity} ($\Delta v$) during any two equal time intervals are themselves equal, we say the object is \emph{moving with uniform acceleration}. If the object is moving with uniform acceleration, we can calculate a quantity we call \textit{the acceleration} ($a$) using this equation:
\begin{equation}\label{eqn:const_accel}
a=\frac{v_f-v_0}{t_f-t_0}
\end{equation}
Like velocity, acceleration can be positive, negative or zero. But unlike velocity, the sign of the acceleration does not just depend upon which direction the object is moving. It also depends upon whether the object's speed is increasing or decreasing. Here's a tip:
\begin{center}
\emph{If speed is increasing, then velocity and acceleration will have the same sign. If speed is decreasing, the signs will be opposite.}
\end{center}
And of course, if the object happens to be moving with constant velocity, then $v_f = v_0$, so $v_f - v_0$ = 0, so its acceleration is zero.
If we draw a graph where the vertical axis is \emph{velocity} and the horizontal axis is time, motion with constant acceleration is a line. The quantity we are calling \textit{the acceleration}, $a$, is equal to the \emph{slope} of that line, its \emph{rise over run}, $\frac{\Delta v}{\Delta t}$. If the horizonal axis starts at $t_0$, then the ``y-intercept'' (where the line hits the vertical axis) is the object's \emph{initial velocity} $v_0$.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/sheet-v-vs-t-00/sheet-v-vs-t-00}
\caption{{Velocity vs. time~ for constant acceleration
{\label{723287}}%
}}
\end{center}
\end{figure}
Examples of velocity vs. time graphs for constant acceleration are shown in Fig. , below, where $t_0$ is set to equal zero.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.92\columnwidth]{figures/vvst/vvst}
\caption{{Velocity vs. time graphs for zero, positive, and
negative~\emph{constant} accelerations. Again, the expression ``= +
const'' means ``equal to a positive constant'' and ``= - const'' means
``equal to a negative constant''.~ ~
{\label{673262}}%
}}
\end{center}
\end{figure}
\subsubsection{Average velocity}
If acceleration is not zero, velocity is changing and \textit{cannot} be calculated using Eqn.\ref{eqn:vx_diffs}. We do have a name for the quantity on the right-hand side of Eqn.\ref{eqn:vx_diffs} however, one that has meaning even if velocity is changing; it's called the \emph{time-weighted average velocity} ($\bar{v}$). To save ink, we'll just use the term \textit{average velocity}. So for \emph{any} motion, we can write:
\begin{equation}\label{eqn:vbar}
\bar{v} = \frac{x_f-x_0}{t_f-t_0}
\end{equation}
\subsubsection{Position vs. time graphs for constant acceleration}
If we draw a graph where the vertical axis is \emph{position} and the horizontal axis is time, motion with constant acceleration is \emph{not} a line, but a \emph{curve}.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/sheet-x-vs-t-01/sheet-x-vs-t-01}
\caption{{Position vs. time for constant acceleration
{\label{828429}}%
}}
\end{center}
\end{figure}
It turns out the curve is \emph{parabola}, and the equation for a parabola is \emph{quadratic}. In math classes, quadratic equations are often written in \emph{standard form} like this:
\begin{equation}\label{eqn:std_quad}
0 = ax^2 + bx + c
\end{equation}
To find \emph{solutions} or \emph{roots} of a quadratic equation, they pointed you to (and probably had you memorize) the famous \emph{quadratic formula}:
\begin{equation}\label{quad_formula}
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\end{equation}
In kinematics, we write the quadratic relationship between position and time like this:
\begin{equation}\label{eq:x}
x=x_0+v_0t+\frac{1}{2} at^2
\end{equation}
Notice the differences: in standard form, the variable is $x$, in the kinematics, it's actually $t$. Plus, the symbol $a$ means two different things in the two equations. So be careful. To help you, here is the quadratic formula adapted to kinematic purposes:
\begin{equation}\label{kine_quad_formula}
t = \frac{-v_0 \pm \sqrt{v_0^2 + 2a(x - x_0)}}{a}
\end{equation}\selectlanguage{english}
\begin{figure}[h]
\begin{center}
\includegraphics[scale=0.5]{figures/pos_vs_t.svg}
\caption{{Position vs. time graphs for zero, positive and negative acceleration.}}
\label{fig:xvsta}
\end{center}
\end{figure}
Since it is continuously changing, we have to be very careful how we define velocity. The inventors of calculus came up with a concept called the \emph{instantaneous velocity}. In brief, it is the ratio $\Delta v/\Delta t$ for a $\Delta t$ so small that, for all practical purposes, the velocity during $\Delta t$ is constant.
At any given instant $t$, the instantaneous velocity is the slope of a line tangent to the curve of position vs. time. If the horizontal axis starts at $t_0$, then the ``y-intercept'' (where the line hits the vertical axis) is still the object's initial position $x_0$.
Examples of position vs. time graphs for constant acceleration are shown in Fig. \ref{579712}, below, where $t_0$ is set to equal zero.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.92\columnwidth]{figures/xvsta/xvsta}
\caption{{Position vs. time graphs for two kinds of constant acceleration. The
labels t1, t2, t3 and t4 , and the dots, are supposed to line up with~
the ``tic marks'' on the time axes.
{\label{579712}}%
}}
\end{center}
\end{figure}
Examples of position vs. time graphs for constant acceleration are shown in Fig.\ref{579712}. The instantaneous velocities at two instants, $t_1$ and $t_2$, are the slopes of the two tangent lines draw for those instants. In the graph on the left, both slopes are positive, meaning both instantaneous velocities are positive (the object is moving in the direction of increasing $x$). But also notice the slope at $t_2$ is greater than the slope at $t_1$. That means the velocity is increasing, so acceleration is positive. Because the velocities and the acceleration have the same sign, we know the \emph{speed} of the object is increasing.
In contrast, the slopes drawn on the graph on the right at $t_3$ and $t_4$ are negative, meaning both instantaneous velocities are negative (the object is moving in the direction of \textit{decreasing} x. Also, the slope at $t_4$ is greater ("less negative") than the slope at $t_3$. Again, this means the velocities are increasing and the acceleration is positive. Because the velocities and the acceleration have opposite signs, we know the \emph{speed} of the object is decreasing.
You can think of the graph on the left in Fig.\ref{579712} as depicting a person getting out of a chair, while the graph on the right depicts the person sitting back down. Tricky.
\selectlanguage{english}
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\end{document}