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% Document Info
\author{So Ty (pastel-galaxies)}
\pdfinfo{
  /Title (distance-displacement-and-position.pdf)
  /Creator (Cheatography)
  /Author (So Ty (pastel-galaxies))
  /Subject (Distance, Displacement, and Position Cheat Sheet)
}

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\begin{multicols}{3}
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        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Distance, Displacement, and Position Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{So Ty (pastel-galaxies)} via \textcolor{DarkBackground}{\uline{cheatography.com/51952/cs/14180/}}}
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}So Ty (pastel-galaxies) \\
  \uline{cheatography.com/pastel-galaxies} \\
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Not Yet Published.\\
   Updated 4th January, 2018.\\
   Page {\thepage} of \pageref{LastPage}.
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  Measure your website readability!\\
  www.readability-score.com
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\begin{multicols*}{3}

\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Distance and Displacement}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Put simply, {\bf{distance}} is the total amount something traveled (measured in centimeters, meters, or kilometers) and {\bf{displacement}} is only how far away it ended from its starting point (simpifies to X\textasciicircum{}F\textasciicircum{} - X\textasciicircum{}I\textasciicircum{}, or final position minus initial position.)% Row Count 6 (+ 6)
} \tn 
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\mymulticolumn{1}{x{5.377cm}}{Distance and displacement are NOT the same thing, and most problems later on in physics will be asking you about {\emph{displacement}}. Make sure you know which one you're looking for!}  \tn 
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\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Average Speed vs. Average Velocity}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{{\bf{Average speed}} is the total {\emph{distance}} traveled divided by the total {\emph{time}} traveled over a certain interval. {\bf{Speed is a scalar measurement}}, which means it has no direction, only a magnitude. \newline % Row Count 4 (+ 4)
{\bf{Average velocity}} is the total {\emph{displacement}} traveled divided by the total {\emph{time}} traveled over a certain interval. {\bf{Velocity is a vector measurement}}, which means it has a magnitude and a direction. It is often written as {\bf{Δx / Δt}}, or {\emph{change in position over change in time}}.% Row Count 10 (+ 6)
} \tn 
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\begin{tabularx}{5.377cm}{x{2.18988 cm} x{2.78712 cm} }
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\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Slope And Its Meanings}}  \tn
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\mymulticolumn{2}{x{5.377cm}}{{\bf{On a position vs. time graph:}}} \tn 
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Positive slope & Object moving forward \tn 
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{\emph{line going down slowly}} & {\emph{object moving backward slowly}} \tn 
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{\emph{line going down quickly}} & {\emph{object moving backward quickly}} \tn 
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Zero slope & Object not moving \tn 
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\mymulticolumn{2}{x{5.377cm}}{To find velocity from a graph like this, find the total displacement over the time traveled during the interval.}  \tn 
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\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Special Cases}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{If an object stops in the exact same place it starts (for example, it travels in a circle or a square), {\bf{the displacement is 0}}. Remember, the displacement is only the difference between the final position and the initial position, so if they are the same point, there is no difference. \newline % Row Count 6 (+ 6)
If the distance is a straight line, the displacement and the distance will be the {\bf{same.}}% Row Count 8 (+ 2)
} \tn 
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\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Interpreting Position vs. Time Graphs}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{Image could not be loaded.} \tn 
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\mymulticolumn{1}{x{5.377cm}}{A {\bf{position vs. time graph}} will have position on the y-axis and time on the x-axis. These graphs can be used to find {\bf{instantaneous speed}} (the speed the object is going at a specific time) and {\bf{average speed}} (about how fast the object was going overall). \newline  \newline Remember that position is measured in m and time is measured in s, so this graph is describing changes in m/s. Since you know that m/s is the unit for velocity, you know that the line is really showing changes in {\bf{velocity.}}}  \tn 
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\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Interpreting Velocity vs. Time Graphs}}  \tn
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\mymulticolumn{1}{p{5.377cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/pastel-galaxies_1515097755_velocityvstimept2.png}}} \tn 
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\mymulticolumn{1}{x{5.377cm}}{In a {\bf{velocity vs. time graph,}} velocity will be plotted on the y-axis and time will be plotted on the x-axis. Remember that velocity is measured in m/s and time is measured in s; therefore, this graph truly shows us a changes in m/s\textasciicircum{}2\textasciicircum{}. This means that the graph is really describing a change in {\bf{acceleration}}.}  \tn 
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\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Example Problems}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{{\emph{Andrew drives 7 kilometers north, then drives 5 kilometers east. What distance did he cover? What was his displacement?}} \newline % Row Count 3 (+ 3)
When solving for {\bf{distance}}, we can just add the 7 km and the 5 km because distance looks for the {\bf{total}} kilometers she traveled. \newline % Row Count 6 (+ 3)
When solving for {\bf{displacement}} we need to find how far away from his starting point he ended. To find this, make a straight line from the beginning point to the end point. This will create a triangle, and then you can use a\textasciicircum{}2\textasciicircum{} + b\textasciicircum{}2\textasciicircum{} = c\textasciicircum{}2\textasciicircum{} to solve for c, which will be the displacement. \newline % Row Count 12 (+ 6)
So, the answers to this problem: \newline % Row Count 13 (+ 1)
{\emph{Distance:}} 12 km \newline % Row Count 14 (+ 1)
{\emph{Displacement:}} √74 km, or about 8.6 km.% Row Count 15 (+ 1)
} \tn 
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\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Tips and Tricks}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{Note: Having graph paper can be extremely helpful when dealing with distance problems. (For problems like these, it's okay to give your answer in the same unit you're given.)% Row Count 4 (+ 4)
} \tn 
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% That's all folks
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