\documentclass[10pt,a4paper]{article}

% Packages
\usepackage{fancyhdr}           % For header and footer
\usepackage{multicol}           % Allows multicols in tables
\usepackage{tabularx}           % Intelligent column widths
\usepackage{tabulary}           % Used in header and footer
\usepackage{hhline}             % Border under tables
\usepackage{graphicx}           % For images
\usepackage{xcolor}             % For hex colours
%\usepackage[utf8x]{inputenc}    % For unicode character support
\usepackage[T1]{fontenc}        % Without this we get weird character replacements
\usepackage{colortbl}           % For coloured tables
\usepackage{setspace}           % For line height
\usepackage{lastpage}           % Needed for total page number
\usepackage{seqsplit}           % Splits long words.
%\usepackage{opensans}          % Can't make this work so far. Shame. Would be lovely.
\usepackage[normalem]{ulem}     % For underlining links
% Most of the following are not required for the majority
% of cheat sheets but are needed for some symbol support.
\usepackage{amsmath}            % Symbols
\usepackage{MnSymbol}           % Symbols
\usepackage{wasysym}            % Symbols
%\usepackage[english,german,french,spanish,italian]{babel}              % Languages

% Document Info
\author{ReSummit}
\pdfinfo{
  /Title (ap-physics-formulas-kinematic.pdf)
  /Creator (Cheatography)
  /Author (ReSummit)
  /Subject (AP Physics Formulas (Kinematic) Cheat Sheet)
}

% Lengths and widths
\addtolength{\textwidth}{6cm}
\addtolength{\textheight}{-1cm}
\addtolength{\hoffset}{-3cm}
\addtolength{\voffset}{-2cm}
\setlength{\tabcolsep}{0.2cm} % Space between columns
\setlength{\headsep}{-12pt} % Reduce space between header and content
\setlength{\headheight}{85pt} % If less, LaTeX automatically increases it
\renewcommand{\footrulewidth}{0pt} % Remove footer line
\renewcommand{\headrulewidth}{0pt} % Remove header line
\renewcommand{\seqinsert}{\ifmmode\allowbreak\else\-\fi} % Hyphens in seqsplit
% This two commands together give roughly
% the right line height in the tables
\renewcommand{\arraystretch}{1.3}
\onehalfspacing

% Commands
\newcommand{\SetRowColor}[1]{\noalign{\gdef\RowColorName{#1}}\rowcolor{\RowColorName}} % Shortcut for row colour
\newcommand{\mymulticolumn}[3]{\multicolumn{#1}{>{\columncolor{\RowColorName}}#2}{#3}} % For coloured multi-cols
\newcolumntype{x}[1]{>{\raggedright}p{#1}} % New column types for ragged-right paragraph columns
\newcommand{\tn}{\tabularnewline} % Required as custom column type in use

% Font and Colours
\definecolor{HeadBackground}{HTML}{333333}
\definecolor{FootBackground}{HTML}{666666}
\definecolor{TextColor}{HTML}{333333}
\definecolor{DarkBackground}{HTML}{FF0000}
\definecolor{LightBackground}{HTML}{FFEFEF}
\renewcommand{\familydefault}{\sfdefault}
\color{TextColor}

% Header and Footer
\pagestyle{fancy}
\fancyhead{} % Set header to blank
\fancyfoot{} % Set footer to blank
\fancyhead[L]{
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{C}
    \SetRowColor{DarkBackground}
    \vspace{-7pt}
    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
    }
\end{tabulary}
\columnbreak
\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{AP Physics Formulas (Kinematic) Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{ReSummit} via \textcolor{DarkBackground}{\uline{cheatography.com/52223/cs/14186/}}}
\end{tabulary}
\end{multicols}}

\fancyfoot[L]{ \footnotesize
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{LL}
  \SetRowColor{FootBackground}
  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}ReSummit \\
  \uline{cheatography.com/resummit} \\
  \end{tabulary}
\vfill
\columnbreak
\begin{tabulary}{5.8cm}{L}
  \SetRowColor{FootBackground}
  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 23rd October, 2020.\\
   Updated 23rd October, 2020.\\
   Page {\thepage} of \pageref{LastPage}.
\end{tabulary}
\vfill
\columnbreak
\begin{tabulary}{5.8cm}{L}
  \SetRowColor{FootBackground}
  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}}  \\
  \SetRowColor{white}
  \vspace{-5pt}
  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
\end{tabulary}
\end{multicols}}




\begin{document}
\raggedright
\raggedcolumns

% Set font size to small. Switch to any value
% from this page to resize cheat sheet text:
% www.emerson.emory.edu/services/latex/latex_169.html
\footnotesize % Small font.

\begin{multicols*}{2}

\begin{tabularx}{8.4cm}{x{3.6 cm} x{4.4 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Kinematics 2D Motion}}  \tn
% Row 0
\SetRowColor{LightBackground}
V = V`0` + at & V`0` = Initial velocity of object\{\{nl\}\}V = Final velocity of object\{\{nl\}\}a = Acceleration of object\{\{nl\}\}t = Time \tn 
% Row Count 6 (+ 6)
% Row 1
\SetRowColor{white}
V\textasciicircum{}2\textasciicircum{} = V`0`\textasciicircum{}2\textasciicircum{} + 2aΔx\{\{nobreak\}\} & V`0` = Initial velocity of object\{\{nl\}\}V = Final velocity of object\{\{nl\}\}a = Acceleration of object\{\{nl\}\}Δx / Δy = Change in position \tn 
% Row Count 13 (+ 7)
% Row 2
\SetRowColor{LightBackground}
Δx = V`0`t + $\frac{1}{2}$at\textasciicircum{}2\textasciicircum{} & Δx / Δy = Change in position\{\{nl\}\}V`0` = Initial velocity\{\{nl\}\}t = Time\{\{nl\}\}a = Acceleration \tn 
% Row Count 18 (+ 5)
% Row 3
\SetRowColor{white}
F = ma & F = Force from object\{\{nl\}\}m = Mass of object\{\{nl\}\}a = Acceleration of object \tn 
% Row Count 22 (+ 4)
% Row 4
\SetRowColor{LightBackground}
F`f` = μN & F`f` = Force of friction\{\{nl\}\}μ = Coefficient of friction\{\{nl\}\}N = Normal force \tn 
% Row Count 26 (+ 4)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{8.4cm}}{Note: Some formulas may involve BOTH the x and y directions, as well as incorporate other formulas outside kinematics.}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{x{3.76 cm} x{4.24 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Momentum}}  \tn
% Row 0
\SetRowColor{LightBackground}
FΔt = Δp = m`V` - m`V0`\{\{nobreak\}\} & FΔt = Δp = Impulse\{\{nl\}\}m`V` = Final momentum\{\{nl\}\}m`V0` = Initial momentum\{\{nobreak\}\} \tn 
% Row Count 5 (+ 5)
% Row 1
\SetRowColor{white}
\mymulticolumn{2}{x{8.4cm}}{m`Vbefore` - m`V0before` = m`Vafter` - m`V0after`\{\{ac\}\}} \tn 
% Row Count 7 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{8.4cm}}{Note: Momentum is ALWAYS conserved. You may need to note that the momentum before is equal to the momentum after.}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{x{4 cm} x{4 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Energy}}  \tn
% Row 0
\SetRowColor{LightBackground}
W = Fd & W = Work done\{\{nl\}\}F = Force applied\{\{nl\}\}d = Distance travelled \tn 
% Row Count 4 (+ 4)
% Row 1
\SetRowColor{white}
W = ΔKE = $\frac{1}{2}$mV\textasciicircum{}2\textasciicircum{} - $\frac{1}{2}$mV`0`\textasciicircum{}2\textasciicircum{}\{\{nobreak\}\} & W = Work done\{\{nl\}\}m = Mass of object\{\{nl\}\}V = Final velocity\{\{nl\}\}V`0` = Initial velocity\{\{nobreak\}\} \tn 
% Row Count 10 (+ 6)
% Row 2
\SetRowColor{LightBackground}
U`g` = mgh & U`g` = Work done by gravity\{\{nl\}\}m = Mass\{\{nl\}\}g = Gravity\{\{nl\}\}h / d = Height or distance traveled \tn 
% Row Count 15 (+ 5)
% Row 3
\SetRowColor{white}
F`S` = kx & F`S` = Force of spring (Restored Force)\{\{nl\}\}k = Spring coefficient\{\{nl\}\}x = Distance from equilibrium \tn 
% Row Count 21 (+ 6)
% Row 4
\SetRowColor{LightBackground}
W`S` = U`S` = $\frac{1}{2}$kx\textasciicircum{}2\textasciicircum{} & W`S` = Work done by spring\{\{nl\}\}k = Spring coefficient\{\{nl\}\}x = Distance from equilibrium \tn 
% Row Count 26 (+ 5)
% Row 5
\SetRowColor{white}
KE = $\frac{1}{2}$mV\textasciicircum{}2\textasciicircum{} & KE = Kinetic Energy\{\{nl\}\}m = Mass\{\{nl\}\}v = Velocity of object \tn 
% Row Count 30 (+ 4)
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{x{4 cm} x{4 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Energy (cont)}}  \tn
% Row 6
\SetRowColor{LightBackground}
KE + U`g` + U`S` =\{\{nl\}\}KE + U`g` +U`S` + W & KE = Kinetic Energy (is the object moving?)\{\{nl\}\}U`g` = Work done by gravity (is the object above where you set x = 0?)\{\{nl\}\}U`S` = Work done by spring (is a spring involved?)\{\{nl\}\}W = Friction (did energy go to friction?) \tn 
% Row Count 12 (+ 12)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{8.4cm}}{Note: Energy is {\bf{SOMETIMES}} conserved depending on the situation. {\bf{Inelastic}} collisions cannot apply the conservation of energy because of the loss of energy. However, you can apply the conservation of energy for {\bf{elastic}} collisions.}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{x{4 cm} x{4 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Rotational Motion}}  \tn
% Row 0
\SetRowColor{LightBackground}
ω = ω`0` + αt & ω`0` = Angular initial velocity\{\{nl\}\}ω = Angular final velocity\{\{nl\}\}α = Angular acceleration\{\{nl\}\}t = Time \tn 
% Row Count 6 (+ 6)
% Row 1
\SetRowColor{white}
ω\textasciicircum{}2\textasciicircum{} = ω`0`\textasciicircum{}2\textasciicircum{} + 2αθ\{\{nobreak\}\} & ω`0` = Angular initial velocity\{\{nl\}\}ω = Angular final velocity\{\{nl\}\}α = Angular acceleration\{\{nl\}\}θ = Angular change in position \tn 
% Row Count 13 (+ 7)
% Row 2
\SetRowColor{LightBackground}
θ = ω`0`t + $\frac{1}{2}$αt\textasciicircum{}2\textasciicircum{} & θ = Angular change in position\{\{nl\}\}ω`0` = Angular initial velocity\{\{nl\}\}t = Time\{\{nl\}\}α = Angular acceleration \tn 
% Row Count 19 (+ 6)
% Row 3
\SetRowColor{white}
V`T` = rω & V`T` = Tangential (Linear) velocity\{\{nl\}\}r = Radius \{\{nl\}\}ω = Angular final velocity \tn 
% Row Count 24 (+ 5)
% Row 4
\SetRowColor{LightBackground}
a`T` = rα & a`T` = Tangential (Linear) acceleration\{\{nl\}\}r = Radius\{\{nl\}\}α = Angular acceleration \tn 
% Row Count 29 (+ 5)
% Row 5
\SetRowColor{white}
a`C` = V`T`\textasciicircum{}2\textasciicircum{} / r & a`C` = Centripetal acceleration\{\{nl\}\}V`T` = Tangential (Linear) velocity\{\{nl\}\}r = Radius \tn 
% Row Count 34 (+ 5)
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{x{4 cm} x{4 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Rotational Motion (cont)}}  \tn
% Row 6
\SetRowColor{LightBackground}
a`r` = rω\textasciicircum{}2\textasciicircum{} & a`r` = Radial Acceleration\{\{nl\}\}r = Radius\{\{nl\}\}ω = Angular velocity \tn 
% Row Count 4 (+ 4)
% Row 7
\SetRowColor{white}
τ = F`⊥`d & τ = Torque\{\{nl\}\}F`⊥` = Perpendicular Forces\{\{nl\}\}d= Distance from Pivot Point \tn 
% Row Count 8 (+ 4)
% Row 8
\SetRowColor{LightBackground}
I = Σmr\textasciicircum{}2 & I = Moment of Inertia (Rotational Moment / Rotational Intertia)\{\{nl\}\}Σmr\textasciicircum{}2\textasciicircum{} = Total of each Mass x Radius Squared \tn 
% Row Count 14 (+ 6)
% Row 9
\SetRowColor{white}
KE`C` = 1/2(I)ω\textasciicircum{}2\textasciicircum{} & KE`C` = Kinetic Circular Energy\{\{nl\}\}I = Moment of Inertia (Rotational Moment / Rotational Intertia)\{\{nl\}\}ω = Angular velocity \tn 
% Row Count 21 (+ 7)
% Row 10
\SetRowColor{LightBackground}
τ = Iα & τ = Torque\{\{nl\}\}I = Moment of Inertia (Rotational Moment / Rotational Intertia)\{\{nl\}\}α = Angular acceleration \tn 
% Row Count 27 (+ 6)
% Row 11
\SetRowColor{white}
KE`R` = 1/2 I`P`ω\textasciicircum{}2\textasciicircum{} = 1/2(I`COM` + mh\textasciicircum{}h\textasciicircum{})ω\textasciicircum{}2\textasciicircum{}\{\{nl\}\}=1/2(m(V`COM`)\textasciicircum{}2\textasciicircum{}) + 1/2Iω\textasciicircum{}2\textasciicircum{} & KE`R` = Kinetic Rolling Energy\{\{nl\}\}1/2(m(V`COM`)\textasciicircum{}2\textasciicircum{}) = Sliding Equation\{\{nl\}\}1/2Iω\textasciicircum{}2\textasciicircum{} = Rotation Equation \tn 
% Row Count 33 (+ 6)
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{x{4 cm} x{4 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Rotational Motion (cont)}}  \tn
% Row 12
\SetRowColor{LightBackground}
l = mrω & l = Momentum of a particle \tn 
% Row Count 2 (+ 2)
% Row 13
\SetRowColor{white}
L = Iω & L = Momentum of a rigid body (not a particle) \tn 
% Row Count 5 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{8.4cm}}{NOTE:  \newline - You may need to consider that ω = dθ / dt and α = dω / dt. \newline - Account for all objects rotating the pivot point when calculating I. \newline - Momentum is {\bf{ALWAYS}} conserved.}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}


% That's all folks
\end{multicols*}

\end{document}