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Motion Basics.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=0.5cm]{geometry} \begin{document} {\large Motion Basics} \begin{align*} \text{Notations:}\quad&x\quad\text{Displacement (different from distance travelled)}\\ &\dot{x}\quad\text{Velocity (directional - one dimension only in this context);}\quad v=\dot{x}=\tfrac{dx}{dt}\\ &\ddot{x}\quad\text{Acceleration (change of Velocity - directional as well);}\quad a=\ddot{x}=\tfrac{dv}{dt}=\tfrac{d^2 x}{dt^2}\\ \\ &\text{It is generally assumed that when $t=0, x(0)=0$ and $v(0)=u$ unless stated otherwise. It follows that}\\ &\big[x(t)\big]^x_0=x=\int_0^t v(t)\:dt\qquad\text{and}\qquad\big[v(t)\big]^v_u=v-u=\int_0^t a(t)\:dt\\ \\ &\text{Other ``unconventional'' notations or assumptions:}\\ &\quad\text{Overloading of the same symbol as dummy variables:}\\ &\qquad\text{e.g. In $x=\int_0^t v(t)\:dt$\:, the $t$ in $\int_0^t$ is the parameter of function $x(t)$,}\\ &\qquad\qquad\text{while the $t$ in $v(t)\:dt$ is a dummy variable for the integration form.}\\ &\qquad\qquad\text{(But writing \tiny $x=\int_0^t v(u)\:du$ \normalsize would be even more confusing.)}\\ &\quad\text{Same measurement but different parameters:}\\ &\qquad\text{e.g. $v(t)$ is velocity in terms of time, while $v(x)$ is velocity at a position.}\\ \\ &\text{Simple Facts}\\ &\qquad\text{Either $u\neq 0$ or $\ddot{x}\neq 0$ is required to kick start the motion.}\\ &\qquad\text{$v=0$ and $\ddot{x}=0$ means the particle remain stationary, which is different from}\\ &\qquad\qquad\text{being momentarily stationary when $v=0$ but $\ddot{x}\neq 0$.}\\ \\ \text{Notes on Physics:}\quad&F=ma=m\ddot{x}\quad\text{(Newton's Second Law) is the only formula of Physics involved in this context.}\\ &\text{(Newton's First Law: No force, no acceleration. \tiny The particle remains stationary or travel at constant speed along a straight line.\normalsize )}\\ &\text{All other relations will be given in terms of $x, \dot{x}, \ddot{x}$ and $t$. Some may have physical meaning}\\ &\text{(e.g. $a(x)$ is a force field), while others in this context may not reflect how the nature works.}\\ &\text{Some measurements are encapsulated in constants to suit this level of mathematics.}\\ &\text{However, mass $m$ is usually left out so that it can be cancelled in $F=ma$. e.g. $R=mkv$}\\ \\ \text{``One on One'':}\quad&\text{In many motion problems, it is required to express $x, v, a$ and $t$ in terms of one another.}\\ \text{Find $a$ from $v(x)$:}\quad&\boxed{\ddot{x}=v\cdot\frac{dv}{dx}=\frac{d}{dx}\left(\frac{\:1\:}{2}v^2\right)}\qquad\ddot{x}=\frac{dv}{dt}=\frac{dx}{dt}\cdot\frac{dv}{dx}=v\cdot\frac{dv}{dx}\\ \\ \text{Find $v^2$ from $a(x)$:}\quad&a(x)=\frac{d}{dx}\left(\frac{\:1\:}{2}v^2\right)\:,\quad\int_0^x a(x)\:dx=\left[\frac{\:1\:}{2}v^2\right]^v_u\:,\quad v^2=u^2+2\int_0^x a(x)\:dx\\ \text{Alternatively,}\quad&a(x)=v\cdot\frac{dv}{dx}\:,\quad\int_0^x a(x)\:dx=\int_0^xv\cdot\frac{dv}{dx}\:dx=\int_u^v v\:dv=\left[\frac{\:1\:}{2}v^2\right]^v_u\:,\quad v^2=u^2+2\int_0^x f(x)\:dx\\ \\ \text{Find $v$ from $a(v)$:}\quad&\frac{dv}{dt}=a(v)\:,\quad\frac{dt}{dv}=\frac{\:1\:}{a(v)}\:,\quad t=\int_u^v\frac{dv}{a(v)}\:,\quad\ldots\\ \\ \text{Find $x$ from $a(v)$:}\quad&a(v)=v\cdot\frac{dv}{dx}\:,\quad\frac{dx}{dv}=\frac{v}{a(v)}\:,\quad x=\int_u^v\frac{v}{a(v)}\:dv\:,\quad\ldots\\ \end{align*} \end{document}