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Resistance V.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \begin{document} {\large Resistance Proportional to Speed} \begin{align*} \text{Basic Equation:}\quad&R \propto v\quad\text{but in opposite direction.}\quad\therefore R=\lambda v\:,\;\text{where }\lambda<0\\ &\text{Physically, $R$ is determined by many factors, but let's (artificially) define $k=-\tfrac{\lambda}{m}>0$,}\\ &\text{so $R=\lambda v=-mkv$}\quad\text{(opposite to $v$)}.\\ &\text{$R=-mkv$ is downwards went the particle is going up ($R<0$ when $v>0$),}\\ &\text{and upwards when it is going down.}\\ &m\ddot{x}=-W+R=-mg-mkv\quad\text{($W=mg>0$ but is always a downward force; $g>0$ here)}\\ \therefore\quad&\boxed{\ddot{x}=-g-kv}\qquad\text{and}\qquad\boxed{R=-mkv\:,\quad\text{where }k>0}\\ \\ &\text{Given a vertical initially upwards projectile with initial speed $u$, i.e. where $t=0,\:x=0,\:\dot{x}=u>0$,}\\ \\ \text{$v$ in terms of $t$:}\quad&\boxed{v=ue^{-kt}-\frac{g}{k}\left(1-e^{-kt}\right)=-\frac{g}{k}+\frac{g+ku}{k}\cdot e^{-kt}}\\ &\frac{dv}{dt}=\ddot{x}=-g-kv\:,\quad\frac{dt}{dv}=\frac{\:1\:}{-g-kv}\:,\quad t=\int_u^v\frac{dv}{-g-kv}=-\frac{\:1\:}{k}\big[\ln(g+kv)\big]^v_u=-\frac{\:1\:}{k}\ln\left(\frac{g+kv}{g+ku}\right)\\ &e^{-kt}=\frac{g+kv}{g+ku}\:,\quad g+kv=(g+ku)e^{-kt}\:,\quad v=\frac{\:1\:}{k}\left(ge^{-kt}+kue^{-kt}-g\right)=ue^{-kt}-\frac{g}{k}\left(1-e^{-kt}\right)\\ \text{Likewise,}\quad&\boxed{t=\frac{\:1\:}{k}\ln\left(\frac{g+ku}{g+kv}\right)}\\ \\ \text{$x$ in terms of $v$:}\quad&\boxed{x=\frac{\:1\:}{k}\left[u-v+\frac{g}{k}\ln\left(\frac{g+kv}{g+ku}\right)\right]}\\ &v\cdot\frac{dv}{dx}=-g-kv\:,\quad\frac{dx}{dv}=\frac{v}{-g-kv}\:,\quad x=\int_u^v\frac{v}{-g-kv}\:dv\\ &x=-\frac{\:1\:}{k}\int_u^v 1-\frac{g}{g+kv}\:dv =-\frac{\:1\:}{k}\left[(v-u)-\frac{g}{k}\left[\ln(g+kv)\right]^v_u\right] =\frac{\:1\:}{k}\left[u-v+\frac{g}{k}\ln\left(\frac{g+kv}{g+ku}\right)\right] \\ \\ \text{$x$ in terms of $t$:}\quad&\boxed{x=-\frac{g}{k}t+\left(\frac{g+ku}{k^2}\right)\left(1-e^{-kt}\right)}\\ &x=\int_0^t v\:dt=\int_0^t\left(-\frac{g}{k}+\frac{g+ku}{k}\cdot e^{-kt}\right)\:dt=-\frac{g}{k}t+\left(\frac{g+ku}{k^2}\right)\cdot\left(1-e^{-kt}\right)\\ \\ \text{Max height $H$ at $T$:}\quad&\boxed{T=\frac{\:1\:}{k}\ln\left(\frac{g+ku}{g}\right)=\frac{u-kH}{g}}\\ &\text{At max, }t=T\text{ and }\frac{dx}{dt}=v=0\:,\quad\text{i.e. }-\frac{g}{k}+\frac{g+ku}{k}\cdot e^{-kT}=0\\ &\left(g+ku\right)e^{-kT}=g\:,\quad e^{-kT}=\frac{g}{g+ku}\:,\quad -kT=\ln\left(\frac{g}{g+ku}\right)\:,\quad T=\frac{\:1\:}{k}\ln\left(\frac{g+ku}{g}\right)\\ &\text{When $x=H, v=0$ , so}\\ &H=x(T) =\frac{\:1\:}{k}\left[u-0+\frac{g}{k}\ln\left(\frac{g+k\cdot 0}{g+ku}\right)\right] =\frac{\:1\:}{k}\left[u+\frac{g}{\not{k}}\left(-\not{k}T\right)\right] =\frac{u-gT}{k}\\ &T=\frac{u-kH}{g}\\ \end{align*} % % % \begin{align*} \text{\large \bf Free Fall:}\quad&\text{Based on the above formulae, but with $u=0$ and the $x$-axis is downward,}\\ &\text{so the signs of the $x, v$ and $\ddot{x}$ variables need to be negated in those formulae.}\\ &\text{e.g. }-\ddot{x}=-g-k(-v)\:,\quad\boxed{\ddot{x}=g-kv}\\ \\ \text{$v$ in terms of $t$:}\quad&\boxed{v=\frac{g}{k}\left(1-e^{-kt}\right)}\qquad\text{and}\qquad \boxed{t=\frac{\:1\:}{k}\ln\left(\frac{g}{g-kv}\right)}\\ \text{Terminal Velocity:}\quad&v_T=\lim_{t\to+\infty} v=\lim_{t\to+\infty}\frac{g}{k}\left(1-e^{-kt}\right)=\frac{\:g\:}{k}\\ &\text{Also, }\ddot{x}=g-kv_T=0\:,\quad\therefore\boxed{v_T=\frac{g}{k}}\\ \\ \text{$x$ in terms of $v$:}\quad&\boxed{x=-\frac{v}{k}-\frac{g}{k^2}\ln\left(1-\frac{kv}{g}\right)}\\ \\ \text{$x$ in terms of $t$:}\quad&\boxed{x=\frac{g}{k}t-\frac{g}{k^2}\left(1-e^{-kt}\right)}\\ \end{align*} \end{document}