Answer to Question #221683 in Differential Equations for Unknown346307

Question

A certain particle falls under gravity in a resisting medium whose resistance varies with velocity.Find the relation between distance and velocity if the initially the particle starts from rest.

Answer 1

$\textsf{If } v \textsf{ is the velocity of the particle falling under}\\ \textsf{gravity, then the equation of motion of the moving}\\ \textsf{particle by Newton's second law is given by}\\ \hspace{0.1cm}\\ mg-mR=ma\\ \hspace{0.1cm}\\ \textsf{where } R\textsf{ represents the resistance of the }\\ \textsf{resisting medium}\\ \hspace{0.1cm}\\ \textsf{ Since}\textsf{ R }\textsf{is proportional to } v,\\\hspace{0.1cm}\\ R=kv\\\hspace{0.1cm}\\ mg-mkv=ma\\ g-kv=a\\ \hspace{0.1cm}\\ a=\frac{\mathrm{d}v}{\mathrm{d}t}\\ \textsf{Using chain rule,}\\ a=\frac{\mathrm{d}v}{\mathrm{d}s}\times\frac{\mathrm{d}s}{\mathrm{d}t}\\ a=\frac{\mathrm{d}v}{\mathrm{d}s}\times v\\ a=v\frac{\mathrm{d}v}{\mathrm{d}s}\\ \hspace{0.1cm}\\ \textsf{Hence,}\\ g-kv=v\frac{\mathrm{d}v}{\mathrm{d}s}\\ \mathrm{d}s=(\frac{v}{g-kv})\mathrm{d}v\\ \textsf{By separating into partial fractions, we have}\\ \mathrm{d}s=(\frac{-1}k+\frac{g}{k(g-kv)})\mathrm{d}v\\ \mathrm{d}s=\frac{-\mathrm{d}v}{k}+\frac{g}{k}\frac{\mathrm{d}v}{g-kv}\\\hspace{0.1cm}\\ \textsf{Integrating, we have,}\\ \hspace{0.1cm}\\ s+c=\frac{-v}{k}+\frac{g}{k}(\frac{-1}{k})ln(g-kv)\\ s+c=\frac{-v}{k}-\frac{g}{k^2}ln(g-kv)\\ \hspace{0.1cm}\\ \textsf{Initially, } s=0, v=0.\, \textsf{Hence,}\\\hspace{0.1cm}\\ c=\frac{-g}{k^2}ln(g)\\\hspace{0.1cm}\\ \textsf{Hence,}\hspace{0.1cm}\\ s-\frac{g}{k^2}ln(g)=\frac{-v}{k}-\frac{g}{k^2}ln(g-kv)\\ s=-\frac{v}{k}-\frac{g}{k^2}ln(\frac{g-kv}{g})\\ s=-(\frac{v}{k}+\frac{g}{k^2}ln(\frac{g-kv}{g})$