Answer to Question #203615 in Classical Mechanics for Aratunde Tobiloba

Question

A body of mass (m) falls from rest through a medium which exact a frictional drag force (bv) proportional to a applied force (f). 1. Expand the velocity in power series up to t^4

Answer 1

$$ma=m\frac{dv}{dt}=mg-bv+f$$ $$\frac{dv}{dt}=g-\frac{b}{m}v+\frac{f}{m}$$ $$bdt=\frac{dv}{\frac{mg}{b}+\frac{f}{b}-v}$$ $$v(t)=\frac{mg+f}{b}+c\exp{(-bt)}$$ $$v(0)=0=\frac{mg+f}{b}+c$$

So,

$$v(t)=\frac{mg+f}{b}(1-\exp{(-bt)})$$

Terminal velocity:

$$v(\infty)=\frac{mg+f}{b}$$