Question #94097
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. Find it's velocity in terms of time t 2. Find it's terminal velocity
1
Expert's answer
2019-09-09T11:20:04-0400
ma=mdvdt=mgbv+fma=m\frac{dv}{dt}=mg-bv+f

dvdt=gbmv+fm\frac{dv}{dt}=g-\frac{b}{m}v+\frac{f}{m}

bdt=dvmgb+fbvbdt=\frac{dv}{\frac{mg}{b}+\frac{f}{b}-v}

v(t)=mg+fb+cexp(bt)v(t)=\frac{mg+f}{b}+c\exp{(-bt)}

v(0)=0=mg+fb+cv(0)=0=\frac{mg+f}{b}+c

So,


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

Terminal velocity:


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


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