# Deriving terminal velocity

by spacetimedude
Tags: deriving, terminal, velocity
 P: 6 Hello PF, I have once simple (well, not so simple for me) question. I'm trying to derive an equation for the velocity of a falling body with accordance to terminal velocity. The equation incorporates drag proportional to the speed. m*dv/dt=mg-bv and mg/b=terminal velocity vt So the steps I took were: m*dv/dt+bv=mg (m/b)*(dv/dt)+v=vt dv/dt=(b/m)(vt-v) dv/(vt-v)=(b/m)dt Integrating both sides would give ln[(vt-v)/(vt)]=(b/m)t But the textbook says that I'm supposed to get negative (b/m)t on the left side. Have I made a mistake on the integration part? Any help will be deeply appreciated.
 P: 6 hmm. I seem to have gotten the answer if I just divided the entire equation by -b in the beginning without bringing the bv to the left side. Have I made a mistake on the integration part?
 P: 134 You just missed the minus sign while integrating (chain rule)
P: 6

## Deriving terminal velocity

pshh. I can't believe I missed that. Thanks so much king vitamin!

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