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Deriving terminal velocity 
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#1
Jun2013, 12:24 AM

P: 8

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=mgbv and mg/b=terminal velocity v_{t} So the steps I took were: m*dv/dt+bv=mg (m/b)*(dv/dt)+v=v_{t} dv/dt=(b/m)(v_{t}v) dv/(v_{t}v)=(b/m)dt Integrating both sides would give ln[(v_{t}v)/(v_{t)}]=(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. 


#2
Jun2013, 12:38 AM

P: 8

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?



#3
Jun2013, 01:00 AM

P: 180

You just missed the minus sign while integrating (chain rule)



#4
Jun2013, 01:13 AM

P: 8

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



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