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Deriving terminal velocity

by spacetimedude
Tags: deriving, terminal, velocity
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spacetimedude
#1
Jun20-13, 12:24 AM
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.
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spacetimedude
#2
Jun20-13, 12:38 AM
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?
king vitamin
#3
Jun20-13, 01:00 AM
P: 172
You just missed the minus sign while integrating (chain rule)

spacetimedude
#4
Jun20-13, 01:13 AM
P: 6
Deriving terminal velocity

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


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