# Deriving terminal velocity

1. Jun 20, 2013

### spacetimedude

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.

2. Jun 20, 2013

### spacetimedude

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. Jun 20, 2013

### king vitamin

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

4. Jun 20, 2013

### spacetimedude

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