How Do You Calculate Terminal Velocity Involving Drag?

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SUMMARY

The discussion focuses on calculating terminal velocity for a falling body, incorporating drag proportional to speed. The equation derived is m*dv/dt = mg - bv, leading to the terminal velocity formula vt = mg/b. The user initially struggles with integration, specifically obtaining the correct sign during the integration process. The resolution involves recognizing the missed negative sign in the integration step, which is crucial for accurate results.

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  • Understanding of differential equations
  • Familiarity with concepts of drag force in physics
  • Knowledge of integration techniques
  • Basic principles of mechanics, specifically Newton's laws
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spacetimedude
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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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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?
 
You just missed the minus sign while integrating (chain rule)
 
pshh. I can't believe I missed that. Thanks so much king vitamin!
 

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