A question about linear drag force

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The discussion clarifies the relationship between velocity (v) and position (r) in the context of linear drag force in classical mechanics. It emphasizes that while velocity is defined as the derivative of position, it does not depend on position in a functional sense for the equations of motion being considered. The drag force is expressed as f = -bv, leading to a first-order differential equation that simplifies the analysis. The key takeaway is that velocity is treated as a separate variable that influences motion but does not directly depend on position in this context. Understanding this distinction is crucial for solving problems involving drag forces in projectile motion.
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My classical mechanics textbook says that, for a projectile, the linear drag force is given by f = -bv and the second law is written as m\ddot{r} = mg - bv (a second order differential equation) which can be rewritten as m\dot{v} = mg - bv (a first order differential equation) because the forces depend only on v and not on r. But I can't figure out why this is the case. Doesn't v depend on r?
 
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No. Velocity doesn't depend on location as such, only to the extent that location may effect some of the parameters.
 
##\mathbf v ## doesn't "depend" on ##\mathbf r## in the sense that it is some (unknown) function of ##\mathbf r## and probably some other variables as well.

The point is that ##\mathbf v## is just another name for ##\mathbf{\dot r}##, (that's what "velocity" means!) and differentiating, ##\mathbf{\dot v}## is identically equal to ##\mathbf{\ddot r}##.
 
Thank you, AlephZero. I believe I understand. I mean, I know that v is just another name for \dot{r} and the like, I just thought it could be rewritten in terms of v for that reason. No one ever explained that this is true specfically because v did not "depend" on r... is it ever the case that v does depend on r?
 
sorry, didn't realize I wasn't bolding \dot{r}.
 
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