Noob doubt about signs in equations

Question

I wanted to solve some simple free fall with friction models so I wrote (with $x$ axis orientated down for convenience) $$m\ddot x=mg -k\dot x$$ which I turned in $$m\dot v=mg -kv$$ with $m,g,k$ positive constants. The solution is a costant term plus a transient term that decays exponentially. Then I tried with a quadratic friction and wrote $$m\dot v=mg -kv^2$$ The solution is an hyperbolic tangent. So far so good. The problem arises when I tried to write the equations with the positive direction being up. The gravity is now negative but friction is still opposite to velocity so the first equation becomes $$m\dot v=-mg -kv$$ while the second $$m\dot v=-mg -kv^2$$ The first is right while the second is wrong (the solution is a tangent). It turns out that the correct way to write the second equation is $$m\dot v=-mg +kv^2$$ And this is why I am puzzled, because if I write $$m\dot v=-mg +kv$$ in the first one I obtain a wrong solution (a negative exponential). It seems like the "reversed" equations are obtained by substituting "-v" in the "direct" ones. So my question is, what is the right way to obtain the equations? I naively thought that the LHS should always be $m \ddot x$ while RHS, the forces (being positive or negative with respect to the choice of the system of reference), would dictate the sign of the solution. What I am mistaking?

Answer 1

Friction of a falling object always points upwards. If the constant $k$ is positive then, in an upward-pointing reference frame, the contribution of friction to acceleration is $+kv$ (or $+kv^2$). Remember that your terminal velocity is negative in that frame!

Answer 2

In more than one dimension of $x$ and $v$, it always holds for the quadratic friction (air type) that $$ m\ddot v = -mg - k·|v|·v $$ so that the force of friction is always opposed to the current velocity.