I have a question regarding John Taylor's treatment of a particle falling down in a medium that exerts a linear drag force (Chapter 2 of his book "Classical Mechanics").First some background, he begins by writing the force equation for the relevant vertical forces $$m\dot{v} = mg - bv$$ where the positive direction is defined as downwards (towards the ground). $m$ is the mass of the particle, $g$ the gravitational constant, $b$ a constant, and $v$ the speed of the particle. After defining the terminal speed $$mg - bv_{ter} = 0\Rightarrow v_{ter} \equiv \frac{mg}{b}$$ He rewrites the first expression as $$\dot{v} = \frac{b}{m} \left(\frac{mg}{b} - v\right)\Rightarrow \dot{v} = \frac{b}{m}\left(v_{ter}-v\right)$$ This differential equation is solvable via a substitution ($u = v_{ter} - v$) and separation of variables. Doing so yields the following expression for $v(t)$. $$v(t) = v_{ter} + Ae^{-\frac{t}{\tau}}$$ where $A$ is an arbitrary constant and $\tau \equiv \frac{m}{b}$. If we allowed time to tend towards infinity ($t\rightarrow \infty$), then the velocity will be equal to $v_{ter}$.

I now try and repeat this derivation by taking the positive direction as upwards (away from the ground). The force equation now becomes $$m\dot{v} = -mg + bv$$ Performing the same process as above, I am able to write the following differential equation. $$\dot{v} = \frac{b}{m} \left(v -v_{ter}\right)$$ Solving this differential equation results in the following solution $$v(t) = v_{ter} + Ae^{\frac{t}{\tau}}$$ Clearly, this is different from Taylor's solution as my exponential term does not fade to zero as time grows large. I have been taught previously that the choice of coordinate systems should yield the same physical result (at least for problems like this). Therefore, I am confused about this result and would appreciate any clarification.

This is my first post on this site, so hopefully I did not violate any rule/norm and would also appreciate feedback if I did!

Thank you,


  • $\begingroup$ You need to change the sign on all your terms, including your acceleration term. $\endgroup$ – Bill Watts Mar 1 at 7:41
  • $\begingroup$ Or you can just change to $-m g$, and leave the others unchanged because your drag term has to be opposite in sign to your acceleration term, because the drag force always decelerates the particle. $\endgroup$ – Bill Watts Mar 1 at 7:46
  • $\begingroup$ Thanks for your response, but 1.) the acceleration term is a scalar and 2.) wouldn't that mean that the drag and weight are pointing in the same direction--meaning it can accelerate infinitely? $\endgroup$ – ATF Mar 1 at 8:02
  • $\begingroup$ If your $m\dot{v}$ term is positive, your $b v$ term must be negative. $\endgroup$ – Bill Watts Mar 1 at 8:18

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