I am trying to solve for a coefficient of drag that creates a certain situation.

The situation is at a high speed, so the drag at high velocity equation applies:

$$ F_D = \frac{1}{2}\rho v^2 C_d A $$

If I write down:

$$ \begin{align} da &= (\frac{d}{dt}\frac{F_D}{m})dt\\ a &= \int(\frac{d}{dt}\frac{F_D}{m})dt\\ v &= v_0 + \int{a}dt\\ x &= x_0 + \int{v}dt\\ \end{align} $$

Then I am left with a recursion through the chain rule, since $F$ depends on $v$, which depends on $a$, etc.

How do I deal with this situation in the context of wind resistance? And I am not sure which tags to add here...

  • $\begingroup$ Why are you differentiating acceleration? $\endgroup$ Jun 15, 2017 at 0:15
  • $\begingroup$ @probably_someone because force changes with respect to time. $\endgroup$
    – Chris
    Jun 15, 2017 at 0:17
  • $\begingroup$ Yeah, but $F=ma$, so you've got your acceleration right there. $\endgroup$ Jun 15, 2017 at 0:18
  • $\begingroup$ @probably_someone yes, so a = F/m... and F = ... $\endgroup$
    – Chris
    Jun 15, 2017 at 0:19
  • $\begingroup$ @probably_someone and mass changes with time, in this particular problem. So I have to differentiate all the way down. There should be another way to construct this, so I don't get an endless differentiation, but I can't remember how. $\endgroup$
    – Chris
    Jun 15, 2017 at 0:21

1 Answer 1


Letting $k=\rho C_D A/2$, and letting $m=m_0 - bt$ for constant $b$, we have

$a=\frac {dv}{dt}=\frac {kv^2}{m_0-bt}$.

This is a first-order ODE in $v$, with solution

$v=\frac {b}{k\log (m_0-bt)-C}$,

where $C$ satisfies $v(0)=v_0$.

  • $\begingroup$ cleaned up my comments here $\endgroup$
    – Chris
    Jun 13, 2018 at 22:04
  • $\begingroup$ What is funny is that I don’t remember why I was confused—was probably thinking in terms of code. Didn’t mean to come off the way i did. $\endgroup$
    – Chris
    Jun 13, 2018 at 22:08
  • $\begingroup$ @bordeo No problem - fighting the backfire effect is always difficult. $\endgroup$ Jun 13, 2018 at 22:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.