# How to account for velocity term in force component of equations of motion?

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...

• Why are you differentiating acceleration? – probably_someone Jun 15 '17 at 0:15
• @probably_someone because force changes with respect to time. – donlan Jun 15 '17 at 0:17
• Yeah, but $F=ma$, so you've got your acceleration right there. – probably_someone Jun 15 '17 at 0:18
• @probably_someone yes, so a = F/m... and F = ... – donlan Jun 15 '17 at 0:19
• @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. – donlan Jun 15 '17 at 0:21

## 1 Answer

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 $v(0)=v_0$.

• cleaned up my comments here – donlan Jun 13 '18 at 22:04
• 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. – donlan Jun 13 '18 at 22:08
• @bordeo No problem - fighting the backfire effect is always difficult. – probably_someone Jun 13 '18 at 22:11