If F be the independent force acting on the object and D be the velocity dependent force acting in the opposite direction of F. The net force accelerating the object is;

$$F − D = ma$$

where $D = \frac{1}{2}CA\rho v^2$, if both the object and air are in motion and in opposite direction then $D = \frac{1}{2}A\rho(v_{air}^2 + v_{object}^2$) resulting into;

$$\frac{F}{m} - \frac{[CA\rho(v_{air}^2 + v_{object}^2)]}{2m} = \frac{\mathrm{d}v_{object}}{\mathrm{d}t}$$

Is this a correct model? and how to manipulate this expression for $v(t)$ ?


2 Answers 2


You should have $(v_\textit{air} + v_\textit{object})^2$, not $v_\textit{air}^2 + v_\textit{object}^2$. This is because the force only cares about the relative speed between the object and the air, which is $v_\textit{air} + v_\textit{object}$. You're not adding two different forces.

Given this, and assuming $v_\textit{air}$ is constant, I would change to a new variable $u = v_\textit{air} + v_\textit{object}$, and the equation becomes

$$\frac{du}{dt} = \frac{F}{m} - \frac{CA\rho}{2m} u^2.$$

This is the standard equation for an object subject to a constant force with quadratic drag, and it's separable:

$$\frac{du}{\frac{F}{m} - \frac{CA\rho}{2m} u^2} = dt.$$

Integrating both sides, the left hand side can be done with partial fractions. You should get something along the lines of $\log\left(\frac{1+u}{1-u}\right) = t$ (with a lot of constants everywhere). You can finally solve this for $u$, and you should find that as $t \to \infty$, $u$ approaches a terminal velocity.

  • $\begingroup$ Actually I am working on a study that the object is a valve (disk) which is guided by a valve seat that the friction is minimum and negligible. The velocity of air is constant and you really did it right, thank you so much sir. But one more thing that I'm worry about is that do I have to include the stagnation pressure that acts on the valve surface in my model? resulting to; du/dt = F/m − PA/m - [(C + 1)ρAu^2]/2m, where P + (ρu^2)/2 is the stagnation pressure and I will get something along the lines of log(1+u/1−u)=t but still I need your opinion sir. Thank you once again! $\endgroup$
    – Solarte
    Commented Dec 1, 2018 at 4:03

Actually I just want to leave comments but my reputation is too low... Im in favor with your idea of considering stag pressure in equation. In fact, during my undergrad I also raised that same question regarding the equation of motion of a parachute. On the otherhand, with regards to your original equation, I think it must be;

$$\frac{F}{m} - \frac{[CAρ(v_{air}^2 + v_{object}^2)]}{2m} = \frac{\mathrm{d}v_{object}}{\mathrm{d}t}$$

Since your object is a valve which is usually used for closing therefore the velocity of air will not remained constant. Anyway either the velocity of air is constant or not, the velocity of your object in still air must be known so that;

$$v_{object} = v_{object\ still\ air} - v_{air}$$

I don't know exactly your situation, but incorporating this equation to your original equation which will lead you to partial frac and with known values you can determine v(t).


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