# Equations for an object moving linearly but with air resistance taken into account?

I know (from Kinematics) that for an object moving linearly with an acceleration and without air resistance the following equations can be used to determine v(velocity) or x(position of the object) at any time:

$v=v_0+at$

$x=x_0+v_0t+\frac{1}{2}at^2$

Where $x_0$ is the position of the object at the start of accelerated movement, $v_0$ is the velocity at the start of the motion and a is the acceleration.

Now if I want to add air drag, say we take the formula for air drag as:

$D=0.5CA\rho v^2$

$C$ is a coefficent, $A$ is the reference Area, $\rho$ is the density and $v$ the velocity. Since $D$ is a force the air drag results in an acceleration, so in other words: The resulting acceleration is $a-D/m$, however $D$ depends on $v$, which depends on the acceleration. And that's where my problem is, I simply can't wrap my mind around that.

How would I go about getting the Equations for $x$ and $v$ including air drag (and don't forget there's acceleration too)?

Let $F$ be the independent force acting on the object. Let $D$ be the velocity dependent force acting in the opposite direction of $F$. The net force accelerating the object is just the difference. We have:

$F - D = ma$

Since $D$ is velocity dependent, the equation is a differential equation for the velocity.

$\dot v + \dfrac{CA\rho}{2m}v^2 = \dfrac{F}{m}$

This can be solved for $v(t)$ and which can then be integrated to find $x(t)$.

To solve this equation for constant independent force $F$, first note that there is a terminal velocity which can be found by setting the acceleration to zero, $\dot v = 0$:

$v^2_{term} = \dfrac{2F}{CA\rho}$

We can now rewrite the differential equation:

$\dfrac{1}{1-v^2/v^2_{term}}dv = \dfrac{F}{m}dt$

We can now straightforwardly integrate both sides to get:

$\tanh^{-1}(v/v_{term}) = \dfrac{F}{mv_{term}}t + C$

For zero initial velocity, we can finally write an expression for $v(t)$:

$v(t) = v_{term} \tanh(\frac{F}{mv_{term}}t)$

Thus, the velocity increases rapidly at first and then much more slowly, asymptotically approaching the terminal velocity which we can see from a plot of $\tanh$: • Hmm... not bad. I should be able to get the formula for x now, also I found it really cool that you gave me a formula where I could directly insert the terminal velocity, that could come in real handy later. – Wingblade Sep 26 '12 at 17:07
• Okay, got it working now! Thank you so much. I managed to figure out a formula for x(t) too. Btw, am I right that the C in tanh−1(v/vterm)=F*t/m*vterm+C is the initial velocity, which would be why it isnt in the formula for an initial velocity of zero? – Wingblade Sep 27 '12 at 11:07
• The integration constant $C$ in my next to last equation is related to the initial velocity but doesn't equal it. Setting $t=0$ in that equation gives: $C = \tanh^{-1}(V_0/v_{term})$. This has the effect of shifting the graph to the left (or the right if the initial velocity is opposite the force). Don't forget that C goes inside the parenthesis in the final equation. – Alfred Centauri Sep 27 '12 at 12:10
• Thank you very much. If you ever need something, ask me. Cause I sure owe you one. However I don't think there is anything I could help you with, except maybe javascript or something along those lines. – Wingblade Sep 27 '12 at 16:00
• You're welcome, thanks, and one final thing. I just recalled that this solution assumes that any initial velocity is in the direction of the force so that the drag opposes the force. For negative velocities, where the drag is in the same direction as the force, you must used $\tan$ instead of $\tanh$. I may add that to the answer for completeness. – Alfred Centauri Sep 27 '12 at 17:53

To build on Alfred's answer, taking into account the OP's premise of no other force than drag acting on the object, the equations can then be rewritten as

$$-\frac{dv}{v^2} = \frac{CA\rho}{2m}dt$$

$$\frac{1}{v} = \frac{CA\rho}{2m}t + \frac{1}{v_0}$$

$$v=\frac{dx}{dt}=\frac{1}{\frac{CA\rho}{2m}t + \frac{1}{v_0}}$$

$$x = \frac{2m}{CA\rho}\log(\frac{CA\rho}{2m}t + \frac{1}{v_0}) + x_0$$

So with a drag force quadratically dependent on speed, the object will never quite stop, as $v\rightarrow0$ as $t\rightarrow\infty$, and what is more relevant the movement is unbounded, as $x\rightarrow\infty$ as $t\rightarrow\infty$.

If you had chosen viscous damping, i.e. drag proportional to $v$, not $v^2$, then the resulting movement would have been bounded, with $x\rightarrow X$ as $t\rightarrow\infty$.

• Tell me if I'm wrong but the object also has an acceleration of its own, not just the air drag. And I dont see a(the acceleration) anywhere in your formula. Also you said no other force than drag, so I guess you must have missed that my movement is accelerated, which is somewhat like a force. Still thank you for your efforts. PS: I'd be eternally grateful if you could also include acceleration. – Wingblade Sep 25 '12 at 16:33
• @ValentinKrummenacher: what do you mean by 'acceleration of its own'? Is it being acted upon by a constant force? A non-constant force? Several? – Jerry Schirmer Sep 25 '12 at 17:22
• My description probably wasnt clear enough: It is being accelerated with a constant acceleration, which could be resulting from a constant force, and it is being acted upon by the air drag force. I'm sorry if I wasnt clear enough – Wingblade Sep 25 '12 at 17:23
• @ValentinKrummenacher, if the acceleration is constant, the equations for constant acceleration you have in your question still apply. Constant acceleration implies constant net force. Without drag, the independent force is thus constant. With drag, the independent force must continually increase, because of the continually increasing drag, in order to have constant acceleration. It occurs to me that you're conflating the notions of force and acceleration. – Alfred Centauri Sep 25 '12 at 19:16
• Man... why do I always get missunderstood. What I am trying to say is: I have my object, it has a constant force (Fc) pulling it in one direction, and it has air drag (Fd), therefore the resulting force will be Fc-Fd. Imagine maybe a rocket flying in a direction, thanks to its engine it has a certain acceleration, but it also has air drag slowing it down as it moves. You get me now? – Wingblade Sep 25 '12 at 19:28

## protected by Community♦Apr 15 '13 at 20:22

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).