# Displacement with Air Drag

1. Jan 29, 2009

### Espen

This is not a homework problem, just an idea I had.

If we have a sphere which is being accelerated in space by a force $$\vec{F}$$ and we take the drag into account, what would the displacement be after t seconds? If the drag force is $$\vec{D}=-c_1v-c_2v^2$$, then the acceleration would be $$\vec{a}=\frac{\vec{F}+\vec{D}}{m} \,,\, a=\frac{F-c_1v-c_2v^2}{m}$$ such that the displacement $$s=\frac{1}{2}\left(\frac{F-c_1v-c_2v^2}{m}\right)t^2$$. Is this correct? Could someone help me solve this differential equation, if it has a solution?

Thanks for any help.

2. Jan 29, 2009

### 3DM@rk

No, that's not the correct formula for the displacement. The well-known $$s=\frac{1}{2}at^{2}$$ holds only if the acceleration is constant. In this case, the acceleration is a quadratic function of velocity, so we have one scalar differential equation:
$$a=\frac{dv}{dt}=A-Bv-Cv^{2}$$
where A= F/m, B=C1/m and C=C2/m are constants. The equation can be solved using separation of variables, as
$$(1) \int dt=\int\frac{dv}{A-Bv-Cv^{2}}$$

I can solve this easily if either B or C is equal to zero.
• $$D=-c_{1}v$$
Then by solving (1), and then another integration with respect to time, we get (for v(0) = 0)
$$s(t)=\frac{F}{c_{1}}\left(t+\frac{exp(-\frac{c_{1}}{m}t)-1}{\frac{c_{1}}{m}}\right)$$
where s(t) is the displacement after time t.
• $$D=-c_{2}v^{2}$$

$$v_{m}=\sqrt{\frac{F}{c_{2}}}$$
is the terminal velocity (where D=F). Then using partial fraction decomposition on (1), integration and then another integration with recpect to time, we get (again for zero velocity at the begining):
$$s(t)=\frac{v_{m}m}{\sqrt{Fc_{2}}}ln cosh\left(t\frac{\sqrt{Fc_{2}}}{m}\right)$$
where s(t) is the displacement at time t.

I'm too lazy at the moment to solve the problem for generally non-zero C1 and non-zero C2, which should be possible, but some difficulties might be encountered when inverting the function. I hope you got the picture how it works. Also, you can use numerical solution of the equation, which is the common approach in physics.

And finally, I apologize for any inconvenience caused by my English or incorrect English math terms.

3. Jan 31, 2009

### Espen

I solved the integral to

$$\int \frac{dv}{A-Bv-Cv^2}=-\frac{2\tan^{-1}\left(\frac{B+2Cv}{\sqrt{-B^2-4AC}}\right)}{\sqrt{-B^2-4AC}}$$, so then $$t=-\frac{2\tan^{-1}\left(\frac{B+2Cv}{\sqrt{-B^2-4AC}}\right)}{\sqrt{-B^2-4AC}}$$

Working with this, I get that

$$v=\frac{\sqrt{-B^2-4AC}\tan\left(\frac{\sqrt{-B^2-4AC}}{2}t\right)-B}{2C}$$

If this is correct, I want to integrate it in order to get an expression for the displacement over time.

$$s=\int \frac{\sqrt{-B^2-4AC}}{2C}\cdot\tan\left(\frac{\sqrt{-B^2-4AC}}{2}t\right)-\frac{B}{2C}\rm{d}t$$

I will simplify it be setting $$\frac{\sqrt{-B^2-4AC}}{2}=k_1$$ and get

$$s=\frac{k_1}{C}\cdot\int\tan\left(k_1\cdot t\right)\rm{d}t-\int\frac{B}{2C}\rm{d}t=\frac{k_1}{C}\left(-\frac{\log\left(\cos\left(k_1t\right)\right)}{k_1}\right)-\frac{B}{2C}t+s_0$$

How does this look?

4. Feb 1, 2009

### 3DM@rk

By $$tan^{-1}$$ you mean $$tanh^{-1}$$, right? Also note that the expression $$\sqrt{-B^2-4AC}$$ would be imaginary, since constants A, B and C are positive.
My solution of the integral is:

$$t=\frac{1}{\sqrt{B^2+4AC}}ln\left(\frac{\sqrt{B^2+4AC}+B+2Cv}{\sqrt{B^2+4AC}-B-2Cv}\right)$$

which could be written like yours considering the fact that:
$$\frac{1}{2}ln\left(\frac{1+x}{1-x}\right)=tanh^{-1}$$

The rest seems fine, it is even consistent with my solution for B=0.

Last edited: Feb 1, 2009
5. Feb 1, 2009

### arildno

Do remember, though, that the proper quadratic law for the air drag is $$-cv|v|$$, rather than $$-cv^{2}$$ for some constant c.

6. Feb 1, 2009

### 3DM@rk

But since the drag force acts in opposite direction to the velocity, for an accelerating force F we can use scalar equations with $$F_{d}=-cv^{2}$$. So the above calculations are correct.

7. Feb 1, 2009

### arildno

Well, that would depend upon the problem you were to model with this.
If you think of, say, launched projectile problem within a constant gravity field, then the general form is necessary.

The force law on the way up (velocity "positive") wll go as -cv^2, whereas on its way down (velocity "negative") as +cv^2.

8. Feb 1, 2009

### 3DM@rk

The better general formula then would be
$$\vec{F_{d}}=-cv^{2}\vec{\hat{v}}$$,
where $$\vec{\hat{v}}$$ is a unit vector codirectional with velocity.
But anyway, with some common sense you can always decide the correct direction of the drag force, so it doesn't accelerate the projectile.
Simple $$}-cv^{2}$$ is a much better function to calculate with, than some expression with an absolute value.
I agree with you and understand your point. Thanks for the comment.