# Acceleration problem

1. Dec 26, 2007

### shiftless

Hi folks, I have a problem that I am trying to figure out that I hope you can help me with. I don't want to be spoon fed as I like discovering things for myself, but this one has me stumped and I'd appreciate some guidance.

The end result I am trying to obtain is a function which allows me to calculate the displacement of a land vehicle with respect to time, given its mass (constant, for the purposes of this simplified calculation), thrust from its jet engine (same), no friction, and here is the kicker, atmospheric drag. The problem as I am sure you are well aware is that drag increases as a square of velocity. The velocity function is a derivative of displacement so I am looking at a function which includes its own derivative.

I have studied Calculus I and II in college and I don't believe they alone provide me with the tools I need to solve this problem. It's been a while though so it's possible I've simply forgotten something that could help me. I did some Google searching and it appears that differential equations are what I need to know. However, I don't want to have to go out and buy a book on differential equations and study it for 6 weeks just to solve this one problem; if someone could just give me a somewhat detailed summary of what I'm up against and some tips and ideas on how I should go about tackling it, that would be very helpful.

Thanks in advance!

Last edited: Dec 26, 2007
2. Dec 26, 2007

### adventure

Dear friend!
hi
please write the equation which you want to solve. I will help u.

best,

3. Dec 26, 2007

### Staff: Mentor

Hi Shiftless, welcome to PF. Yes you are looking into solving a differential equation of the form:

x'' = a - b x'^2

This may look fairly simple, but it is actually a pretty advanced equation to solve. I don't even think that an equation of this form would be covered by the first semester of differential equations. You will probably have to look it up in a reference book. The only way I could solve it is with Mathematica.

EDIT: actually, this is the same as the differential equation

v' = a - b v^2

which is a first order non-linear differential equation

Last edited: Dec 26, 2007
4. Dec 26, 2007

### Rainbow Child

I assume you mean atmospheric resistance instead of atmospheric drag.

Let's call m the mass of the vehicle, $$f_j$$ the thrust from its jet engine and $$f_a=-k\,v^2$$ the atmospheric resistance where $$k$$ is a constant and $$v=\frac{d\,x}{d\,t}$$ the velocity of the vehicle.
From Newton's 2nd law we have

$$m\,\frac{d\,v}{d\,t}=f_j+f_a\Rightarrow\frac{d\,v}{d\,t}=\frac{f_j}{m}-\frac{k}{m}\,v^2$$

This is actually the DE DaleSpam, mentioned. This is a fisrt order DE that is separatable (if my English are correct! ). In order to solve it you must do what the name suggests ...separate it! That is to bring all the $v$ dependence is in one side of the equation and the $t$ dependence on the other, i.e.

$$\frac{d\,v}{\frac{f_j}{m}-\frac{k}{m}\,v^2}=d\,t\Rightarrow\int\frac{d\,v}{\frac{f_j}{m}-\frac{k}{m}\,v^2}=t+\tau \quad (1)$$

where $\tau$ is the constant of intergration. Having found the function $$v(t)$$ the only thing you to do in order to find the displacement of the vehicle $$x(t)$$ is to integrate it once more, i.e.

$$x(t)=\int v(t)d\,t+x_o$$

where $x_o$ is a second constant of integration.

P.S.1 If you have trouble with the integration of (1), try to split the fraction into two simpler ones
$$\frac{1}{\frac{f_j}{m}-\frac{k}{m}\,v^2}=\frac{A}{\sqrt{\frac{f_j}{m}}-\sqrt{\frac{k}{m}}\,v}+\frac{B}{\sqrt{\frac{f_j}{m}}+\sqrt{\frac{k}{m}}\,v}$$

P.S.2 If you insist on atmospheric drag replace $$k$$ with $$-k$$, in which case you have an integral of the form
$$\int \frac{d\,z}{a^2+b^2\,z^2}$$
which is resolved with the help of arc tangent function.

5. Dec 26, 2007

### shiftless

Hey, thanks! That's exactly what I was looking for. After posting this thread I went and wrote a simple C program to solve the problem by brute force; i.e. an iterative solver. It works, and it gave me an accurate solution, but not quite as elegant as doing it your way.

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