# Both Quadratic and Linear Drag

1. Jan 27, 2008

### Col.Buendia

An object is coasting on the horizontal axis, in the positive direction and is subject to a drag force f = -bv - cv$$^{2}$$.
Write down Newton's 2nd Law and solve for v using separation of variables.

So first I wrote out Newton's law as:
F= m(dv/dt) = -bv - cv$$^{2}$$

Solving the integral: dt = $$\frac{dv}{-bv-cv^{2}}$$, with boundaries from 0 to t and v(0) to v

I got: t = $$\frac{-m}{b}$$ ln$$\frac{v}{1+\frac{c}{b}v^{2}}$$

Note: I haven't put in my boundaries on v yet. However, Once I put in my boundaries on v and try to rearrange to solve for v, I can't get anywhere. Any suggestions or help would be greatly appreciated.

Thx, CB

2. Jan 28, 2008

### CompuChip

Are you sure about the ln? If I differentiate it, I first get
$$\frac{ 1 + (c/b) v^2 }{ v }$$
times the derivative of that (chain rule!) which will be something like (... - ...) / v^2
in total giving me something containing 1/v^3.

If you want to do it analytically, you could try writing
$$\frac{-dv}{v} \frac{1}{b + c v}$$
as
$$\left( \frac{A}{v} + \frac{B}{b + c v}\right) dv$$
with A and B constants (which, in this case, works). Then split the integral and solve both parts separately.

3. Jan 28, 2008

### Col.Buendia

Well looks like I made an error in my calculations: the integral turns out to be :
t =$$\frac{-m}{b} \frac{1}{b + c v}$$ ln$$\frac{v}{1+\frac{c}{b}v}$$

And yes, using partial fractions will also solve this integral, and yes a ln function will turn out to be there.

Anyway, then the solution is solved for just rearranging for v in terms of t.