Given a known initial speed, how can I determine how far my car will coast before it stops, taking wind resistance into account? I'm assuming air resistance is the only force acting on my car (IE: no rolling resistance, no engine drag, no gravity, no traffic).  I'm also assuming that the car is a magic levitating sphere with a radius such that the air resistance formula gets simplified to:
$$ a = \frac {v^2} 2 $$
My end goal is to create a function, $D(v_0)$, that will tell me how far the car will coast before it stops, where $v_0$ is the speed that it starts at.
I started out with this formula:
$$D(v_0) = \int _ 0 ^ T V(t) \ \mathrm d t \text ,$$
where $T$ is the amount of time it would take for the car to come to a complete stop.
The car comes to a complete stop when $V(t) = 0$, so I can find $T$ by setting $V(t) = 0$ and solving for $t$. Then I can plug $T$ back into $D(v_0)$ to get the formula that I'm after!
I've worked out that the velocity at a given time is:
$$V(t) = V_0 - \int _ 0 ^ t \text {(air resistance at time $t'$)} \ \mathrm d t' \text .$$
The issue is that the air resistance depends on the current velocity.  When I try to expand it, I get:
$$V(t) = V_0 - \int _ 0 ^ t \frac {V(t')^2} 2 \ \mathrm d t' \text .$$
Now I have $V(t)$ defined in terms of itself, which isn't much help. Where do I go from here?
How can I find the integral of a function that references itself like this?
 A: In short, you can't define a function $\Delta x(v_o)$.
What you should is start with Newton's Second law:
$F = m*a$
or $F = m\frac{\mathrm{d}v(t)}{\mathrm{d}t}$
The only force you have is air resistance, so you'll have the following.
$-\frac{v(t)^2}{2} = m\frac{\mathrm{d}v(t)}{\mathrm{d}t}$
Now what you can do is treat the derivatives as "fractions" and multiply both sides of the equation by $\mathrm{d}t$ and then divide both sides by $\frac{v(t)^2}{2}$:
$-\mathrm{d}t = \frac{2*m}{v(t)^2}\mathrm{d}v(t)$
You can now integrate both sides:
$-\int\mathrm{d}t = 2*m\int v^{-2}\mathrm{d}v$
Which results in:
$ -t + C = -2*m*v(t)^{-1} $
Substituting $t = 0$ and $v(0)=v_o$ we can calculate the value of C:
$ C = -2*m*v_o^{-1} $
And now, we can isolate $v(t)$
$ -t -2*m*v_o^{-1} = -2*m*v(t)^{-1}$
$ v(t) = \frac{v_o}{1 + \frac{t*v_o}{2*m}} $
We can see an issue that arises in this problem's formulation, the equation for the final has no zeroes, and only approaches zero in the limit of $t \to \infty$. This happens because as the velocity gets smaller, the deceleration will get smaller and will never be enough to cease the movement. Since you can't calculate the time at which the car will stop, you can't define a function $\Delta x(v_o)$.
