Nice to see some ship questions around here, I'm a naval engineer!
So, you're looking for a simple, ball-park number for a question that is in reality pretty complicated. Johannes answer might give reasonable results since it's constantly updating the number; but I want to point out some assumptions made here which might affect the result's accuracy.
Background Info: First is that Johannen's $C_d$ (which in Naval Arch is usually named $C_t$ since it would correspond to the total drag coefficient) is actually described as $C_t = f_1(\frac{V^2}{gL})+f_2(\frac{VL}{v})$, where $f_1$ and $f_2$ represent the wave-making (residuary) resistance coefficient ($C_r$) and frictional resistance coefficient ($C_f$) accordingly, $V$ is the ship's speed, $L$ is the ship's length, $g$ is gravitational acceleration, and $v$ is water's viscosity. As you can see, it's far from constant and changes from ship to ship, strongly dependent on their length. So to have an accurate result for your computer algorithm, you would need the chart for the boat's $C_t$. But even if you had this, it would still be off (but on the conservative side) since ships fouling strongly affect $C_f$.
Answering your Question: If your speed readings updated a little quiker, you could approximate the "instantanous" $C_t$ by approximating it with a Taylor's expansion, and then setting a system of equations with Johannes third equation. However, even with a first order approximation, you would need 3 samples or 1.5 minutes to get your first reading. And this might mean that the your "accuracy" might be lagging by the same amount. So, it might be that without any prior information of the ships (and no fancy smart/learning algorithms saving/estimating information of the ships from past data), the best you could do is Johannes approach, with some few modifications so that you can get the information you are asking for:
Quick-and-Dirty Method:
First (sorry for any Kosher mathematicians out there), consider that:
$$\frac{\partial^2 x}{\partial t^2} = \frac{\partial }{\partial x}\left(\frac{\partial t}{\partial x} \right ) = \frac{\partial V}{\partial t}\left ( \frac{\partial x}{\partial x} \right ) = V\left ( \frac{\partial V}{\partial x} \right )$$
Substituting this to Johannes third equation, and integrating using separation of variables (let's assume that Johannes $L$ is actually constant, and let's name it $\alpha$) with limits of integration $(0-x_{end})$ and $(V_0-\delta )$ for the $x$ and $V$ accordingly, we get:
$$x_{end} = \alpha\ln\left(\frac{V_0}{\delta}\right)$$
where $V_0$ would be your initial speed (in your case, your current speed), $\delta$ is the speed your going to end at, and $\alpha$ you assume to be a constant (but in reality you'll updated at each time step). You mentioned you want $\delta$ to be zero, but as you can see this is not possible, your result would be infinity (classic example of Zeno's Paradox, as Johannes result more clearly illustrates).
You have many options to estimate $\alpha$. If you get erratic results with the most basic option I'm going to present here, I recommend you look into derivative smoothing. The most basic option would be to use a numerical derivative in Johannes third equation, $$\frac{V_t-V_{t-1}}{\Delta t} = \frac{-1}{\alpha_t} V_t^2$$
Solving for $\alpha$, $$\alpha_t = \frac{V_t^2\Delta t }{V_{t-1}-V_t}$$
To make this obvious, you'll calculate at each time step $\alpha_t$, and apply it on
$$x_{end} = \alpha_t\ln\left(\frac{V_t}{\delta}\right)$$
Now $\delta$ would have to be a speed you'll reach when you're at $x_{end}$ (this result will be very ballpark, for the reasons I commented above). You mentioned zero, so a speed you'll consider small enough to be zero... perhaps 0.02 knots? But let's be real, in a river you'll have currents so you'll never really get to zero unless you're going upstream or you're facing some strong winds. So you'll have to play around with $\delta$ until you get results that seem useful to you (and probably conservative as well).
