Velocity at a given time with known force, drag, area and mass I am trying to plot the speed of an object of known mass, area and drag coefficient in a given density of fluid.
I have correctly calculated the terminal velocity by
$$v_t=\sqrt{\frac{2[Force]}{\rho[Area][C_d]}} $$
However I have tried a few methods to determine the velocity between t=0 and a given time.
They all involve gravity and mass for the force and when I swap out for a given force, it all goes wrong.
Hopefully somebody can assist me.

Edit
Some more info, the object (a boat) is initially at rest before a constant thrust force is applied.   $$f = ma$$ rearranged gives acceleration and applying delta-t gives me the speed at any given time, however it all goes wrong when i apply a drag equation,
$$F_d = [0.5 * (Rho * V^2)]  *  [Cd] *  [Area]$$
I can correctly calculate the increase in drag with speed and initially tried an iterative formula in excel which solved the applied force minus the drag force to give net force and thus acceleration, applying a dT step and using that for V, and having excel loop until a steady V was found where Fd balanced the applied thrust.   It worked to a point however it never successfully found a point which matched a terminal velocity calculation, which i was using to check the incremental time step calculation.
The closest i have got is using a free falling object calculation in a spreadsheet called "Falling Motion under Gravity: Resistance as Velocity Squared" by Michael Fowler, University of Virginia:
$$V_t = [Vprev] + (g-((Cd/m)*[Vprev]^2)*deltaT$$
But (and in answer to a question) my calculus is no where near good enough to put all of this into a pot and come up with a way to remove gravity and add in the Rho of water and drag area of the boat.   We are ignoring air as the V will be low, less than 3m/sec so air resistance is going to be negligible.
From the terminal velocity calculation I get a sensible value for the final speed, i just want to be able to plot speed with regards to time and end up with a flat line representing the terminal velocity.
Many thanks :)
 A: The aerodynamic drag force can be generalized as $F = -( \tfrac{1}{2} \rho A C_D) v^2$ and assuming there is some other constant thrust load $T$ applied (or weight $T=m g$), the equation of motion are
$$ T -( \tfrac{1}{2} \rho A C_D) v^2 = m  a \tag{1}$$
You correctly identified the terminal velocity $v_f$ from the above by setting $a=0$ and solving for $v$
$$ v_f = \sqrt{ \frac{2 T}{\rho A C_D} } \tag{2} $$
Now the acceleration value as a function of any speed $v \leq v_f$ can be written as
$$ a = \frac{T}{m} \left( 1 - \frac{v^2}{v_f^2} \right) \tag{3}$$
Using calculus and the relationship ${\rm d}v = a\,{\rm d}t$, the time to reach a certain speed $v$, starting from $v_i$ is found using direct integration
$$ \Delta t = \int \frac{1}{a}\,{\rm d}v = \int_{v_i}^v \frac{1}{\frac{T}{m} \left( 1 - \frac{v^2}{v_f^2} \right)}\,{\rm d}v$$
Carry out the integral to get
$$ \Delta t = \frac{m v_f}{2 T} \ln \left( \frac{ \frac{v+v_f}{v_i+v_f} }{ \frac{ v-v_f}{v_i-v_f} } \right)  $$
and invert to solve for $v(t)$
$$ \large \boxed{ v(t) = \frac{2 v_f (v_f+v_i)}{(v_f-v_i) e^{-\frac{2 T}{m v_f} t} + v_f+v_i} - v_f } \tag{4} $$
Using a similar treatment you find the distance traveled from the relationship ${\rm d}v = \tfrac{a}{v} {\rm d}x$ or in integral from $\Delta x = \int \frac{v}{a}\,{\rm d}v$
The result is
$$ \Delta x = \frac{m v_f^2}{2 T} \ln \left( \frac{v_f^2-v_i^2}{v_f^2-v^2} \right) \tag{5} $$
to be used after speed is known from (4) to find distance.
A: I have programmed the formula from John Alexiou into excel, available here for anybody in the future, link will directly download the file:
https://drive.google.com/uc?export=download&id=199grgekATxTvB2e1NhPs5iV0B3WZEMj5
Many thanks for the assistance :)
