# 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 :)

• Your third paragraph is unclear. Please provide more details. $F=ma$ gives you a differential equation to solve. Sep 20, 2022 at 17:05
• How familiar are you with calculus? Sep 20, 2022 at 20:12
• Sep 20, 2022 at 20:12
• What forces are applied to the object? It is straight-up gravity, or thrust, or a combination? Sep 21, 2022 at 0:22

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.

• Hi, Thanks for your detailed answer - im inputting to excel and just sorting out my brackets etc - it was not working out as expected, but i note you have just edited so ill update to match your edit and let you know how i get on. $$((2*[Vf]*([Vf]-[Vi])) / (([Vf]+[Vi])*EXP(((2*[T])/([m]*[Vf]))*[t]) + ([Vf]-[Vi]))) + [Vf]$$ Sep 21, 2022 at 13:07
• Here is an example plot for speed vs time, when $v_i=1$, $v_f=5$ and the exponential is $e^{(-t/\pi)}$. Sep 21, 2022 at 13:13
• So i had come up with a method whereby i calculated the drag for a range of speeds, then separately calculated the acceleration based on the mass and the thrust minus drag. Then combined the delta_speed step with the avaerage accelerations between the two speed steps. This then gives me a table of times at which certain speeds were reached. I've programmed your solution as follows: <typing> $$=((2*[Vf]*([Vf]+[Vi])) / (([Vf]-[Vi])*EXP(-((2*T) / ([m]*[Vf]))*[t]) + ([Vf]+[Vi]))) - [Vf]$$ & then plotted our solutions: ibb.co/J54PCcR Looks good! Sep 21, 2022 at 13:18
• @StuartGraham - [new] make sure you have the negative sign in the exponent. Sep 21, 2022 at 13:18
• This would work for an approximation for a short range of speeds. If you are doing this in Excel, you can do a mini simulation, where each row is a change in speed, and you calculate the time step and distance step from that. Use $\Delta t = \frac{1}{a} \Delta v$ and $\Delta x = \frac{v}{a} \Delta v$ and use (3) for the acceleration values. Sep 21, 2022 at 13:21

I have programmed the formula from John Alexiou into excel, available here for anybody in the future, link will directly download the file: