# How do you calculate velocity reduction over time due to drag, when drag is also a function of velocity?

The drag equation is: $$F_D=\frac12 \rho C_d A v^2$$, assuming constant values let's simplify to $$F_D=Cv^2$$

Assuming there are no other forces to consider, I'm really not sure where to go from here. I tried plugging it into $$F=ma$$, but now I just have an acceleration that is dependent on instantaneous velocity. What I would like is the ability to plug in an initial velocity along with the travel time, and get out the final velocity to see how much it's slowed due to drag. I'm having difficulty conceptualising this problem because at every step the amount by which the velocity is reduced depends on the velocity at that given moment, it's self referential. This leads me to believe calculus must be involved, but I'm not sure how to set up the problem. It seems like there should be an analytical solution.

edit: Since my post was marked as a duplicate of "Why do rain drops fall with a constant velocity?" - No, I am not asking about terminal velocity. Suppose we fire a spherical ball of known dimensions horizontally at a speed of 500m/s and let it fly for exactly 5 seconds, I am asking if there is an analytical way to solve exactly how much the horizontal speed will be reduced by (ignoring any other external forces). After that 5 second period, will it be going 250m/s, 300m/s, 350?? This would be easy for me to do were the force constant, I could just apply the kinematics formula: $$v_f = v_i+at$$. My predicament is that the acceleration is not constant, and I don't know how to account for that.

• @MichaelSeifert the comment that was pointed out in Roger Vadim's comment spells out (in somewhat suspect notation) the separation of variables and resulting integral in the $v^2$ drag case. Nov 9, 2022 at 17:53
As you have figured out, assuming that the velocity $$v$$ of the object is positive, the velocity of the object satisfies $$m \frac{dv}{dt} = -C v^2 \quad \Rightarrow \quad \frac{dv}{dt} = -\alpha v^2,$$ where $$\alpha \equiv C/m$$. This is what's called an ordinary differential equation (ODE), since it relates the function $$v(t)$$ to its derivative with respect to $$t$$. The solution will be a function $$v(t)$$ which satisfies this relationship at every point and (in addition) has the correct initial velocity (i.e., $$v(0) = v_0$$ for whatever $$v_0$$ the object starts with.
• Next, I did try this actually, solving the ode using separation of variables, here was my method: $\int\frac{1}{v^2}\:dv=\int-A\:dt\\$, ==> $-\frac{1}{v}+C_1=-At+C_2\\$, ==> $\frac{1}{v}=At+C_3\\$, ==> $v=\frac{1}{At+C_3}$, I'm not sure what I've accomplished here though. I've seemingly ended up with a linear function, where I know the reduction in velocity can not be linear. I also have an unknown constant. Nov 10, 2022 at 10:23
• @Scott: The negative sign is there because a drag force is in the opposite direction of the motion; so if $v$ is positive, $dv/dt$ must be negative. Your answer looks correct to me, and it's not linear; linear would mean that (for example) $\Delta v$ between $t = 0$ and $t = 1$ would be the same as $\Delta v$ between $t = 1$ and $t=2$, but it's not hard to show that this is not the case. Finally, the constant $C_3$ is determined by the condition that $v(0) = v_0$; in this case, this implies that $C_3 = 1/v_0$. Nov 10, 2022 at 12:41