Terminal velocity and acceleration

[physicals]

Homework Statement

How is acceleration affected by drag force in terminal velocity?

Relevant Equations

F=ma

Hey there,
I was just wondering how acceleration is affected in terminal velocity. The way I understood it was that acceleration decreased over time a constant rate since drag force increases at a constant rate. But after doing some research I have some doubts. Does drag force really increase at a constant rate? The image i have provided is from a google search and clearly it has shown that the rate at which acceleration decreases is not constant meaning the rate at which drag force increases, increases over time.

So the real question is, is the rate at which drag force increases really constant?

 

[PeroK]

In general, there may be both linear and quadratic components of the drag force. The linear component increases in proportional to the speed $v$. And the quadratic component increases in proportion to the speed squared $v^2$. The full differential equation would look like:
$$ma = g - k_1v - k_2v^2$$Where $k_1, k_2$ are constants.

 

[haruspex]

drag force increases at a constant rate

It depends what you mean by rate. Generally it means in relation to increase in time, but it can be used in relation to other parameters.
Drag varies with speed, but in a complicated way. At low speeds it is roughly linear, so you could say it increases at a constant rate wrt speed. At higher speeds it is roughly quadratic. The boundary depends mostly on attributes of the fluid medium.
It certainly does not increase uniformly wrt time (which would lead to the object reversing direction!).

 

[physicals]

so if drag force doesn't increase at a constant rate then the acceleration time graph wont be linear. So is the image provided right about the acceleration time graph for terminal velocity.

 

[PeroK]

so if drag force doesn't increase at a constant rate then the acceleration time graph wont be linear. So is the image provided right about the acceleration time graph for terminal velocity.

It's already been pointed out that "constant rate" normally refers to constant with respect to time. Drag increases with speed, not time.

 

[haruspex]

is the image provided right about the acceleration time graph for terminal velocity.

Yes.

 

[Chestermiller]

"Drag force increasing at constant rate" is poor terminology, but (reading the questioner's mind), it means that the drag force is proportional to the velocity of the object.

 

[kuruman]

To @physicals:
I will summarize what has been said.

The drag force must go to zero when the velocity is zero, otherwise the object will reverse direction after coming instantaneously to rest¹. One can write in general the acceleration due to drag as a poynomial in $v$, $$a(v)=\frac{1}{m}F_{\text{drag}}(v)=C_1v+C_2v^2 +\dots +C_nv^n$$ where the $C_i$ are known constants. Clearly, $a(0)=0$. The relevant rate of change is $\dfrac{da}{dv}$. When only $C_1$ is non-zero, this rate of change is constant as @Chestermiller pointed out.

As far as the terminal velocity is concerned, it does not affect the acceleration. By definition, terminal velocity is relevant when the acceleration is zero, i.e. the velocity of the mass is not changing. When, in addition to the drag force, you have external force $F_0$ acting on an object and you want to find the terminal velocity $v_{\text{ter}}$, you write
$$\begin{align} & a(v)=\frac{1}{m}F_0+C_1v+C_2v^2 +\dots +C_nv^n \nonumber \\
& 0=\frac{1}{m}F_0+C_1v_{\text{ter}}+C_2v_{\text{ter}}^2 +\dots +C_nv_{\text{ter}}^n \nonumber
\end{align}$$ and solve the polynomial equation for $v_{\text{ter}}$.

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¹This stipulation is swept under the rug when one says that the force of kinetic friction is constant $(f_k=\mu_k~N)$ but accepts that it goes to zero when the sliding mass stops moving.