# Average Velocity ($\vec{\bar{v}}$) Intuition and Analogy for Non-Uniform Acceleration

## Background

So I was trying to make as much sense out of kinematics through intuition after having taken my first semester of university physics, and I've stumbled onto a dillema that I can't seem to work around.

Specifically, I'm stuck on the issue of average velocity. From what I understand, in kinematics, there are two definitions through which everything else can be derived ($$\Delta{t}\neq 0$$): \begin{align} \vec{\bar{v}} &= \frac{\Delta{\vec{x}}}{\Delta{t}} \tag{1}\label{1}\\ \vec{\bar{a}} &= \frac{\Delta{\vec{v}}}{\Delta{t}} \tag{2}\label{2} \end{align}

There are, if my knowledge doesn't fail me, several methods of deriving equations past this point. One is via calculus; integrating to solve for equations describing displacement in terms of a constant acceleration, for example.

The other is via the assumption that, whenever $$\vec{\bar{a}}$$ is constant (in magnitude and direction, correct me if I am wrong), that: $$\vec{\bar{v}} = \frac{\vec{v}(t)+\vec{v}_i}{2}\tag{3}\label{3}$$

Through this you can combine equations algebraically to arrive at the same equations you would otherwise attain via calculus.

## Question 1

That got me thinking, however - what's the intuition behind this uniform acceleration-specific equation \eqref{3}?

If my Cal I doesn't fail me, for any given function's interval, the average value is: $$\bar{f} = \frac{\int_{a}^{b}f(x)dx}{b-a} \tag{4}$$

If we apply such idea to average velocity (assuming we can extend the logic to vectors), we get the following: $$\vec{\bar{v}} = \frac{\int_{t_1}^{t_2} \vec{v}(t)dt}{\Delta{t}} \tag{5}\label{5}$$ (Note: not sure if average velocity is considered a function of time, if so, I'm assuming it's denoted $$\vec{\bar{v}}(t)$$. Again, correct me if I am wrong, please.)

From \eqref{2} we can solve for $$\vec{v}(t)$$ and plug into \eqref{5} (otherwise the integral reduces to \eqref{1}).

As a result, we get: \begin{aligned} \vec{\bar{v}} &= \frac{\int_{t_1}^{t_2} (\vec{v}_i+\vec{\bar{a}}\Delta{t})dt}{\Delta{t}}\\ &= \frac{\vec{v}_i\Delta{t}+\frac{1}{2}\vec{\bar{a}}\Delta{t^2}}{\Delta{t}}\\ &= \vec{v}_i+\frac{1}{2}\vec{\bar{a}}\Delta{t} \end{aligned}\tag{6} By replacing $$\vec{\bar{a}}$$ with \eqref{2}, we end up with \eqref{3}, which is exactly what we intended to begin with. But why does this work?

## Question 2

My second inquiry involves whether a similar equation exists for average velocity beyond \eqref{1}, preferably in terms of $$\vec{v}_i$$ and $$\vec{v}(t)$$ with non-uniform acceleration. Either specifically in the case of jerk, defined as $$\vec{\bar{j}} = \frac{\Delta\vec{{a}}}{\Delta{t}}\tag{7}$$ or, even better, generalized to any velocity function $$\vec{v}(t)$$!

## Progress so Far

I've thought of three different possibilities for finding that elusive generalized average velocity function, each hypothesis less desirable than the last. The first is that perhaps it is of the form $$\vec{\bar{v}} = \frac{\vec{v}(t)+\vec{v}_i}{k}$$ where k is a unitless constant that might depend on the specific velocity function of the particle or object at play.

The second is that it's a lot more complicated, likely not of the aforementioned form and perhaps not expressable without adding more variables (like $$\vec{a}_i?$$) into the mix.

The last is that no such function exists, I'm guessing maybe due to $$\vec{v}(t)$$ not being linear, as $$\vec{a}(t)$$ is no longer constant (zero included).

Help would be greatly appreciated! Additionally, please feel free to correct any and all of my notation, as this is my first time posting and I need improvement.

Well,

$$\int_{t_1}^{t_2}v(t)\text dt=\Delta x$$

so that's why that all works.

In general

$$\int_{t_1}^{t_2}\frac{\text df}{\text dt}\text dt=\Delta f=f(t_2)-f(t_1)$$

so this works for any function and its derivative.

Usually in physics we deal with instantaneous values with derivatives rather than average values with difference quotients.

$$v=\frac{\text dx}{\text dt}$$ $$a=\frac{\text dv}{\text dt}=\frac{\text d^2x}{\text dt^2}$$

This is how to handle scenarios where the acceleration is not uniform.

• Yes, this all makes sense, but I still fail to understand why it happens to be that when acceleration is a non-zero constant function of time, the average velocity is equal to the average between its initial and final. Is there no extension of said formula to non-uniform acceleration? – Naganite Jun 6 '20 at 3:11
• @Naganite that’s because in the constant acceleration case, velocity is a linear function of time (constant slope). This means the instantaneous slope at each point is equal to average slope between two points. – Superfast Jellyfish Jun 6 '20 at 4:12
• @SuperfastJellyfish Ah, so then no extension of that function in terms of the initial and final velocity would be possible with for example jerk because the curve is quadratic? – Naganite Jun 6 '20 at 4:23
• @Naganite: "no extension of that function in terms of the initial and final velocity would be possible with for example jerk because the curve is quadratic?" Indeed! – Gert Jun 6 '20 at 6:48