I am using the American dialect of the English language. The following definitions distinguish my question from the similar, previously asked question:
https://ahdictionary.com/word/search.html?q=approximation
- Mathematics An inexact result adequate for a given purpose.
https://ahdictionary.com/word/search.html?q=average
- Mathematics a. A number that typifies a set of numbers of which it is a function.
My question: is there an exact expression for the time-average of acceleration in terms of projectile position coordinates measured at instants in time separated by a constant, known time period?
I am not asking how to approximate acceleration based on this data. Though I am interested in that matter as well.
I am writing a short discussion of projectile motion, primarily for my own edification. I had originally intended to give exact expressions for both average velocity and average acceleration, based on my hypothetical input data. After playing with the math a bit, it occurred to me that it is not possible to give such an exact formulation of average acceleration. I wish to say as much in my essay, but it seems prudent to verify the fact before committing it to an affirmative statement.
By an exact expression for the time average of acceleration, I mean an expression similar to that used below to define the time-average of velocity.
The following is my method of approximating the acceleration near the times $t_{p}:$
Two-dimensional projectile position coordinates are recorded at instants in time denoted by
$$ t_{p}=p\Delta t, $$
where $p$ is an integer index increasing with time, and $\Delta t$ is a fixed time increment. These known coordinates are written
$$ \mathfrak{r}_{p}=\begin{bmatrix}y_{p}\\ z_{p} \end{bmatrix}. $$
We assume projectile coordinates have some unknown functional dependence on time, denoted by $$ \mathfrak{r}\left[t\right]=\begin{bmatrix}y\left[t\right]\\ z\left[t\right] \end{bmatrix},\text{ and }\mathfrak{r}_{p}=\mathfrak{r}\left[t_{p}\right]; $$
where $\mathfrak{r}\left[t\right]$ is sufficiently differentiable. Projectile displacement during a time interval $\Delta t_{p}=\left[\left(p-1\right)\Delta t,p\Delta t\right]=\left[t_{p-1},t_{p}\right]$ is determined from these measurements to be $$ \Delta\mathfrak{r}_{p}=\mathfrak{r}_{p}-\mathfrak{r}_{p-1}. $$ The time-average of velocity over a time interval $\Delta t_{p}$ is defined by the equation
$$ \left\langle \mathfrak{v}_{p}\right\rangle \equiv\frac{\Delta\mathfrak{r}_{p}}{\Delta t}. $$
First, suppose $\Delta t$ is large compared to changes in the actual velocity; which we denote as the unknown differentiable function of time, $$ \mathfrak{v}\left[t\right]=\frac{d\mathfrak{r}}{dt}\left[t\right]. $$
Hypothetically, velocity could be zero over most of an interval $\Delta t_{p},$ then change rapidly during some small final sub-interval, so that the entire displacement $\Delta\mathfrak{r}_{p}$ occurs in that final period. In such a case the time-average of velocity over the entire interval $\Delta t_{p}$ will not equal the numerical average of the velocities at the endpoints:
$$ \left\langle \mathfrak{v}_{p}\right\rangle \ne\frac{\mathfrak{v}_{p-1}+\mathfrak{v}_{p}}{2}=\mathfrak{v}_{p-1}+\frac{\Delta\mathfrak{v}_{p}}{2}. $$
Alternatively, if velocity were to change with time at a constant rate
$$ \frac{d\mathfrak{v}}{dt}=\mathfrak{a}\text{, so that} $$
$$ \mathfrak{v}\left[t\right]=\mathfrak{v}_{0}+t\mathfrak{a},\text{ and} $$
$$ \mathfrak{v}_{p}-\mathfrak{v}_{p-1}=\Delta t\mathfrak{a}=\Delta\mathfrak{v}_{p}=\Delta\mathfrak{v}, $$
we would find that
$$ \left\langle \mathfrak{v}_{p}\right\rangle =\mathfrak{v}_{p-1}+\frac{\mathfrak{v}_{p}-\mathfrak{v}_{p-1}}{2}=\mathfrak{v}_{p-1}+\frac{\Delta t}{2}\mathfrak{a} $$
$$ =\mathfrak{v}_{p}-\Delta\mathfrak{v}_{p}+\frac{\Delta t}{2}\mathfrak{a}=\mathfrak{v}_{p}-\frac{\Delta t}{2}\mathfrak{a}. $$ In such a case the exact relationships
$$ \left\langle \mathfrak{v}_{p}\right\rangle =\mathfrak{v}\left[t_{p}-\frac{\Delta t}{2}\right]=\mathfrak{v}_{p}-\frac{\Delta t}{2}\mathfrak{a},\text{ and} $$
$$ \left\langle \mathfrak{v}_{p+1}\right\rangle =\mathfrak{v}\left[t_{p}+\frac{\Delta t}{2}\right]=\mathfrak{v}_{p}+\frac{\Delta t}{2}\mathfrak{a} $$
would hold. Thus, in the case of constant acceleration, $$ \frac{\left\langle \mathfrak{v}_{p+1}\right\rangle -\left\langle \mathfrak{v}_{p}\right\rangle }{\Delta t}=\mathfrak{a}=\mathfrak{a}_{p}. $$
Still assuming constant acceleration; position as a function of time with constants $\mathfrak{r}_{0},\mathfrak{v}_{0},\mathfrak{a},$is written
$$ \mathfrak{r}\left[t\right]=\mathfrak{r}_{0}+\mathfrak{v}_{0}t+\frac{1}{2}\mathfrak{a}t^{2},\text{ and} $$
$$ \mathfrak{r}_{p}=\mathfrak{r}_{0}+\mathfrak{v}_{0}p\Delta t+\frac{1}{2}\mathfrak{a}\left(p\Delta t\right)^{2} $$
$$ =\mathfrak{r}_{p-1}+\mathfrak{v}_{p-1}\Delta t+\frac{1}{2}\mathfrak{a}\Delta t^{2}.\text{ So that} $$
$$ \Delta\mathfrak{r}_{p}=\mathfrak{r}_{p}-\mathfrak{r}_{p-1}=\left(\mathfrak{v}_{p-1}+\frac{\Delta t}{2}\mathfrak{a}\right)\Delta t $$
$$ =\left(\mathfrak{v}_{p}-\frac{\Delta t}{2}\mathfrak{a}\right)\Delta t,\text{ and} $$
$$ \Delta\mathfrak{r}_{p+1}=\mathfrak{r}_{p+1}-\mathfrak{r}_{p}=\left(\mathfrak{v}_{p}+\frac{\Delta t}{2}\mathfrak{a}\right)\Delta t. $$
Taking the difference $\Delta\mathfrak{r}_{p+1}-\Delta\mathfrak{r}_{p},$ then dividing both sides by $\Delta t;$ upon applying the definition of average velocity we arrive at
$$ \left\langle \mathfrak{v}_{p+1}\right\rangle -\left\langle \mathfrak{v}_{p}\right\rangle =\frac{\Delta\mathfrak{r}_{p+1}-\Delta\mathfrak{r}_{p}}{\Delta t} $$
$$ =\left(\mathfrak{v}_{p}+\frac{\Delta t}{2}\mathfrak{a}\right)-\left(\mathfrak{v}_{p}-\frac{\Delta t}{2}\mathfrak{a}\right) $$
$$ =\Delta t\mathfrak{a}. $$
Dividing by $\Delta t$ again gives us an expression for the acceleration in terms of our known data:
$$ \mathfrak{a}_{p}=\mathfrak{a}=\frac{\left\langle \mathfrak{v}_{p+1}\right\rangle -\left\langle \mathfrak{v}_{p}\right\rangle }{\Delta t} $$
$$ =\frac{1}{\Delta t}\left(\frac{\Delta\mathfrak{r}_{p+1}}{\Delta t}-\frac{\Delta\mathfrak{r}_{p}}{\Delta t}\right) $$
$$ =\frac{1}{\Delta t^{2}}\left(\begin{bmatrix}y_{p+1}-y_{p}\\ z_{p+1}-z_{p} \end{bmatrix}-\begin{bmatrix}y_{p}-y_{p-1}\\ z_{p}-z_{p-1} \end{bmatrix}\right) $$
$$ =\frac{1}{\Delta t^{2}}\left(\mathfrak{r}_{p+1}+\mathfrak{r}_{p-1}-2\mathfrak{r}_{p}\right). $$
Now we consider the case in which the acceleration changes with time, but the time increment $\Delta t$ is small in comparison to the acceleration. Our assumption of sufficient differentiability for $\mathfrak{r}$ insures us that the velocity near $\mathfrak{r}_{p}$ is linear to the first order in time. Thus our previous result provides a good approximation for the acceleration at $t_{p}:$
$$ \mathfrak{a}_{p}\approx\frac{1}{\Delta t^{2}}\left(\mathfrak{r}_{p+1}+\mathfrak{r}_{p-1}-2\mathfrak{r}_{p}\right). $$
This is an approximation of acceleration, not a precisely defined average.
