# Can average acceleration be exactly determined from discrete points separated by equal time intervals? [duplicate]

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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

1. Mathematics An inexact result adequate for a given purpose.

https://ahdictionary.com/word/search.html?q=average

1. 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.

## marked as duplicate by ja72, user191954, Kyle Kanos, Aaron Stevens, ZeroTheHeroOct 19 '18 at 23:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• I don't know the answer, but it is possible to prove it for simple examples. Take a function $x(t)$ and discretize it, and calculate. I think there is something interesting with velocity, but calculations are needed. – santimirandarp Oct 16 '18 at 4:56
• Possible duplicate of How to approximate acceleration from a trajectory's coordinates? – ja72 Oct 16 '18 at 18:56

## 3 Answers

With three points you must assume a constant acceleration. Then this problem becomes a problem of fitting a parabola through three points. The general solution is given by the Lagrange interpolation where the 3-point acceleration is

$$a=2\dfrac{\frac{x_3-x_2}{t_3-t_2}-\frac{x_2-x_1}{t_2-t_1}}{t_3-t_1}=\dfrac{v_2-v_1}{\frac{t_3-t_1}{2}}$$

This matches the solution in your question.

Also a very practical analytic and graphical solution for experimental data points is given by Desmos.

You have a series of $$x$$ and $$y$$ values in a table (as well as time), and you want to calculate a cubic spline through the points. This will give you the velocities and acceleration (components) at each time frame.

Then use the .Derivative() method to get what you want.

Ref this post

The difference between an approximation of acceleration and an exact formulation of average acceleration can be seen by observing that thee data points are sufficient to approximate instantaneous acceleration; a formulation of average acceleration requires a minimum of four data points.

That I have provided distinct expressions for these distinct quantities should be sufficient to convince the qualified reader of the distinction between these similar concepts.

The answer is: Yes, it is possible to formulate an exact expression which can reasonably be called average acceleration. What I have called the approximate acceleration at time $$t_p$$ is an exactly defined value derived from two average velocities. Applying the same method to two consecutive "approximate" accelerations gives the two forms

$$\left\langle \mathfrak{a}_{p}\right\rangle \equiv\frac{\mathfrak{a}_{p}-\mathfrak{a}_{p-1}}{\Delta t}$$

$$=\frac{1}{\Delta t^{2}}\left(\left\langle \mathfrak{v}_{p+1}\right\rangle -2\left\langle \mathfrak{v}_{p}\right\rangle +\left\langle \mathfrak{v}_{p-1}\right\rangle \right)$$

$$=\frac{1}{\Delta t^{3}}\left(\mathfrak{r}_{p+1}+3\left(\mathfrak{r}_{p-1}-\mathfrak{r}_{p}\right)-\mathfrak{r}_{p-2}\right).$$

I'm not convinced of the utility of this expression, but it is an exact formulation consisting entirely of the input data, having a form similar to that used for average velocity.

I strongly urge anyone wishing to borrow this result to double-check my algebra and reasoning.