A point mass moves along a straight path. We have recorded its trajectory data in the table below:

| time  $(\mathrm{s})$ | distance  $(\mathrm{m})$ | time  $(\mathrm{s})$ | distance  $(\mathrm{m})$ | time  $(\mathrm{s})$ | distance  $(\mathrm{m})$ | time  $(\mathrm{s})$ | distance  $(\mathrm{m})$ | time  $(\mathrm{s})$ | distance  $(\mathrm{m})$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 0.0 | 0.000 | 0.5 | 0.241 | 0.9 | 0.916 | 1.3 | 1.837 | 1.7 | 2.692 |
| 0.1 | 0.006 | 0.6 | 0.369 | 1.0 | 1.137 | 1.4 | 2.064 | 1.8 | 2.886 |
| 0.2 | 0.028 | 0.7 | 0.527 | 1.1 | 1.368 | 1.5 | 2.282 | 1.9 | 3.078 |
| 0.3 | 0.072 | 0.8 | 0.710 | 1.2 | 1.604 | 1.6 | 2.492 | 2.0 | 3.274 |
| 0.4 | 0.142 |  |  |  |  |  |  |  |  |

What would be the minimum measurement error of the distance if these results came from the observation of a uniformly accelerating body? (Consider the error in time measurement negligible.)
Solution. It is known that the distance-time function of a body undergoing such motion is a second-degree polynomial, with the usual notation $s(t)=\frac{a}{2} t^{2}+v_{0} t+s_{0}$. The question, therefore, is how well we can approximate the data in the table with the values of a second-degree polynomial.

If we subtract the values of any second-degree polynomial $P(t)$ from the distance data in the table, the data in the new table can be approximated as well by a polynomial $Q(t)$ as the original data by $P(t)+Q(t)$.

We would like to find a polynomial $P(t)$ such that when we subtract it from the data in the table, the new table will be "prettier," making it easier to recognize a well-approximating polynomial $Q(t)$. Looking at our table, we can observe that the data corresponding to 2.0 s is 1 greater than twice the data corresponding to 1.0 s,

| time  $(\mathrm{s})$ | distance  $(\mathrm{m})$ | time  $(\mathrm{s})$ | distance  $(\mathrm{m})$ | time  $(\mathrm{s})$ | distance  $(\mathrm{m})$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 0.0 | 0.0000 | 0.7 | $-0.1639$ | 1.4 | 0.1922 |
| 0.1 | $-0.0627$ | 0.8 | $-0.1196$ | 1.5 | 0.2015 |
| 0.2 | $-0.1194$ | 0.9 | $-0.0623$ | 1.6 | 0.1928 |
| 0.3 | $-0.1641$ | 1.0 | $-0.0000$ | 1.7 | 0.1641 |
| 0.4 | $-0.1928$ | 1.1 | 0.0623 | 1.8 | 0.1194 |
| 0.5 | $-0.2025$ | 1.2 | 0.1196 | 1.9 | 0.0627 |
| 0.6 | $-0.1932$ | 1.3 | 0.1639 | 2.0 | 0.0000 |

subtracting half of the square of the time from them, it will be exactly twice that (0.637, and 1.274, respectively). Therefore, let's subtract the polynomial $P(t)=\frac{1}{2} t^{2}+0.637 t$ from the table. The differences obtained are contained in the new table, and let's also plot the points in a coordinate system. The resulting point system is essentially symmetric about the point $(1 ; 0)$ (there is a 0.001 deviation at 2-2 points), so it can be well approximated by a line passing through the point of symmetry. The slope of the line is determined so that the error is the same at the points $(0 ; 0)$ and $(0.6 ;-0.1932)$. The error, denoted by $h$, can be written from similar triangles as

$$
\frac{0.1932-h}{h}=\frac{0.4}{1}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_188a2e1179c89d432fb7g-1.jpg?height=773&width=1797&top_left_y=1538&top_left_x=153)

from which $h=0.138$ is obtained. Thus, the equation of the line is $Q(t)=0.138 t-0.138$, the function that approximately describes the motion is $s^{*}(t)=\frac{1}{2} t^{2}+0.775 t-0.138$, and the error is less than $0.138$ at every point.

A lower bound for the error can be obtained if we only want to find the best approximating second-degree function $f(t)=a t^{2}+b t+c$ for the points with abscissae $t_{1}=0 ; t_{2}=0.6 ; t_{3}=1.4$ and $t_{4}=2.0$. For these $t_{1}, t_{2}, t_{3}, t_{4}$ numbers (and generally for any such quartet of numbers), the numbers $\lambda_{1}, \lambda_{2}, \lambda_{3}, \lambda_{4}$ can be determined such that the following equalities hold:

$$
\sum_{i=1}^{4} \lambda_{i} t_{i}^{n}=0, \quad n=0,1,2
$$

This is a system of three equations for the unknowns $\lambda_{1}, \lambda_{2}, \lambda_{3}, \lambda_{4}$. It can be proven that the system of equations always has (infinitely many) solutions. In our case, $\lambda_{1}=2, \lambda_{2}=-5, \lambda_{3}=5, \lambda_{4}=-2$ is a solution (the calculation is not detailed). Let $s_{i}$ denote the value measured at $t_{i}$, and let $\Delta_{i}=s_{i}-f\left(t_{i}\right)$ be the error of the approximation at $t_{i}$. Then

$$
\sum_{i=1}^{4} \lambda_{i} \Delta_{i}=\sum_{i=1}^{4} \lambda_{i}\left(s_{i}-a t_{i}^{2}-b t_{i}-c\right)=\sum_{i=1}^{4} \lambda_{i} s_{i}-a \sum_{i=1}^{4} \lambda_{i} t_{i}^{2}-b \sum_{i=1}^{4} \lambda_{i} t_{i}-c \sum_{i=1}^{4} \lambda_{i}
$$

In (2), the last three terms on the right-hand side of the equality are 0 according to (1), so

$$
\sum_{i=1}^{4} \lambda_{i} \Delta_{i}=\sum_{i=1}^{4} \lambda_{i} s_{i}
$$

Taking the absolute value of both sides, and using the fact that the absolute value of a sum is not greater than the sum of the absolute values of the terms,

$$
\left|\sum_{i=1}^{4} \lambda_{i} s_{i}\right|=\left|\sum_{i=1}^{4} \lambda_{i} \Delta_{i}\right| \leq \sum_{i=1}^{4}\left|\lambda_{i}\right| \cdot\left|\Delta_{i}\right| \leq\left(\max _{i=1, \ldots, 4} \Delta_{i}\right) \cdot \sum_{i=1}^{4}\left|\lambda_{i}\right|
$$

from which we get

$$
\max _{i=1, \ldots, 4}\left|\Delta_{i}\right| \geq \frac{\left|\sum_{i=1}^{4} \lambda_{i} s_{i}\right|}{\sum_{i=1}^{4}\left|\lambda_{i}\right|}
$$

Applying formula (3) to the selected quartet of points and the already determined $\lambda_{i}$, we obtain

$$
\max _{i=1, \ldots, 4}\left|\Delta_{i}\right| \geq 0.1376
$$

It is obvious that the error of the parabola approximating the entire point system can only be larger than this. However, we have already found a parabola that approximates with an error of 0.138, so we can state that the measurement error of the distance is at least 0.138 units. (There is no point in looking for a parabola with an error of 0.1376 - although we could find one - since the original data was only accurate to three decimal places.)