A new type of aircraft flew over a 378.6 km long test flight route between two finish lines with a time difference of 693.5 seconds. What was its speed? - What are the limits within which the speed lies if we also take into account that the exact length of the route can deviate by up to 30 meters from the stated value, and the exact time difference can deviate by up to 1.5 seconds from the given data? - How does the first result change if we take into account that the aircraft flew at an altitude of $10 \mathrm{~km}$? (Take the radius of the Earth as $6370 \mathrm{~km}$.)
I. Assuming the data ( $s$ distance and $t$ time) are precise, the speed, immediately converted to the practical km/h unit, is

$$
c=\frac{s}{t}=\frac{378.6 \mathrm{~km}}{693.5 \mathrm{sec}}=0.5459 \frac{\mathrm{km}}{\mathrm{sec}}=1965 \frac{\mathrm{km}}{\text { h }}
$$

Since the airplane was traveling at a distance of $6380 \mathrm{~km}$ from the Earth's center, and the designated segment is only $6370 \mathrm{~km}$ away from it, the distance traveled and the speed are greater, and we must multiply by the ratio of the distances

$$
\frac{6380}{6370}=1+\frac{1}{637}-\text { approx. }
$$

Thus, $c=1968 \mathrm{~km} /$ h.

II. According to the given measurement accuracies, we know that the length of the segment is not greater than 378.63 km and not less than 378.57 km, and the elapsed time is not longer than 695 sec and not shorter than 692 sec. Any number within these limits can be the correct value for the length of the segment and the time, so the value of $c$ can be greater or smaller than the above. $c$ increases if we increase the numerator and decrease the denominator, but it cannot be greater than the value of $3600 \cdot 378.63 / 692$, where we considered the largest possible value for the distance and the smallest possible value for the time. Similarly, it cannot be less than the value of $3600 \cdot 378.57 / 695$. Calculating these and rounding the upper limit upwards and the lower limit downwards, in any case,

$$
1960 \frac{\mathrm{km}}{\mathrm{h}}<c<1970 \frac{\mathrm{km}}{\mathrm{h}}
$$

Sándor Burszán (Budapest, Fazekas M. High School II. Class)

![](https://cdn.mathpix.com/cropped/2024_05_02_eb3f4953c3b23b6116fdg-1.jpg?height=602&width=686&top_left_y=1187&top_left_x=730)

Remark. The usual time-distance graph of the motion, based on the precise data, is the diagonal $O C$ of the rectangle $A O B C$ in the figure. Considering the possibility of errors, any point on the segment $A_{1} A_{2}$ on the time axis and the segment $B_{1} B_{2}$ on the distance axis can be the positions of points $A$ and $B$. Considering every possible pair, any point inside or on the boundary of the rectangle $C_{11} C_{21} C_{22} C_{12}$ can be the position of point $C$.

In such graphs, the magnitude of the speed is characterized by the slope of the segment $O C$. Any ray within the sector $C_{21} O C_{12}$ can replace the ray $O C$. The smallest slope and speed are given by considering the point $C_{21}$, and the largest values by considering the point $C_{12}$. (The figure is distorted!)

József Váradi (Budapest, Ságvári E. High School II. Class)