Two straight highways intersect each other perpendicularly. On one highway, a car starts from a distance of $l_{1}=80 \mathrm{~km}$ from the intersection and heads towards the intersection at a constant speed of $v_{1}=80 \mathrm{~km} / \mathrm{h}$. On the other highway, another car starts at the same time from a distance of $l_{2}=60 \mathrm{~km}$ and heads towards the intersection at a speed of $v_{2}=60 \mathrm{~km} / \mathrm{h}$. At what rate does the distance between the cars change? What is the situation if the initial distances are $l_{1}=l_{2}^{\prime}=80 \mathrm{~km}$? How much time after the start will the two cars be closest to each other, and what is this distance in both cases?
In the first case, within 1 hour, both cars arrive exactly at the intersection, at which point their distance is $0 \mathrm{~km}$. From the theorem of parallel chords, it follows that their distance decreases uniformly from the initial $\sqrt{l_{1}^{2}+l_{2}^{2}}=100 \mathrm{~km}$ to zero (Figure 1), thus the speed of the distance change is $100 \mathrm{~km} /$ hour.

In the second case, the line connecting the two cars does not move parallel to itself (Figure 2), so the speed of the distance change is not constant. The distance between the cars as a function of time, based on the Pythagorean theorem, is:

$$
d(t)=\sqrt{\left(l_{1}-v_{1} t\right)^{2}+\left(l_{2}^{\prime}-v_{2} t\right)^{2}}=100 \sqrt{(t-1.12)^{2}+0.0256}
$$

(measuring distance in kilometers and time in hours). This expression is minimized when the squared term is zero, i.e., $t=1.12$ hours. The distance at this point is $d_{\min }=16 \mathrm{~km}$. The speed of the distance change between the two cars as a function of time is

$$
v(t)=\frac{100(t-1.12)}{\sqrt{(t-1.12)^{2}+0.0256}}(\mathrm{~km} / \text{ hour })
$$

This relationship can be determined by differentiating (1) with respect to time. (We also accepted as correct those papers that only determined the average speed for a specific segment. For example, from the start to the point of minimum distance, $v=86.7 \mathrm{~km} /$ hour is the average speed.)