Four people want to complete a journey of length $s$, but they only have one motorcycle. Their plan is as follows: they all start at the same time, $A$ and $B$ walk at a speed of $g$, while $C$ and $D$ ride the motorcycle at a speed of $m$ to a certain distance $d$. From there, $C$ continues on foot, and $D$ turns back to pick up $B$, bringing them forward until they catch up with $C$. Then $B$ and $C$ continue walking together, while $D$ turns back to pick up $A$ and brings them forward until they catch up with the others. At this point, they rest for $T$ time, and by repeating the previous plan twice, they exactly reach their destination. How did they choose the length of the $d$ segment? What average speed did the group plan for? - During the first rest, it turned out that the walking speed was only $g$ when two people walked together, but it was only $g_{1}(<g)$ when walking alone; furthermore, the motorcycle only reached the speed of $m$ without a passenger, but with a passenger, the speed was $m_{2}(<m)$. Did this affect the length of the $d$ segment in the repeated plan?

What fraction of the journey did each traveler complete on foot?

What was the average speed of the group?

Numerical data: $s=60 \mathrm{~km}, g=5, g_{1}=4, m=30, m_{2}=25 \mathrm{~km} /$ hour.
I. Let's first consider the actual course of the journey up to the first rest. Of course, we assume that $m_{2}>g$, as the task only makes practical sense this way. Let the duration of the time intervals between the start, the 1st, 2nd, 3rd, 4th direction change, and the arrival at the rest stop be $t_{1}, t_{2}, t_{3}, t_{4}, t_{5}$ respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_d68d1e353cbef2cd47eag-1.jpg?height=718&width=1196&top_left_y=300&top_left_x=451)

When the motion graphs of the 4 travelers are plotted in the same coordinate system, it is evident from the alignment of the slopes (speeds) that the graph is symmetric about the midpoint of the line segment representing $B$'s motor travel, hence $t_{4}=t_{2}$, and $t_{5}=t_{1}$. Furthermore, $A$ and $C$ travel the same distance on the motor, while $B$ travels a shorter distance. The distance traveled alone by $A$ and $C$ is also the same, as well as the distance they travel together with $B$.

$C$ travels the planned distance $d$ on the motor in $t_{1}=\frac{d}{m_{2}}$ time. In the same time, the distance traveled together by $A$ and $B$ is $g t_{1}=d \cdot \frac{g}{m_{2}}$, so the lead of $D$ and $C$ when the motor first turns back is

$$
e_{1}=d-d \cdot \frac{g}{m_{2}}=d \cdot \frac{m_{2}-g}{m_{2}}
$$

Then $D$ and the $A, B$ pair approached each other at a speed of $m+g$ and the lead became 0 at the meeting point, so

$$
t_{2}=\frac{e_{1}}{m+g}=\frac{d\left(m_{2}-g\right)}{m_{2}(m+g)} \quad\left(=t_{1} \cdot \frac{m_{2}-g}{m+g}\right)
$$

Meanwhile, $C$ and $D$ moved away from each other at a speed of $m+g_{1}$, so $C$'s lead at the end of this period is

$$
e_{2}=\left(m+g_{1}\right) t_{2}=e_{1} \cdot \frac{m+g_{1}}{m+g}=d \cdot \frac{m_{2}-g}{m_{2}} \cdot \frac{m+g_{1}}{m+g} \quad\left(2
$$

furthermore, $x_{C, 3}$ does not depend on $g_{1}$. However, $x_{C, 2}$, according to (2), can be larger or smaller than planned depending on the value of $g_{1}$; for example, if $g_{1}=g$, then $x_{C, 2}=x_{C, 3}$, but if $g_{1}=0$, then $x_{C, 2}=0$; therefore, the distance $x=d+x_{C, 2}+x_{C, 3}$ can be larger or smaller than $s / 3$.

Thus, the value of $d^{\prime}$ to be used in the two iterations of the program should be chosen such that - using the new values in the same way but with the comma used above - the following holds:

$$
d^{\prime}+x_{C, 2}^{\prime}+x_{C, 3}^{\prime}=\frac{s-x}{2}
$$

From this, $d^{\prime}$ can be expressed as a function of $s$ and the four speeds using the above.

For the numerical example, the distance to be traveled in each of the two iterations is 20.11 km, so

$$
d^{\prime}=d \frac{20.11}{19.78}=12.95 \text{ km}
$$

the increase is about 1.7%.

The time used in the iterations of the program also increases in this proportion, to 2 hours and 6 minutes. Allowing for half-hour breaks, they reach their destination in about 7 hours and 16 minutes from the start, and their average speed is about 8.3 km/h.

V. The proportion of the distance traveled on foot can be determined even from the results of the first part of the journey before the rest. We will only calculate this for the numerical example. For $C$ and $A$, the ratio is 7.05 / 19.78 = 0.356. $B$'s distance traveled on foot is 2 $x_{C, 3}=8 \text{ km}$, and the ratio is 0.405. With this, we declare the solution complete.

Remark. The partial results of the plan can, of course, be determined much more simply, but they cannot make the calculation of the above or other versions of the realized variant unnecessary.

For example, if the distance traveled together by $A$ and $B$ is denoted by $z$, then $D$ travels a distance of $2 d-z$ until picking up $B$, so from the ratio of the distances and speeds

$$
(2 d-z): z=m: g, \quad z=d \cdot \frac{2 g}{m+g}
$$

Thus, from the fact that the length of the first segment of the program is

$$
d+2 z=d\left(1+\frac{4 g}{m+g}\right)=\frac{s}{3}
$$

(5) follows.