A truck started from city $A$ to city $B$ at midnight, and a car started from city $B$ to city $A$ (on the same route) at $t_{1}$. They met at $t_{2}$. The car arrived at its destination $r$ hours later than the truck. - After finishing their business, they started back, and met again at $t_{3}$, and finally arrived home at the same time. When did they arrive? -

Numerical data: $t_{1}=2^{\mathrm{h}} 40^{\mathrm{m}}, t_{2}=4^{\mathrm{h}}, r=0^{\mathrm{h}} 40^{\mathrm{m}}, t_{3}=14^{\mathrm{h}}$.
In the solutions, we will use the following notations: Assuming the speeds of the vehicles are constant, as is customary in such problems, let the speed of the truck be $v$, the speed of the car be $V$, the distance between $A$ and $B$ be $s$, and the common time of arrival at home be $x$.

I. Solution. Up to the first meeting, the truck travels $v t_{2}$, and the car travels $V\left(t_{2}-t_{1}\right)$, and together this is the entire distance

$$
s=v t_{2}+V\left(t_{2}-t_{1}\right)
$$

The travel time of the truck is $t_{1}-r$ hours more than that of the car, because the latter left $A$ $t_{1}$ hours later and arrived at $B$ $r$ hours later than the truck, i.e.,

$$
\frac{s}{v}=\frac{s}{V}+t_{1}-r
$$

Eliminating the distance $s$ from these two equations,

$$
t_{2}+\frac{V}{v}\left(t_{2}-t_{1}\right)=\frac{v}{V} t_{2}+t_{2}-r
$$

and rearranging this for the ratio $\frac{v}{V}$, we get the equation

$$
t_{2}\left(\frac{v}{V}\right)^{2}-r\left(\frac{v}{V}\right)-\left(t_{2}-t_{1}\right)=0
$$

The positive root of this equation is

$$
\frac{v}{V}=\frac{1}{2 t_{2}}\left(r+\sqrt{r^{2}+4 t_{2}\left(t_{2}-t_{1}\right)}\right)
$$

For $t_{2}>t_{1}(\geq 0)$, the root is always real. The negative root is not meaningful.

After the second meeting, both vehicles were on the road for $x-t_{3}$ more time, and during this time, they collectively covered the distance $s$ again, i.e.,

$$
v\left(x-t_{3}\right)+V\left(x-t_{3}\right)=s=v t_{2}+V\left(t_{2}-t_{1}\right)
$$

Dividing by $V+v$ and rearranging,

$$
x=t_{3}+t_{2}-\frac{1}{\frac{v}{V}+1} \cdot t_{1}
$$

With the numerical data

$$
\frac{v}{V}=\frac{2}{3} \quad \text { and } \quad x=16 \frac{2}{5} \text { hours, }
$$

i.e., both vehicles arrived home at 16 hours and 24 minutes.

Remarks. 1. The following consideration leads directly to a slightly rearranged form of equation (1). Up to the first meeting, the truck traveled $v t_{2}$, and the car traveled $V\left(t_{2}-t_{1}\right)$, and at that point, each had as much distance left as the other had already traveled. The truck took $\frac{V\left(t_{2}-t_{1}\right)}{v}$ time to cover the remaining distance, and the car took $\frac{v t_{2}}{V}$ time, but the latter arrived at the destination $r$ hours later, so

$$
\frac{V\left(t_{2}-t_{1}\right)}{v}+r=\frac{v t_{2}}{V}
$$

2. Similarly, $x$ can also be determined. From the start of the return trip to the second meeting, the truck traveled the distance $V\left(x-t_{3}\right)$ in front of the car. This took $\frac{V\left(x-t_{3}\right)}{v}$ time. Similarly, the car started the return trip $\frac{v\left(x-t_{3}\right)}{V}$ time before the meeting. The difference between these two times is also the difference in the times required to travel the entire distance. According to the data for the outbound trip, this time difference is $t_{1}-r$, i.e.,

$$
\frac{V\left(x-t_{3}\right)}{v}-\frac{v\left(x-t_{3}\right)}{V}=t_{1}-r
$$

from which $x$ can be calculated knowing the ratio $\frac{v}{V}$.

II. Solution. Let the first meeting point be $C$. The ratio of the travel times of the car and the truck on the $A B$ route is the same as on the $A C$ route. Let the travel times on the $A B$ route be $T$ and $t$, respectively, so

$$
T: t=\left(t_{2}-t_{1}\right):\left(t-t_{2}\right)
$$

Furthermore, we have seen that

$$
t-T=t_{1}-r
$$

Eliminating $T$,

$$
t^{2}-\left(2 t_{2}-r\right) t+t_{2}\left(t_{1}-r\right)=0
$$

from which the larger root is

$$
t=t_{2}+\frac{1}{2}\left(-r+\sqrt{r^{2}+4 t_{2}\left(t_{2}-t_{1}\right)}\right)
$$

(since it is obvious that $t>t_{2}$, the other root cannot give the solution to the problem).

Now we can calculate the time $x-t_{3}$ from the second meeting to the arrival home. In this time, the truck covers the fraction $\left(x-t_{3}\right) / t$ of the entire distance, and the car covers the fraction $\left(x-t_{3}\right) / T$ of the entire distance, and together they cover the entire distance (1 times the distance), i.e.,

$$
\begin{gathered}
\frac{x-t_{3}}{t}+\frac{x-t_{3}}{T}=1 \\
x=t_{3}+\frac{1}{\frac{1}{t}+\frac{1}{T}}=t_{3}+\frac{t}{1+\frac{t-t_{2}}{t_{2}-t_{1}}}=t_{3}+\frac{t\left(t_{2}-t_{1}\right)}{t-t_{1}}
\end{gathered}
$$

With the numerical data, $t=6, x=16.4$ is obtained.

III. Solution. We can also solve the problem using a calculation based on a sketch of the time-distance graph of the movements.

![](https://cdn.mathpix.com/cropped/2024_05_02_0d0c9579a5365d70b052g-2.jpg?height=413&width=1023&top_left_y=1062&top_left_x=537)

Each car's round trip (separately) is represented by segments with equal slopes, so the endpoints of these segments, as well as the points representing the meetings, form similar triangles, and the projections of the meeting points cut off proportional segments. With the notations of the first figure,

$$
\frac{B_{1} B_{2}}{B_{1} B_{3}}=\frac{A_{2} A_{1}}{A_{2} A}=\frac{B_{7} B_{6}}{B_{7} B_{5}}=\frac{B_{7} B_{6}}{B B_{3}}
$$

Let $A_{1} A_{2}=u$ be the time it takes for the car to reach $A$ after the first meeting. This is obtained as the positive root of the quadratic equation derived from the equality of the first two ratios:

$$
\begin{aligned}
& \frac{t_{2}-t_{1}}{t_{2}+u-r-t_{1}}=\frac{u}{t_{2}+u} \\
& u^{2}-r u-t_{2}\left(t_{2}-t_{1}\right)=0 \\
& u=\frac{1}{2}\left(r+\sqrt{r^{2}+4 t_{2}\left(t_{2}-t_{1}\right)}\right)
\end{aligned}
$$

Based on this, the desired time point is derived from the equality of the fourth and second ratios:

$$
\begin{gathered}
\frac{x-t_{3}}{t_{2}+u-r}=\frac{u}{t_{2}+u} \\
x=t_{3}+u \frac{t_{2}+u-r}{t_{2}+u}=t_{3}+u\left(1-\frac{r}{t_{2}+u}\right)
\end{gathered}
$$

Numerically, $u=8 / 3$ hours, and $x=16.4$ hours.