13.326. Three cars are dispatched from $A$ to $B$ at equal time intervals. They arrive in $B$ simultaneously, then proceed to point $C$, which is 120 km away from $B$. The first car arrives there one hour after the second. The third car, upon arriving at $C$, immediately turns back and meets the first car 40 km from $C$. Determine the speed of the first car, assuming that the speed of each car was constant throughout the route.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1207.jpg?height=518&width=618&top_left_y=76&top_left_x=93)

Fig. 13.16
Solution.

Let $V_{1}, V_{2}, V_{3}$ be the speeds of the cars. Compare the time intervals (see Fig. 13.16), expressing them as the ratio of distance to speed:

$A P=\frac{x}{V_{1}}, M P=\frac{x}{V_{2}}, N P=\frac{x}{V_{3}}$.

By the condition:

$$
\frac{x}{V_{1}}-\frac{x}{V_{2}}=\frac{x}{V_{2}}-\frac{x}{V_{3}}, \text { hence }
$$

$\frac{1}{V_{1}}+\frac{1}{V_{3}}=\frac{1}{V_{2}}$ (1). Further, $D E=\frac{80}{V_{1}}$ and $D E=\frac{80+40}{V_{3}}$, from which $\frac{80}{V_{1}}=\frac{120}{V_{3}}$ (2); $D K=\frac{120}{V_{1}} ; D F=\frac{120}{V_{2}}$, from which $\frac{120}{V_{1}}-\frac{120}{V_{2}}=1$ (3). Solving the system of equations (1), (2), (3), we find $V_{1}=30 \mathrm{km} / \mathrm{h}$.

Answer: 30 km/h.