1. Detective Podberezyakov is pursuing Maksim Detochkin (each driving their own car). At the beginning, both were driving on the highway at a speed of 60 km/h, with Podberezyakov lagging behind Detochkin by 2 km. Upon entering the city, each of them reduced their speed to 40 km/h, and upon exiting the city, finding themselves on a good highway, each increased their speed to 70 km/h. When the highway ended, at the border with a dirt road, each had to reduce their speed to 30 km/h again. What was the distance between them on the dirt road?
Answer: 1 km.

Solution: At the moment Detochkin entered the city, Podberezyakov was 2 km behind him. Podberezyakov covered this distance in 2/60 hours, during which time Detochkin traveled $\frac{2}{60} \cdot 40$ km. That is, the distance was multiplied by 40/60. Similarly, when exiting the city and transitioning to the dirt road, the distance is multiplied by 70/40 and then by 30/70. In the end, $2 \times \frac{40}{60} \times \frac{70}{40} \times \frac{30}{70}=1$ km.