1. From points $A$ and $B$, which are 70 km apart, a pedestrian and a cyclist set off simultaneously at speeds of 5 km/h and 15 km/h, respectively. What distance will be between them after 3 hours?
1. In this problem, the direction of movement of the cyclist and the pedestrian is not specified, so there are four different cases.

Case 1. Moving towards each other:

![](https://cdn.mathpix.com/cropped/2024_05_21_e53d4ff4f2278a1c0d12g-074.jpg?height=266&width=945&top_left_y=581&top_left_x=561)

1) $5+15=20($ km/h) - closing speed;
2) $20 \times 3=60$ (km) - they will have closed this distance in 3 hours;
3) $70-60=10$ (km) - they will be this distance apart after 3 hours.

Case 2. Moving in the same direction:

![](https://cdn.mathpix.com/cropped/2024_05_21_e53d4ff4f2278a1c0d12g-074.jpg?height=260&width=928&top_left_y=1195&top_left_x=564)

1) $5 \times 3=15$ (km) - the pedestrian walked in 3 hours;
2) $15 \times 3=45$ (km) - the cyclist rode in 3 hours;
3) $70-45=25$ (km) - the cyclist will have left to reach point $A$;
4) $25+15=40$ (km) - they will be this distance apart after 3 hours.

Case 3. Moving in opposite directions:

![](https://cdn.mathpix.com/cropped/2024_05_21_e53d4ff4f2278a1c0d12g-074.jpg?height=257&width=928&top_left_y=1887&top_left_x=564)

1) $5 \times 3=15$ (km) - the pedestrian walked in 3 hours;
2) $15 \times 3=45$ (km) - the cyclist rode in 3 hours;
3) $15+45=60$ (km) - they will have moved apart;
4) $60+70=130$ (km) - they will be this distance apart after 3 hours.

Case 4. Moving in the same direction:

![](https://cdn.mathpix.com/cropped/2024_05_21_e53d4ff4f2278a1c0d12g-075.jpg?height=255&width=930&top_left_y=318&top_left_x=563)

1) $5 \times 3=15$ (km) - the pedestrian walked in 3 hours;
2) $15 \times 3=45$ (km) - the cyclist rode in 3 hours;
3) $70-15=55$ (km) - the pedestrian will have left to reach $B$;
4) $55+45=100$ (km) - they will be this distance apart after 3 hours.

Answer: They may be 10, 40, 100, or 130 km apart.