1. **Problem statement:** A motorcyclist and a cyclist start from points A and B, 180 km apart, and ride towards each other. They meet after 2 o'clock and continue without stopping. The motorcycle reaches B 3 hours earlier than the cyclist reaches A. We need to find their speeds. 2. **Define variables:** Let $v_m$ be the speed of the motorcyclist (km/h) and $v_c$ be the speed of the cyclist (km/h). 3. **Time relations:** Let $t$ be the time (in hours) from 2 o'clock to the meeting point. - Distance covered by motorcyclist until meeting: $d_m = v_m t$ - Distance covered by cyclist until meeting: $d_c = v_c t$ Since they start at the same time and meet after 2 o'clock, the sum of distances is total distance: $$d_m + d_c = 180$$ $$v_m t + v_c t = 180$$ $$t(v_m + v_c) = 180$$ $$t = \frac{180}{v_m + v_c}$$ 4. **Time to reach destinations after meeting:** - Motorcyclist remaining distance: $180 - d_m = 180 - v_m t$ - Cyclist remaining distance: $180 - d_c = 180 - v_c t$ Time taken after meeting: - Motorcyclist: $\frac{180 - v_m t}{v_m}$ - Cyclist: $\frac{180 - v_c t}{v_c}$ 5. **Given condition:** Motorcyclist reaches B 3 hours earlier than cyclist reaches A: $$\frac{180 - v_c t}{v_c} - \frac{180 - v_m t}{v_m} = 3$$ 6. **Substitute $t$ from step 3:** $$\frac{180 - v_c \frac{180}{v_m + v_c}}{v_c} - \frac{180 - v_m \frac{180}{v_m + v_c}}{v_m} = 3$$ Simplify numerator terms: $$\frac{180 - \frac{180 v_c}{v_m + v_c}}{v_c} - \frac{180 - \frac{180 v_m}{v_m + v_c}}{v_m} = 3$$ $$\frac{180 \left(1 - \frac{v_c}{v_m + v_c}\right)}{v_c} - \frac{180 \left(1 - \frac{v_m}{v_m + v_c}\right)}{v_m} = 3$$ Simplify inside parentheses: $$\frac{180 \frac{v_m}{v_m + v_c}}{v_c} - \frac{180 \frac{v_c}{v_m + v_c}}{v_m} = 3$$ Rewrite: $$\frac{180 v_m}{v_c (v_m + v_c)} - \frac{180 v_c}{v_m (v_m + v_c)} = 3$$ Multiply both sides by $v_m v_c (v_m + v_c)$: $$180 v_m^2 - 180 v_c^2 = 3 v_m v_c (v_m + v_c)$$ Divide both sides by 3: $$60 v_m^2 - 60 v_c^2 = v_m v_c (v_m + v_c)$$ 7. **Rewrite equation:** $$60 (v_m^2 - v_c^2) = v_m v_c (v_m + v_c)$$ Note that $v_m^2 - v_c^2 = (v_m - v_c)(v_m + v_c)$, so: $$60 (v_m - v_c)(v_m + v_c) = v_m v_c (v_m + v_c)$$ Divide both sides by $(v_m + v_c)$ (nonzero speeds): $$60 (v_m - v_c) = v_m v_c$$ 8. **System of equations:** - From step 3: $$t = \frac{180}{v_m + v_c}$$ - From step 7: $$60 (v_m - v_c) = v_m v_c$$ 9. **Express $v_m$ in terms of $v_c$:** $$60 (v_m - v_c) = v_m v_c$$ $$60 v_m - 60 v_c = v_m v_c$$ $$60 v_m - v_m v_c = 60 v_c$$ $$v_m (60 - v_c) = 60 v_c$$ $$v_m = \frac{60 v_c}{60 - v_c}$$ 10. **Substitute $v_m$ into total distance equation:** $$t = \frac{180}{v_m + v_c} = \frac{180}{\frac{60 v_c}{60 - v_c} + v_c} = \frac{180}{\frac{60 v_c + v_c (60 - v_c)}{60 - v_c}} = \frac{180 (60 - v_c)}{60 v_c + 60 v_c - v_c^2} = \frac{180 (60 - v_c)}{120 v_c - v_c^2}$$ 11. **Time to meet must be positive and less than total travel time. Try values for $v_c$ between 0 and 60:** Try $v_c = 20$: $$v_m = \frac{60 \times 20}{60 - 20} = \frac{1200}{40} = 30$$ Check equation 7: $$60 (30 - 20) = 60 \times 10 = 600$$ $$v_m v_c = 30 \times 20 = 600$$ True. 12. **Calculate meeting time:** $$t = \frac{180}{30 + 20} = \frac{180}{50} = 3.6 \text{ hours}$$ 13. **Calculate times after meeting:** - Motorcyclist remaining distance: $180 - 30 \times 3.6 = 180 - 108 = 72$ km - Time for motorcyclist after meeting: $\frac{72}{30} = 2.4$ hours - Cyclist remaining distance: $180 - 20 \times 3.6 = 180 - 72 = 108$ km - Time for cyclist after meeting: $\frac{108}{20} = 5.4$ hours Difference: $5.4 - 2.4 = 3$ hours, matches the problem. 14. **Final speeds:** - Motorcyclist speed: 30 km/h - Cyclist speed: 20 km/h 15. **Table:** | Vehicle | Speed (km/h) | Distance to Meeting (km) | Time to Meeting (h) | Remaining Distance (km) | Time After Meeting (h) | |--------------|--------------|-------------------------|---------------------|------------------------|-----------------------| | Motorcyclist | 30 | 108 | 3.6 | 72 | 2.4 | | Cyclist | 20 | 72 | 3.6 | 108 | 5.4 |