1. **Problem statement:** We have ports A(34°N, 16°W), B(34°N, 24°E), C(26°S, 16°W), and towns P(30°N, 45°W), R(30°N, 25°E). We need to find distances between these points and solve time-related questions. 2. **Formula for distance on Earth's surface:** The Earth is approximated as a sphere with radius $R = 6371$ km. For two points with latitudes $\phi_1, \phi_2$ and longitudes $\lambda_1, \lambda_2$, the central angle $\Delta \sigma$ in radians is given by the spherical law of cosines: $$\cos(\Delta \sigma) = \sin(\phi_1) \sin(\phi_2) + \cos(\phi_1) \cos(\phi_2) \cos(\Delta \lambda)$$ where $\Delta \lambda = |\lambda_2 - \lambda_1|$. Distance $d = R \times \Delta \sigma$. Important: Convert degrees to radians: $\text{radians} = \text{degrees} \times \frac{\pi}{180}$. --- ### a) Distance between A and B, and A and C 3. Convert coordinates to radians: - $\phi_A = 34^\circ N = 34 \times \frac{\pi}{180} = 0.593$ rad - $\lambda_A = -16^\circ = -0.279$ rad - $\phi_B = 34^\circ N = 0.593$ rad - $\lambda_B = 24^\circ = 0.419$ rad - $\phi_C = -26^\circ = -0.454$ rad - $\lambda_C = -16^\circ = -0.279$ rad 4. Calculate $\Delta \lambda$: - Between A and B: $\Delta \lambda = |0.419 - (-0.279)| = 0.698$ rad - Between A and C: $\Delta \lambda = |-0.279 - (-0.279)| = 0$ rad 5. Calculate $\cos(\Delta \sigma)$ for A-B: $$\cos(\Delta \sigma) = \sin(0.593) \sin(0.593) + \cos(0.593) \cos(0.593) \cos(0.698)$$ Calculate each term: - $\sin(0.593) = 0.559$ - $\cos(0.593) = 0.829$ - $\cos(0.698) = 0.766$ So, $$\cos(\Delta \sigma) = 0.559^2 + 0.829^2 \times 0.766 = 0.312 + 0.687 \times 0.766 = 0.312 + 0.526 = 0.838$$ 6. Find $\Delta \sigma$: $$\Delta \sigma = \arccos(0.838) = 0.573 \text{ radians}$$ 7. Distance A-B: $$d = 6371 \times 0.573 = 3650 \text{ km}$$ 8. For A-C, since $\Delta \lambda = 0$, use difference in latitude: $$\cos(\Delta \sigma) = \sin(0.593) \sin(-0.454) + \cos(0.593) \cos(-0.454) \times 1$$ Calculate: - $\sin(0.593) = 0.559$ - $\sin(-0.454) = -0.439$ - $\cos(0.593) = 0.829$ - $\cos(-0.454) = 0.899$ So, $$\cos(\Delta \sigma) = 0.559 \times (-0.439) + 0.829 \times 0.899 = -0.245 + 0.745 = 0.5$$ 9. Find $\Delta \sigma$: $$\Delta \sigma = \arccos(0.5) = 1.047 \text{ radians}$$ 10. Distance A-C: $$d = 6371 \times 1.047 = 6670 \text{ km}$$ --- ### b) Ship sailing from A to B at 40 knots 11. Speed: 1 knot = 1.852 km/h, so speed = $40 \times 1.852 = 74.08$ km/h 12. Time to travel A-B: $$t = \frac{3650}{74.08} = 49.26 \text{ hours}$$ 13. Time zones: - Longitude difference between A and B is $24 - (-16) = 40^\circ$ - Each 15° longitude corresponds to 1 hour time difference - Time difference = $\frac{40}{15} = 2.67$ hours = 2 hours 40 minutes 14. i) Local time at B when ship left A: - A's local time: Monday 13:30 - B is ahead by 2h40m, so local time at B when ship left A is: $$13:30 + 2:40 = 16:10 \text{ Monday}$$ 15. ii) Arrival time at B: - Travel time = 49.26 hours = 2 days 1 hour 16 minutes - Departure Monday 13:30 + 2 days 1h16m = Wednesday 14:46 (A's local time) - Convert to B's local time by adding 2h40m: $$14:46 + 2:40 = 17:26 \text{ Wednesday}$$ --- ### c) Shortest distance between P(30°N, 45°W) and R(30°N, 25°E) 16. Convert to radians: - $\phi_P = \phi_R = 30^\circ = 0.524$ rad - $\lambda_P = -45^\circ = -0.785$ rad - $\lambda_R = 25^\circ = 0.436$ rad 17. $\Delta \lambda = |0.436 - (-0.785)| = 1.221$ rad 18. Calculate $\cos(\Delta \sigma)$: $$\cos(\Delta \sigma) = \sin(0.524)^2 + \cos(0.524)^2 \cos(1.221)$$ Calculate: - $\sin(0.524) = 0.5$ - $\cos(0.524) = 0.866$ - $\cos(1.221) = 0.342$ So, $$\cos(\Delta \sigma) = 0.5^2 + 0.866^2 \times 0.342 = 0.25 + 0.75 \times 0.342 = 0.25 + 0.257 = 0.507$$ 19. $\Delta \sigma = \arccos(0.507) = 1.038$ rad 20. Distance in km: $$d = 6371 \times 1.038 = 6613 \text{ km}$$ 21. Distance in nautical miles: - 1 nautical mile = 1.852 km $$d_{nm} = \frac{6613}{1.852} = 3571 \text{ nautical miles}$$ --- ### d) Distance between two points on equator at longitudes 102°W and 43°E 22. Latitude = 0°, so $\phi = 0$ rad 23. $\Delta \lambda = |43 - (-102)| = 145^\circ = 2.530$ rad 24. On equator, distance is: $$d = R \times \Delta \lambda = 6371 \times 2.530 = 16120 \text{ km}$$ --- **Final answers:** - a) Distance A-B = 3650 km, A-C = 6670 km - b) i) Local time at B when ship left A = 16:10 Monday ii) Arrival at B = 17:26 Wednesday - c) Distance P-R = 6613 km or 3571 nautical miles - d) Distance on equator = 16120 km