Question

A ship leaves port at noon and has a bearing of S 24° W. The ship sails at 20 knots. At 6:00 P.M., the ship changes course to due west.

a. How many nautical miles south and how many nautical miles west will the ship have traveled by 6:00 P.M.?
b. Find the ship's bearing and distance from the port of departure at 7:00 P.M.

Answer 1

Step-by-step explanation:

1. 28 nautical miles

Answer 2

a) Distance traveled:

South: 104.9 nautical miles

West: 58.2 nautical miles

b) Ship's position at 7:00 PM:

Bearing: 270° West

Distance from port: Approximately 113.5 nautical miles

Step-by-step explanation:

a) Distance traveled:

Total distance: The ship sails for 6 hours at 20 knots, so the total distance traveled is 6 hours * 20 knots/hour = 120 nautical miles.

Bearing components: We can use trigonometry to find the north-south and east-west components of the distance traveled.

South: Sin(24°) * 120 nautical miles ≈ 104.9 nautical miles south

West: Cos(24°) * 120 nautical miles ≈ 58.2 nautical miles west

b) Ship's position at 7:00 PM:

Westward distance: Since the ship sails due west for 1 hour, it covers an additional 20 nautical miles westward.

Resultant distance: Combining the south and west components, the total distance from the port is √(104.9² + (58.2 + 20)² ≈ 113.5 nautical miles.

Bearing: The ship's final direction is due west, which is 270° from the port.

Note: This is a simplified solution assuming the Earth is flat. In reality, Earth's curvature would slightly affect the calculations, but the general approach remains the same.