(a) How many nautical miles south and how many nautical miles west will the ship have traveled by 6:00 p.m.?

Given:

• The ship's bearing is S 24° W, meaning it is traveling at an angle of 24° west of due south.
• Speed is 30 knots, or 30 nautical miles per hour.
• Travel time from noon to 6:00 p.m. is 6 hours.

To find the distances south and west, we break down the ship's velocity into components based on the bearing. The bearing S 24° W means the ship is moving southward and westward.

1. Southward distance: This is the vertical component of the velocity. We can calculate it using the sine function. $\text{South distance} = \text{Speed} \times \sin(\theta)$ where $\theta = 24^\circ$.

   $\text{South distance} = 30 \, \text{knots} \times \sin(24^\circ)$ $\text{South distance} \approx 30 \times 0.4067 \approx 12.2 \, \text{nautical miles}$

2. Westward distance: This is the horizontal component of the velocity. We calculate it using the cosine function. $\text{West distance} = \text{Speed} \times \cos(\theta)$ $\text{West distance} = 30 \, \text{knots} \times \cos(24^\circ)$ $\text{West distance} \approx 30 \times 0.9135 \approx 27.4 \, \text{nautical miles}$

Since the ship has been sailing for 6 hours, we multiply the above distances by 6:

• Southward distance: $12.2 \, \text{nautical miles} \times 6 = 73.2 \, \text{nautical miles}$

• Westward distance: $27.4 \, \text{nautical miles} \times 6 = 164.4 \, \text{nautical miles}$

Answer (a):

• South: 73.2 nautical miles
• West: 164.4 nautical miles

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(b) At 6:00 p.m., the ship changes course to due west. Find the ship's bearing and distance from port at 7:00 p.m.

At 6:00 p.m., the ship has already traveled southward and westward as calculated. The ship changes course to due west at 6:00 p.m., so the bearing will now be S 90° W.

Now, let's calculate the new distance and bearing at 7:00 p.m. (1 hour later).

1. Distance traveled due west from 6:00 p.m. to 7:00 p.m.: The ship is moving at 30 knots due west, so it will travel: $\text{Westward distance} = 30 \, \text{knots} \times 1 \, \text{hour} = 30 \, \text{nautical miles}$

2. Total distance from port: The total distance is the hypotenuse of a right triangle formed by the southward and westward components.

   Using the Pythagorean theorem: $\text{Total distance} = \sqrt{(73.2)^2 + (164.4 + 30)^2}$ $\text{Total distance} = \sqrt{(73.2)^2 + (194.4)^2}$ $\text{Total distance} = \sqrt{5365.44 + 37744.16} = \sqrt{43109.6} \approx 207.7 \, \text{nautical miles}$

3. Bearing: Since the course is now due west (S 90° W), the bearing at 7:00 p.m. remains S 90° W.

Answer (b):

• Bearing: S 90° W
• Distance from port: 207.7 nautical miles

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Let me know if you'd like any further clarification or details! Here are some related questions for further practice:

1. How would the ship's travel be affected if the bearing was instead S 45° W?
2. If the ship had sailed at 25 knots instead of 30, how would the distances and times change?
3. What is the angular difference between a bearing of S 24° W and S 90° W?
4. If the ship had stayed on its initial bearing until 9:00 p.m., how far south and west would it have traveled?
5. How do the Pythagorean theorem and trigonometry apply to navigation?

Tip: When calculating distances in navigation, always break down the motion into its north-south and east-west components using trigonometry.