A ship leaves port at noon and has a bearing of S 24° W. The ship sails at 30 knots (or 30 nautical miles per hour).
(a)
How many nautical miles south and how many nautical miles west will the ship have traveled by 6:00 p.m.? (Round your answers to two decimal places.)
south			
 
 nautical miles
west			
 
 nautical miles
(b)
At 6:00 p.m., the ship changes course to due west. Find the ship's bearing (in degrees west of south) and distance from port (in nautical miles) at 7:00 p.m. (Round your answers to one decimal place.)
bearing			S 
 
 °W
distance			
 nautical miles

### (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**

---

### (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**

---

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.

A ship leaves port at noon and has a bearing of S 24° W. The ship sails at 30 knots (or 30 nautical miles per hour). (a) How many nautical miles south and how many nautical miles west will the ship have traveled by 6:00 p.m.? (Round your answers to two decimal places.) south nautical miles west nautical miles (b) At 6:00 p.m., the ship changes course to due west. Find the ship's bearing (in degrees west of south) and distance from port (in nautical miles) at 7:00 p.m. (Round your answers to one decimal place.) bearing S °W distance nautical miles

This problem involves calculating the ship's travel in terms of southward and westward distances, using trigonometry and vector decomposition. Additionally, the bearing and total distance from port after a course change to due west are determined. Suitable for students in Grades 10-12.

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