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The problem describes a ship's journey, which involves two segments:

1. **First Leg (Noon to 1400H):**  
   The ship travels **due west** at 20 knots for 2 hours. Therefore, the distance covered during this period is:
   \[
   \text{Distance} = 20 \, \text{NM/h} \times 2 \, \text{hours} = 40 \, \text{NM}.
   \]
   This leg of the journey is directly west, along the negative x-axis.

2. **Second Leg (1400H to 1500H):**  
   After 1400H, the ship changes its course to **N 54° W**, and travels at the same speed of 20 knots for 1 hour. The distance covered in this leg is:
   \[
   \text{Distance} = 20 \, \text{NM/h} \times 1 \, \text{hour} = 20 \, \text{NM}.
   \]
   In this direction, the ship's movement can be broken down into two components: one in the **westward direction (x-axis)** and one in the **northward direction (y-axis)**.

   Using trigonometry:
   - Westward component (x-axis):  
     \[
     x_2 = 20 \times \cos(54^\circ)
     \]
   - Northward component (y-axis):  
     \[
     y_2 = 20 \times \sin(54^\circ)
     \]

### Total Distance and Bearing from the Port (at 1500H):

After calculating the components of each leg, we will determine the ship's total displacement (final position) and bearing.

Let me calculate the exact values for the displacement and bearing.The ship's position at 1500H is:

- **Distance from the port**: 54.23 nautical miles (NM)
- **Bearing**: \(72.64^\circ\) west of north (or approximately N 72.64° W).

Would you like further details or have any questions?

Here are five related questions for further exploration:
1. What would the ship's bearing and distance be if it continued on the N 54° W course for another hour?
2. How would the result change if the ship traveled at 15 knots instead of 20 knots?
3. What if the ship changed its course to N 30° W at 1400H? How would this affect the distance and bearing?
4. How does calculating displacement differ for varying speed or angles of travel?
5. Can the bearing be expressed in a different format, such as relative to the east or west directions?

**Tip:** Always break down a ship’s course into components when dealing with non-cardinal directions.

A ship leaves port at noon and heads due west at 20 knots. At 1400H, the ship changes course to N 54° W. Find the ship's bearing and distance from the port of departure at 1500H.

Learn how to calculate the bearing and distance of a ship after changing its course from due west to N 54° W. This problem covers key navigation concepts, including vector components, trigonometric functions, and the Pythagorean theorem. Suitable for Grades 10-12.

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