a ship leaves the port of miami with a bearing of s70e and a speed of 15 knots. after 1 hr, the ship turns 90 degrees toward the south. after 2 hours, maintaining the same speed, what is the bearing to the ship from port?

To solve this problem, we need to calculate the position of the ship at each stage and then find the bearing from the port of Miami to the final location of the ship.

### Step 1: First Leg of the Journey
- **Initial bearing**: \( S70^\circ E \) (which means 70 degrees east of south).
- **Speed**: 15 knots (nautical miles per hour).
- **Time**: 1 hour.

After 1 hour, the ship would have traveled 15 nautical miles. To determine the ship's position relative to the port, we break this distance into its southward and eastward components:

- Southward component: 
  \[
  15 \times \cos(70^\circ) = 15 \times 0.3420 = 5.13 \text{ nautical miles}
  \]
  
- Eastward component: 
  \[
  15 \times \sin(70^\circ) = 15 \times 0.9397 = 14.10 \text{ nautical miles}
  \]

So, after the first leg, the ship has moved 5.13 nautical miles south and 14.10 nautical miles east.

### Step 2: Second Leg of the Journey
After 1 hour, the ship turns 90 degrees and heads due south for 2 hours at the same speed (15 knots). The distance traveled in this leg is:

\[
\text{Distance traveled} = 15 \times 2 = 30 \text{ nautical miles south}
\]

### Step 3: Total Displacement
- **Southward displacement**: From the first leg, the ship has moved 5.13 nautical miles south, and in the second leg, it moves 30 nautical miles south. So, the total southward displacement is:
  \[
  5.13 + 30 = 35.13 \text{ nautical miles south}
  \]

- **Eastward displacement**: From the first leg, the ship has moved 14.10 nautical miles east. There is no further eastward movement in the second leg. So, the total eastward displacement is:
  \[
  14.10 \text{ nautical miles east}
  \]

### Step 4: Bearing Calculation
We now use trigonometry to find the bearing from the port to the ship's final location. The bearing is the angle \( \theta \) east of south.

The tangent of \( \theta \) is given by:
\[
\tan(\theta) = \frac{\text{Eastward displacement}}{\text{Southward displacement}} = \frac{14.10}{35.13} = 0.4013
\]

To find \( \theta \):
\[
\theta = \tan^{-1}(0.4013) = 21.88^\circ
\]

Thus, the bearing from the port to the ship is \( S21.88^\circ E \).

### Final Answer:
The bearing to the ship from the port of Miami is approximately **S21.88°E**.

---

Do you need any further details or clarifications? Here are five related questions for practice:

1. If the ship had traveled at a different speed, how would that affect the final bearing?
2. What if the ship turned 45 degrees toward the south instead of 90 degrees?
3. How would you calculate the ship’s final position using vector notation?
4. How can you convert this bearing into standard compass notation (0° to 360°)?
5. What if the second leg of the journey was due west instead of due south?

**Tip**: Always break vectors into their components when dealing with bearings and directions!

A ship leaves the port of Miami with a bearing of S70°E and a speed of 15 knots. After 1 hour, the ship turns 90 degrees toward the south. After 2 hours, maintaining the same speed, what is the bearing to the ship from port?

Solve the problem of a ship's bearing after changing direction from S70°E and traveling at 15 knots. This solution involves trigonometric calculations, vector components, and bearing determination. Suitable for students in Grades 10-12.

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