: A boat leaves a dock and travels due north at a speed of 10 km/hr. A second boat leaves the same dock at the same time, but travels N50oE at a speed of 15 km/hr. 
How far apart are the two boats after 2 hours?
Assume the second boat broke down after those 2 hours. What direction would the first boat need to travel to pick up the people in the broken down boat.



Let's solve this problem step by step.

### Part 1: Finding the distance between the two boats after 2 hours.

1. **Boat 1 (Northbound boat):**
   - Travels at a speed of 10 km/h due north.
   - After 2 hours, the distance traveled by Boat 1 is:
   \[
   \text{Distance}_1 = 10 \, \text{km/h} \times 2 \, \text{hours} = 20 \, \text{km}.
   \]

2. **Boat 2 (Traveling N50°E):**
   - Travels at a speed of 15 km/h in the direction N50°E.
   - After 2 hours, the distance traveled by Boat 2 is:
   \[
   \text{Distance}_2 = 15 \, \text{km/h} \times 2 \, \text{hours} = 30 \, \text{km}.
   \]

3. **Setting up a coordinate system:**
   - The path of Boat 1 is along the positive y-axis (due north).
   - The path of Boat 2 makes an angle of 50° with the north direction (N50°E).

   Using trigonometry, we can break down the displacement of Boat 2 into components:
   - Northward (y-component) is \( 30 \times \cos(50^\circ) \).
   - Eastward (x-component) is \( 30 \times \sin(50^\circ) \).

   Calculating these components:
   \[
   y_2 = 30 \times \cos(50^\circ) \approx 30 \times 0.6428 = 19.284 \, \text{km},
   \]
   \[
   x_2 = 30 \times \sin(50^\circ) \approx 30 \times 0.7660 = 22.98 \, \text{km}.
   \]

4. **Calculating the distance between the two boats:**
   - Boat 1 is at (0, 20) in the coordinate plane.
   - Boat 2 is at (22.98, 19.284).
   
   The distance between the two boats is the straight-line distance between these two points:
   \[
   \text{Distance} = \sqrt{(22.98 - 0)^2 + (19.284 - 20)^2}
   \]
   \[
   = \sqrt{22.98^2 + (-0.716)^2}
   \]
   \[
   = \sqrt{528.9604 + 0.5122} = \sqrt{529.4726} \approx 23.01 \, \text{km}.
   \]
   
   After 2 hours, the boats are approximately **23.01 km apart**.

### Part 2: Finding the direction the first boat needs to travel to reach the second boat.

To find the direction Boat 1 must travel to reach Boat 2, we need to calculate the angle between the direction of Boat 1's current position (0, 20) and the broken-down boat's position (22.98, 19.284).

1. **Relative position:**
   - The displacement vector from Boat 1 to Boat 2 is:
     - Change in \(x\): \( \Delta x = 22.98 - 0 = 22.98 \).
     - Change in \(y\): \( \Delta y = 19.284 - 20 = -0.716 \).

2. **Finding the angle:**
   - The angle \( \theta \) that Boat 1 must travel relative to due east is given by:
   \[
   \theta = \tan^{-1}\left( \frac{\Delta y}{\Delta x} \right) = \tan^{-1}\left( \frac{-0.716}{22.98} \right).
   \]
   \[
   \theta = \tan^{-1}(-0.0312) \approx -1.79^\circ.
   \]

   This means that Boat 1 must travel **1.79° south of east** to reach Boat 2.

### Final Answer:
- After 2 hours, the boats are **23.01 km** apart.
- Boat 1 must travel **1.79° south of east** to reach Boat 2.

---

Would you like further details on any step or explanation?

Here are five related questions to explore:

1. How would the calculation change if Boat 1 traveled at a different speed?
2. What if Boat 2 had traveled in a different direction, say N30°E?
3. How would you find the time it takes for Boat 1 to reach Boat 2?
4. What is the concept of relative velocity, and how would it apply here?
5. How would you solve this problem if both boats had variable speeds?

**Tip**: Always break down vectors into components when solving problems involving angles and directions. It simplifies calculations immensely.

A boat leaves a dock and travels due north at a speed of 10 km/hr. A second boat leaves the same dock at the same time, but travels N50°E at a speed of 15 km/hr. How far apart are the two boats after 2 hours? Assume the second boat broke down after those 2 hours. What direction would the first boat need to travel to pick up the people in the broken down boat?

Solve a trigonometry-based problem where two boats leave a dock at different angles and speeds. After 2 hours, find how far apart they are using vector decomposition and trigonometric functions. Learn how to calculate the direction for one boat to reach the other. Suitable for high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation