A boat is heading due east at 23 km/hr (relative to the water). The current is moving toward the southwest at 18 km/hr.
(a) Give the vector representing the actual movement of the boat. (Round each component to two decimal places.)

Correct: Your answer is correct.

(b) How fast is the boat going, relative to the ground? (Round your answer to two decimal places.)
16.36

Correct: Your answer is correct.
km/hr

(c) By what angle does the current push the boat off of its due east course? (Round your answer to two decimal places.)
51.1

Correct: Your answer is correct.
°
south of east

Correct: Your answer is correct.

Let's break down the steps to solve each part of the problem.

### Given:
- **Boat's velocity relative to the water:** 23 km/hr due east.
- **Current's velocity relative to the ground:** 18 km/hr towards the southwest.

### Part (a): Vector Representation of the Boat's Actual Movement

1. **Boat's Velocity Vector (Relative to Water):**
   - Due east direction: \( \vec{v}_b = \langle 23, 0 \rangle \) km/hr.

2. **Current's Velocity Vector:**
   - Southwest means 45° south of west. The components are:
     \[
     \vec{v}_c = 18 \cdot \langle -\cos(45^\circ), -\sin(45^\circ) \rangle = 18 \cdot \langle -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \rangle
     \]
     \[
     \vec{v}_c = \langle -12.73, -12.73 \rangle \text{ km/hr}
     \]

3. **Actual Movement Vector of the Boat (Relative to the Ground):**
   \[
   \vec{v}_a = \vec{v}_b + \vec{v}_c = \langle 23, 0 \rangle + \langle -12.73, -12.73 \rangle = \langle 10.27, -12.73 \rangle \text{ km/hr}
   \]

### Part (b): Speed of the Boat Relative to the Ground

- The magnitude of the actual movement vector \( \vec{v}_a \):
  \[
  \text{Speed} = \sqrt{(10.27)^2 + (-12.73)^2} = \sqrt{105.49 + 162.05} = \sqrt{267.54} \approx 16.36 \text{ km/hr}
  \]

### Part (c): Angle at Which the Current Pushes the Boat Off Course

- The angle \( \theta \) south of east can be found using the inverse tangent function:
  \[
  \theta = \tan^{-1}\left(\frac{|\text{South Component}|}{\text{East Component}}\right) = \tan^{-1}\left(\frac{12.73}{10.27}\right) \approx 51.1^\circ
  \]

### Summary of Answers:
- (a) The vector representing the actual movement of the boat is \( \langle 10.27, -12.73 \rangle \) km/hr.
- (b) The speed of the boat relative to the ground is 16.36 km/hr.
- (c) The angle at which the current pushes the boat off its due east course is 51.1° south of east.

Would you like any more details on the steps or have any other questions? Here are some related questions to explore:

1. How would the results change if the boat's speed were faster?
2. What if the current were flowing directly south instead of southwest?
3. How would the angle of deviation change if the current speed increased?
4. What would be the boat's path if there were no current at all?
5. How does the direction of the current affect the boat's actual path?

**Tip:** When working with vectors, always break them into components first to simplify calculations.

Explore how to solve a relative motion problem involving a boat moving eastward against a southwest current. Learn about vector addition, calculating boat speed relative to the ground, and determining angular deviation caused by the current. Suitable for high school students in Grades 10-12.

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