A ship is travelling due east at a speed of `15 km//h.` Find the speed of a boat heading `30^@` east of north if it appears always due north from the ship.

Options

A. `30 km//h`
B. `(15sqrt(3))/2 km//h`
C. `10sqrt(3) km//h`
D. `20 km//h`

Explanation

To solve the problem, we need to find the speed of a boat that is heading 30 degrees east of north, such that it appears to be moving due north from the perspective of a ship traveling due east at a speed of 15 km/h.

### Step-by-Step Solution:

1. **Understand the Directions**:
- The ship is traveling due east at 15 km/h. We can represent this velocity as a vector:
\[ \vec{V}_{\text{ship}} = 15 \hat{i} \text{ km/h} \]
- The boat is heading 30 degrees east of north. This means its direction can be represented in terms of its components.

2. **Components of the Boat's Velocity**:
- Let the speed of the boat be \( V_b \) km/h. The components of the boat's velocity can be calculated as:
- Northward component (y-direction):
\[ V_{b_y} = V_b \cos(30^\circ) \]
- Eastward component (x-direction):
\[ V_{b_x} = V_b \sin(30^\circ) \]

3. **Relative Velocity Condition**:
- From the perspective of the ship, the boat appears to be moving due north. This means that the eastward component of the boat's velocity relative to the ship must be zero:
\[ V_{b_x} - V_{\text{ship}} = 0 \]
- Substituting the expressions for the components:
\[ V_b \sin(30^\circ) - 15 = 0 \]

4. **Solving for the Boat's Speed**:
- Rearranging the equation gives:
\[ V_b \sin(30^\circ) = 15 \]
- Since \(\sin(30^\circ) = \frac{1}{2}\):
\[ V_b \cdot \frac{1}{2} = 15 \]
- Multiplying both sides by 2:
\[ V_b = 30 \text{ km/h} \]

5. **Conclusion**:
- The speed of the boat is \( 30 \text{ km/h} \).

### Final Answer:
The speed of the boat is **30 km/h**.

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