A ship is travelling due east at 10 km/h. A ship heading `30^(@)` east of north is always due north from the first ship. The speed of the second ship in km/h is

Options

A. `20sqrt(2)`
B. `20sqrt(3//2)`
C. `20`
D. `20//sqrt(2)`

Answer

To solve the problem, we need to analyze the motion of both ships and find the speed of the second ship.

### Step-by-Step Solution:

1. **Identify the velocities of the ships**:
- Let Ship 1 (S1) be traveling due east with a speed of \( V_1 = 10 \, \text{km/h} \).
- Ship 2 (S2) is heading \( 30^\circ \) east of north.

2. **Understand the relative motion**:
- The problem states that Ship 2 is always due north from Ship 1. This means that from the perspective of Ship 1, Ship 2 appears to be moving directly north.

3. **Set up the coordinate system**:
- Assume Ship 1 is moving along the positive x-axis (east), and Ship 2 is moving at an angle of \( 30^\circ \) east of north. In a standard coordinate system:
- The x-component (east-west) of Ship 2's velocity can be expressed as \( V_2 \sin(30^\circ) \).
- The y-component (north-south) of Ship 2's velocity can be expressed as \( V_2 \cos(30^\circ) \).

4. **Establish the relationship between the velocities**:
- Since Ship 2 is always moving north from the perspective of Ship 1, the eastward component of Ship 2's velocity must equal the speed of Ship 1:
\[ V_2 \sin(30^\circ) = V_1 \]
- We know \( \sin(30^\circ) = \frac{1}{2} \), so substituting this in gives:
\[ V_2 \cdot \frac{1}{2} = 10 \, \text{km/h} \]

5. **Solve for \( V_2 \)**:
- Rearranging the equation gives:
\[ V_2 = 10 \, \text{km/h} \cdot 2 = 20 \, \text{km/h} \]

6. **Conclusion**:
- The speed of the second ship is \( 20 \, \text{km/h} \).

### Final Answer:
The speed of the second ship is \( 20 \, \text{km/h} \).

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