Ship `A` is travelling with a velocity of `5 km h^-1` due east. A second ship is heading `30^@` east of north. What should be the speed of second ship if it is to remain always due north with respect to the first ship ?

Options

A. `10 km h^-1`
B. `9 km h^-1`
C. `8 km h^-1`
D. `7 km h^-1`

Explanation

To solve the problem, we need to analyze the motion of both ships and find the speed of ship B such that it always appears to move due north with respect to ship A.

### Step-by-Step Solution:

1. **Identify the velocities of the ships:**
- Ship A is moving due east with a velocity \( V_A = 5 \, \text{km/h} \).
- Ship B is moving at an angle of \( 30^\circ \) east of north. We need to find its speed \( V_B \).

2. **Break down the velocity of ship B into components:**
- The angle \( 30^\circ \) east of north means that the angle with respect to the east direction is \( 90^\circ - 30^\circ = 60^\circ \).
- The components of the velocity of ship B can be expressed as:
- \( V_{Bx} = V_B \cos(60^\circ) \) (eastward component)
- \( V_{By} = V_B \sin(60^\circ) \) (northward component)

3. **Set up the condition for ship B to appear due north with respect to ship A:**
- For ship B to appear to move due north with respect to ship A, the eastward component of ship B's velocity must cancel out the velocity of ship A. Thus, we have:
\[ V_{Bx} = V_A \]
- Substituting the expressions for \( V_{Bx} \) and \( V_A \):
\[ V_B \cos(60^\circ) = 5 \, \text{km/h} \]

4. **Calculate \( V_B \):**
- We know that \( \cos(60^\circ) = \frac{1}{2} \). Therefore, we can substitute this value:
\[ V_B \cdot \frac{1}{2} = 5 \, \text{km/h} \]
- To find \( V_B \), multiply both sides by 2:
\[ V_B = 5 \times 2 = 10 \, \text{km/h} \]

5. **Conclusion:**
- The speed of ship B should be \( 10 \, \text{km/h} \) to remain always due north with respect to ship A.

### Final Answer:
The speed of the second ship (ship B) should be \( 10 \, \text{km/h} \).

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