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. a. `10 km h^-1`
B. b. `9 km h^-1`
C. c. `8 km h^-1`
D. d. `7 km h^-1`

Explanation

To solve the problem, we need to determine the speed of the second ship (Ship B) such that it always remains due north with respect to the first ship (Ship A), which is moving due east.

### Step-by-step Solution:

1. **Identify the velocities**:
- Let the velocity of Ship A (V_A) = 5 km/h (due east).
- Let the velocity of Ship B (V_B) be at an angle of 30° east of north.

2. **Break down the velocity of Ship B into components**:
- The northward component of Ship B's velocity (V_B_north) = V_B * cos(30°).
- The eastward component of Ship B's velocity (V_B_east) = V_B * sin(30°).

3. **Establish the condition for Ship B to remain due north of Ship A**:
- For Ship B to remain due north of Ship A, the eastward component of Ship B's velocity must equal the velocity of Ship A. Thus, we have:
\[ V_B \cdot \sin(30°) = V_A \]
- Substituting the known value of V_A:
\[ V_B \cdot \sin(30°) = 5 \text{ km/h} \]

4. **Calculate sin(30°)**:
- We know that sin(30°) = 1/2.

5. **Substitute sin(30°) into the equation**:
\[ V_B \cdot \frac{1}{2} = 5 \]

6. **Solve for V_B**:
- Multiply both sides by 2:
\[ V_B = 5 \cdot 2 = 10 \text{ km/h} \]

### Conclusion:
The speed of the second ship (Ship B) should be **10 km/h** to remain always due north with respect to the first ship (Ship A).

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