A ship A is moving westwards with a speed of `10 km h^(-1)` and a ship B 100 km south of A, is moving northwards with a speed of `10 km h^(-1)` The time after which the distance between them becomes shortest, is

Options

A. 0 h
B. 5 h
C. `5sqrt2 h `
D. `10sqrt2 h`

Answer

To solve the problem, we need to determine the time after which the distance between the two ships becomes shortest. We will use the concept of relative motion and the Pythagorean theorem to find the minimum distance.

### Step-by-Step Solution:

1. **Define the positions of the ships**:
- Let the initial position of ship A be at the point (0, 0) and it moves westwards.
- Ship B starts at the point (0, -100) since it is 100 km south of ship A and moves northwards.

2. **Determine the positions after time T**:
- After time T hours, ship A will have moved westwards by \(10T\) km, so its position will be (-10T, 0).
- Ship B will have moved northwards by \(10T\) km, so its position will be (0, -100 + 10T).

3. **Calculate the distance between the two ships**:
- The distance \(D\) between the two ships at time T can be calculated using the distance formula:
\[ D = \sqrt{((-10T) - 0)^2 + (0 - (-100 + 10T))^2} \]
- Simplifying this, we have:
\[ D = \sqrt{(10T)^2 + (100 - 10T)^2} \]
- Expanding the equation:
\[ D = \sqrt{100T^2 + (100 - 10T)^2} \]
\[ D = \sqrt{100T^2 + (10000 - 2000T + 100T^2)} \]
\[ D = \sqrt{200T^2 - 2000T + 10000} \]

4. **Minimize the distance**:
- To find the minimum distance, we can minimize \(D^2\) instead of \(D\):
\[ D^2 = 200T^2 - 2000T + 10000 \]
- Taking the derivative of \(D^2\) with respect to \(T\) and setting it to zero:
\[ \frac{d(D^2)}{dT} = 400T - 2000 = 0 \]
- Solving for \(T\):
\[ 400T = 2000 \implies T = \frac{2000}{400} = 5 \text{ hours} \]

5. **Conclusion**:
- The time after which the distance between the ships becomes shortest is **5 hours**.