A ship A is moving Westwards with a speed of `10kmh^(-1)` and a ship B 100km South of A is moving northwards with a speed of `10kmh^(-1)`. The time after which the distance between them shortest is

Options

A. 0h
B. 5h
C. `5sqrt(2)h`
D. `10sqrt(2)h`

Explanation

To solve the problem step by step, we will analyze the motion of the two ships and find the time when the distance between them is shortest.

### Step 1: Understand the motion of the ships
- Ship A is moving westward at a speed of \(10 \, \text{km/h}\).
- Ship B is initially \(100 \, \text{km}\) south of Ship A and is moving northward at a speed of \(10 \, \text{km/h}\).

### Step 2: Set up the coordinate system
- Let the position of Ship A at time \(t\) be represented as \(A(t) = (-10t, 0)\) (moving west).
- The initial position of Ship B is \(B(0) = (0, -100)\) and its position at time \(t\) is \(B(t) = (0, -100 + 10t)\) (moving north).

### Step 3: Determine the distance between the two ships
The distance \(D\) between the two ships at time \(t\) can be expressed using the distance formula:
\[ D(t) = \sqrt{(x_A - x_B)^2 + (y_A - y_B)^2} \]
Substituting the positions of the ships:
\[ D(t) = \sqrt{((-10t) - 0)^2 + (0 - (-100 + 10t))^2} \]
This simplifies to:
\[ D(t) = \sqrt{(10t)^2 + (100 - 10t)^2} \]

### Step 4: Simplify the expression for distance
Expanding the equation:
\[ D(t) = \sqrt{100t^2 + (100 - 10t)^2} \]
\[ = \sqrt{100t^2 + (10000 - 2000t + 100t^2)} \]
\[ = \sqrt{200t^2 - 2000t + 10000} \]

### Step 5: Find the minimum distance
To find the time when the distance is shortest, we need to minimize \(D(t)\). This can be done by minimizing the square of the distance \(D^2(t)\):
\[ D^2(t) = 200t^2 - 2000t + 10000 \]
This is a quadratic equation in the form \(at^2 + bt + c\). The minimum value occurs at:
\[ t = -\frac{b}{2a} = -\frac{-2000}{2 \cdot 200} = \frac{2000}{400} = 5 \, \text{hours} \]

### Conclusion
The time after which the distance between the two ships is shortest is \(5 \, \text{hours}\).