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, we need to find the time after which the distance between the two ships is minimized. Here is a step-by-step solution:

### Step 1: Understand the positions 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: Define the positions of the ships over time
- Let \(t\) be the time in hours after the start.
- The position of Ship A after time \(t\) will be:
- \(x_A = -10t\) (moving west, hence negative direction)
- The position of Ship B after time \(t\) will be:
- \(y_B = 100 - 10t\) (moving north, starting from \(100 \, \text{km}\) south)

### Step 3: Set up the distance formula
The distance \(d\) between the two ships can be represented using the Pythagorean theorem:
\[ d^2 = (x_A)^2 + (y_B)^2 \]
Substituting the positions:
\[ d^2 = (-10t)^2 + (100 - 10t)^2 \]
\[ d^2 = 100t^2 + (100 - 10t)^2 \]

### Step 4: Expand the equation
Now, expand the second term:
\[ (100 - 10t)^2 = 10000 - 2000t + 100t^2 \]
Thus, the distance squared becomes:
\[ d^2 = 100t^2 + 10000 - 2000t + 100t^2 \]
\[ d^2 = 200t^2 - 2000t + 10000 \]

### Step 5: Differentiate to find the minimum distance
To find the time when the distance is minimized, we differentiate \(d^2\) with respect to \(t\) and set it to zero:
\[ \frac{d(d^2)}{dt} = 400t - 2000 \]
Setting the derivative equal to zero:
\[ 400t - 2000 = 0 \]
\[ 400t = 2000 \]
\[ t = 5 \, \text{hours} \]

### Step 6: Conclusion
The time after which the distance between the two ships is shortest is \(5 \, \text{hours}\).

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