Two ships, V and W, move with constant velocities `2 ms^(-1)` and `4 ms^(-1)` along two mutually perpendicular straight tracks toward the intersection point O. At the moment `t=0`, the ships V & W were located at distances `100 m` and `200 m` respectively from the point O.
The distance between them will be shortest at t= ........

Options

A. `2bhat(j)`
B. `a hat(j)`
C. `-2bhat(j)`
D. None of these

Explanation

To solve the problem of finding the time at which the distance between ships V and W is shortest, we can follow these steps:

### Step 1: Understand the positions of the ships
At time \( t = 0 \):
- Ship V is 100 m away from point O and moves towards O with a velocity of \( 2 \, \text{ms}^{-1} \).
- Ship W is 200 m away from point O and moves towards O with a velocity of \( 4 \, \text{ms}^{-1} \).

### Step 2: Determine the positions of the ships at time \( t \)
At time \( t \):
- The position of ship V from point O will be:
\[ \text{Position of V} = 100 - 2t \]
- The position of ship W from point O will be:
\[ \text{Position of W} = 200 - 4t \]

### Step 3: Set up the distance formula
The distance \( L \) between the two ships at time \( t \) can be calculated using the Pythagorean theorem, since they are moving along perpendicular paths:
\[ L = \sqrt{(100 - 2t)^2 + (200 - 4t)^2} \]

### Step 4: Simplify the expression for \( L \)
Expanding the squares:
\[ L = \sqrt{(100 - 2t)^2 + (200 - 4t)^2} \]
\[ = \sqrt{(10000 - 400t + 4t^2) + (40000 - 1600t + 16t^2)} \]
\[ = \sqrt{(10000 + 40000) + (4t^2 + 16t^2) - (400t + 1600t)} \]
\[ = \sqrt{50000 - 2000t + 20t^2} \]

### Step 5: Find the minimum distance
To find the minimum distance, we need to differentiate \( L \) with respect to \( t \) and set the derivative to zero:
\[ \frac{dL}{dt} = \frac{1}{2\sqrt{50000 - 2000t + 20t^2}} \cdot (20t - 2000) = 0 \]
This implies:
\[ 20t - 2000 = 0 \implies t = 100 \, \text{s} \]

### Step 6: Verify if this is a minimum
To confirm that this is indeed a minimum, we can check the second derivative or analyze the behavior of \( L \) around \( t = 100 \).

### Final Answer
The distance between the two ships will be shortest at \( t = 100 \, \text{s} \).

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