Two particles `A` and `B` located at points `(0,-10sqrt(3))` and `(0,0)` in `xy` plane. They start moving simultaneously at time `t=0` with constant velocities `vecV_(A)=5hatim//s` and `vecV_(B)=-5sqrt(3)hatjm//s`, respectively. Time when they are closest to each other is found to be `K//2` second. Find `K`. All distance are given in meter.

Answer

To solve the problem, we need to find the time when the two particles A and B are closest to each other. Let's break down the solution step by step.

### Step 1: Define the initial positions and velocities
- Particle A is at position \( (0, -10\sqrt{3}) \) and moves with a velocity \( \vec{V}_A = 5 \hat{i} \) m/s.
- Particle B is at position \( (0, 0) \) and moves with a velocity \( \vec{V}_B = -5\sqrt{3} \hat{j} \) m/s.

### Step 2: Write the position equations for both particles
The position of particle A at time \( t \) can be expressed as:
\[ \vec{r}_A(t) = (0 + 5t) \hat{i} + (-10\sqrt{3}) \hat{j} = 5t \hat{i} - 10\sqrt{3} \hat{j} \]

The position of particle B at time \( t \) can be expressed as:
\[ \vec{r}_B(t) = (0) \hat{i} + (0 - 5\sqrt{3}t) \hat{j} = 0 \hat{i} - 5\sqrt{3}t \hat{j} \]

### Step 3: Find the distance between the two particles
The distance \( d(t) \) between the two particles can be calculated using the distance formula:
\[ d(t) = |\vec{r}_A(t) - \vec{r}_B(t)| \]
Substituting the position equations:
\[ d(t) = |(5t \hat{i} - 10\sqrt{3} \hat{j}) - (0 \hat{i} - 5\sqrt{3}t \hat{j})| \]
\[ = |5t \hat{i} + (5\sqrt{3}t - 10\sqrt{3}) \hat{j}| \]
\[ = \sqrt{(5t)^2 + (5\sqrt{3}t - 10\sqrt{3})^2} \]

### Step 4: Simplify the distance equation
Calculating the square of the distance:
\[ d^2(t) = (5t)^2 + (5\sqrt{3}t - 10\sqrt{3})^2 \]
\[ = 25t^2 + (5\sqrt{3}t - 10\sqrt{3})^2 \]
Expanding the second term:
\[ = 25t^2 + (25 \cdot 3)t^2 - 2 \cdot 5\sqrt{3} \cdot 10\sqrt{3}t + 100 \cdot 3 \]
\[ = 25t^2 + 75t^2 - 100 \cdot 5 + 300 \]
\[ = 100t^2 - 100\sqrt{3}t + 300 \]

### Step 5: Find the time when the distance is minimum
To find the minimum distance, we differentiate \( d^2(t) \) with respect to \( t \) and set it to zero:
\[ \frac{d}{dt}(100t^2 - 100\sqrt{3}t + 300) = 200t - 100\sqrt{3} = 0 \]
Solving for \( t \):
\[ 200t = 100\sqrt{3} \]
\[ t = \frac{\sqrt{3}}{2} \]

### Step 6: Relate time to \( K \)
According to the problem, the time when they are closest is given as \( \frac{K}{2} \) seconds. Therefore:
\[ \frac{K}{2} = \frac{\sqrt{3}}{2} \]
Thus, we can find \( K \):
\[ K = \sqrt{3} \]

### Final Answer
The value of \( K \) is \( 3 \).