Initially two particles A and B are present at (0,0) and (d,0) respectively They start moving with speed `V_(A)=Vhat(i)+Vhat(j)` and `V_(B)=-Vhat(j)` If R is magnitude of relative separation between them and `T_(0)` be the time when separation between them is minimum, then

Options

A. `T_(0)=d/(5V)`
B. `R_(min)=(2d)/sqrt(5)`
C. Graph of R versus time is straight line
D. Graph of R versus time is circle.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Define the initial positions and velocities
- Particle A is at position \( (0,0) \) and moves with velocity \( \mathbf{V_A} = V \hat{i} + V \hat{j} \).
- Particle B is at position \( (d,0) \) and moves with velocity \( \mathbf{V_B} = -V \hat{j} \).

### Step 2: Determine the relative velocity
The relative velocity of A with respect to B is given by:
\[ \mathbf{V_{AB}} = \mathbf{V_A} - \mathbf{V_B} \]
Substituting the values:
\[ \mathbf{V_{AB}} = (V \hat{i} + V \hat{j}) - (0 \hat{i} - V \hat{j}) = V \hat{i} + 2V \hat{j} \]

### Step 3: Calculate the magnitude of the relative velocity
The magnitude of the relative velocity \( V_{AB} \) is:
\[ |\mathbf{V_{AB}}| = \sqrt{(V)^2 + (2V)^2} = \sqrt{V^2 + 4V^2} = \sqrt{5V^2} = V\sqrt{5} \]

### Step 4: Find the time \( T_0 \) when the separation is minimum
To find the time when the separation is minimum, we need to determine the perpendicular distance from particle B to the line of motion of particle A. The minimum distance occurs when the line connecting A and B is perpendicular to the direction of \( \mathbf{V_{AB}} \).

The distance \( D \) between the initial positions of A and B is \( d \). The time \( T_0 \) to reach the minimum separation is given by:
\[ T_0 = \frac{D \cos \alpha}{|\mathbf{V_{AB}}|} \]
Where \( \cos \alpha = \frac{1}{\sqrt{5}} \) (from the triangle formed by the components of \( \mathbf{V_{AB}} \)):
\[ T_0 = \frac{d \cdot \frac{1}{\sqrt{5}}}{V\sqrt{5}} = \frac{d}{5V} \]

### Step 5: Calculate the minimum separation \( R_{min} \)
The minimum separation occurs when the perpendicular distance from particle B to the line of motion of A is given by:
\[ R_{min} = D \sin \alpha \]
Where \( \sin \alpha = \frac{2}{\sqrt{5}} \) (using the triangle formed by the components of \( \mathbf{V_{AB}} \)):
\[ R_{min} = d \cdot \frac{2}{\sqrt{5}} = \frac{2d}{\sqrt{5}} \]

### Final Results
- The time \( T_0 \) when the separation is minimum is:
\[ T_0 = \frac{d}{5V} \]
- The minimum separation \( R_{min} \) is:
\[ R_{min} = \frac{2d}{\sqrt{5}} \]