Two particles A and B are moving with constant velocities `V_(1)=hatj` and `v_(2)=2hati` respectively in XY plane. At time t=0, the particle A is at co-ordinates (0,0) and B is at (-4,0). The angular velocities of B with respect to A at t=2s is (all physical quantities are in SI units)

Options

A. `(1)/(2)rad//s`
B. `2rad//s`
C. `4rad//s`
D. `1rad//s`

Answer

To find the angular velocity of particle B with respect to particle A at time \( t = 2 \) seconds, we can follow these steps:

### Step 1: Determine the positions of particles A and B at \( t = 2 \) seconds.

- Particle A starts at the origin \( (0, 0) \) and moves with a velocity \( \mathbf{V_1} = \hat{j} \) (which means it moves in the positive y-direction).
- The displacement of particle A after \( 2 \) seconds is:
\[ \text{Displacement of A} = \mathbf{V_1} \cdot t = \hat{j} \cdot 2 = (0, 2) \]
- Therefore, the position of particle A at \( t = 2 \) seconds is:
\[ \text{Position of A} = (0, 0) + (0, 2) = (0, 2) \]

- Particle B starts at \( (-4, 0) \) and moves with a velocity \( \mathbf{V_2} = 2\hat{i} \) (which means it moves in the positive x-direction).
- The displacement of particle B after \( 2 \) seconds is:
\[ \text{Displacement of B} = \mathbf{V_2} \cdot t = 2\hat{i} \cdot 2 = (4, 0) \]
- Therefore, the position of particle B at \( t = 2 \) seconds is:
\[ \text{Position of B} = (-4, 0) + (4, 0) = (0, 0) \]

### Step 2: Calculate the relative position vector \( \mathbf{R} \) from A to B.

- The position vector from A to B at \( t = 2 \) seconds is given by:
\[ \mathbf{R} = \text{Position of B} - \text{Position of A} = (0, 0) - (0, 2) = (0, -2) \]

### Step 3: Calculate the velocities of particles A and B.

- The velocity of particle A is:
\[ \mathbf{V_A} = \hat{j} \]
- The velocity of particle B is:
\[ \mathbf{V_B} = 2\hat{i} \]

### Step 4: Calculate the relative velocity \( \mathbf{V_{BA}} \).

- The relative velocity of B with respect to A is:
\[ \mathbf{V_{BA}} = \mathbf{V_B} - \mathbf{V_A} = (2\hat{i}) - (\hat{j}) = (2, -1) \]

### Step 5: Calculate the magnitude of the position vector \( \mathbf{R} \).

- The magnitude of the position vector \( \mathbf{R} \) is:
\[ |\mathbf{R}| = \sqrt{(0)^2 + (-2)^2} = \sqrt{4} = 2 \]

### Step 6: Calculate the angular velocity \( \omega \) of B with respect to A.

- The angular velocity \( \omega \) can be calculated using the formula:
\[ \omega = \frac{|\mathbf{V_{BA}}|}{|\mathbf{R}|} \]
- First, we need to find the magnitude of \( \mathbf{V_{BA}} \):
\[ |\mathbf{V_{BA}}| = \sqrt{(2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]
- Now substituting the values:
\[ \omega = \frac{\sqrt{5}}{2} \]

### Final Answer:
The angular velocity of B with respect to A at \( t = 2 \) seconds is \( \frac{\sqrt{5}}{2} \) radians per second.

---