Two particles A and B start moving with velocities 20 m/s and 30`sqrt(2)` m/s along x-axis and at an angle `45^(@)` with x- axis respectively in xy-plane from origin the relative velocity of B w.r.t A

Options

A. `(10 hat(i) + 30 hat(j))` m/s
B. `(30 hat(i) + 10 hat(j))` m/s
C. `(30 hat(i) - 20 sqrt(2)hat(j))` m/s
D. `(30 sqrt(2) hat(i) + 10sqrt(2)hat(j))` m/s

Answer

To find the relative velocity of particle B with respect to particle A, we can follow these steps:

### Step 1: Identify the velocities of particles A and B.
- Particle A moves along the x-axis with a velocity of \( V_A = 20 \, \text{m/s} \).
- Particle B moves at an angle of \( 45^\circ \) with the x-axis with a velocity of \( V_B = 30\sqrt{2} \, \text{m/s} \).

### Step 2: Break down the velocity of particle B into its components.
Since particle B is moving at an angle of \( 45^\circ \), we can resolve its velocity into x and y components:
- The x-component of \( V_B \) is given by:
\[ V_{B_x} = V_B \cos(45^\circ) = 30\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 30 \, \text{m/s} \]
- The y-component of \( V_B \) is given by:
\[ V_{B_y} = V_B \sin(45^\circ) = 30\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 30 \, \text{m/s} \]

Thus, the velocity vector of particle B can be expressed as:
\[ V_B = 30 \hat{i} + 30 \hat{j} \, \text{m/s} \]

### Step 3: Write the velocity vector of particle A.
Since particle A is moving along the x-axis, its velocity vector is:
\[ V_A = 20 \hat{i} \, \text{m/s} \]

### Step 4: Calculate the relative velocity of B with respect to A.
The relative velocity of B with respect to A, denoted as \( V_{BA} \), is calculated using the formula:
\[ V_{BA} = V_B - V_A \]

Substituting the values we found:
\[ V_{BA} = (30 \hat{i} + 30 \hat{j}) - (20 \hat{i}) = (30 - 20) \hat{i} + 30 \hat{j} = 10 \hat{i} + 30 \hat{j} \, \text{m/s} \]

### Step 5: State the final answer.
The relative velocity of particle B with respect to particle A is:
\[ V_{BA} = 10 \hat{i} + 30 \hat{j} \, \text{m/s} \]

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