A particle moves in xy-plane from position (2m, 4m) to (6m,8m) is 2s. Magnitude and direction of average velocity is

Options

A. `sqrt(2) ms^(-1)` and `45^(@)`
B. `2sqrt(2) ms^(-1)` and `45^(@)`
C. `4sqrt(2) ms^(-1)` and `30^(@)`
D. `3sqrt(2) ms^(-1)` and `60^(@)`

Explanation

To find the magnitude and direction of the average velocity of a particle moving in the xy-plane from position (2m, 4m) to (6m, 8m) in 2 seconds, we can follow these steps:

### Step 1: Determine the initial and final positions
- The initial position \( \mathbf{r_1} \) is given as \( (2 \, \text{m}, 4 \, \text{m}) \).
- The final position \( \mathbf{r_2} \) is given as \( (6 \, \text{m}, 8 \, \text{m}) \).

### Step 2: Calculate the displacement
- Displacement \( \Delta \mathbf{r} \) is calculated as:
\[ \Delta \mathbf{r} = \mathbf{r_2} - \mathbf{r_1} = (6 \, \text{m} - 2 \, \text{m}) \hat{i} + (8 \, \text{m} - 4 \, \text{m}) \hat{j} \]
- This simplifies to:
\[ \Delta \mathbf{r} = 4 \hat{i} + 4 \hat{j} \, \text{m} \]

### Step 3: Calculate the average velocity
- The average velocity \( \mathbf{V_{avg}} \) is given by:
\[ \mathbf{V_{avg}} = \frac{\Delta \mathbf{r}}{\Delta t} \]
- Here, \( \Delta t = 2 \, \text{s} \).
- Thus, we have:
\[ \mathbf{V_{avg}} = \frac{4 \hat{i} + 4 \hat{j}}{2} = 2 \hat{i} + 2 \hat{j} \, \text{m/s} \]

### Step 4: Calculate the magnitude of the average velocity
- The magnitude of \( \mathbf{V_{avg}} \) is calculated using the formula:
\[ |\mathbf{V_{avg}}| = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \, \text{m/s} \]

### Step 5: Calculate the direction of the average velocity
- The direction \( \theta \) can be found using the tangent function:
\[ \tan \theta = \frac{V_y}{V_x} = \frac{2}{2} = 1 \]
- Therefore, \( \theta = \tan^{-1}(1) = 45^\circ \).

### Final Result
- The magnitude of the average velocity is \( 2\sqrt{2} \, \text{m/s} \) and the direction is \( 45^\circ \) with respect to the x-axis.

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