a particle is moving in x-y-plane at `2 m//s` along x-axis. `2` seconds later, its velocity is `4 m//s` in a direction making `60^(@)` with positive x-axis. Its average acceleration for the period of motion is:-

Options

A. `sqrt(5) m//s^(2)`, along y-axis
B. `sqrt(3) m//s^(2)`, along y-axis
C. `sqrt(5) m//s^(2)`, along at `60^(@)` with positive x-axis
D. `3m//s^(2)` at `60^(@)` with positive x-axis.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the initial and final velocities
- The initial velocity \( \mathbf{V_i} \) is given as \( 2 \, \text{m/s} \) along the x-axis. In vector form, this can be represented as:
\[ \mathbf{V_i} = 2 \hat{i} \]

- The final velocity \( \mathbf{V_f} \) is given as \( 4 \, \text{m/s} \) at an angle of \( 60^\circ \) with the positive x-axis. We can break this into its x and y components:
\[ V_{fx} = 4 \cos(60^\circ) = 4 \times \frac{1}{2} = 2 \, \text{m/s} \]
\[ V_{fy} = 4 \sin(60^\circ) = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \, \text{m/s} \]

Thus, the final velocity in vector form is:
\[ \mathbf{V_f} = 2 \hat{i} + 2\sqrt{3} \hat{j} \]

### Step 2: Calculate the change in velocity
The change in velocity \( \Delta \mathbf{V} \) is given by:
\[ \Delta \mathbf{V} = \mathbf{V_f} - \mathbf{V_i} \]
Substituting the values:
\[ \Delta \mathbf{V} = (2 \hat{i} + 2\sqrt{3} \hat{j}) - (2 \hat{i}) = 2\sqrt{3} \hat{j} \]

### Step 3: Calculate the average acceleration
The average acceleration \( \mathbf{a_{avg}} \) is given by the change in velocity divided by the time interval. The time interval is \( 2 \, \text{s} \):
\[ \mathbf{a_{avg}} = \frac{\Delta \mathbf{V}}{\Delta t} = \frac{2\sqrt{3} \hat{j}}{2} = \sqrt{3} \hat{j} \, \text{m/s}^2 \]

### Step 4: Final Result
Thus, the average acceleration of the particle is:
\[ \mathbf{a_{avg}} = \sqrt{3} \hat{j} \, \text{m/s}^2 \]