Velocity of a particle at time `t=0` is `2 hati` m/s. A constant acceleration of `2 m/s^2` acts on the particle for `2 s` at an angle of `60^@` with its initial velocity. Find the magnitude of velocity and displacement of particle at the end of `t=2s.`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the initial conditions
The initial velocity \( \mathbf{u} \) of the particle at time \( t = 0 \) is given as:
\[ \mathbf{u} = 2 \hat{i} + 0 \hat{j} \, \text{m/s} \]
This means the particle has a velocity of 2 m/s in the x-direction and 0 m/s in the y-direction.

### Step 2: Determine the acceleration components
The constant acceleration \( \mathbf{a} \) is given as \( 2 \, \text{m/s}^2 \) at an angle of \( 60^\circ \) with respect to the initial velocity. We can break this acceleration into its components:
\[ \mathbf{a} = a_x \hat{i} + a_y \hat{j} \]
where:
- \( a_x = 2 \cos(60^\circ) \)
- \( a_y = 2 \sin(60^\circ) \)

Calculating these values:
\[ a_x = 2 \cdot \frac{1}{2} = 1 \, \text{m/s}^2 \]
\[ a_y = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \, \text{m/s}^2 \]
Thus, the acceleration vector is:
\[ \mathbf{a} = 1 \hat{i} + \sqrt{3} \hat{j} \, \text{m/s}^2 \]

### Step 3: Calculate the final velocity after 2 seconds
Using the equation of motion:
\[ \mathbf{v} = \mathbf{u} + \mathbf{a} t \]
Substituting the values:
\[ \mathbf{v} = (2 \hat{i} + 0 \hat{j}) + (1 \hat{i} + \sqrt{3} \hat{j}) \cdot 2 \]
This simplifies to:
\[ \mathbf{v} = 2 \hat{i} + 0 \hat{j} + 2 \hat{i} + 2\sqrt{3} \hat{j} \]
\[ \mathbf{v} = (2 + 2) \hat{i} + (0 + 2\sqrt{3}) \hat{j} = 4 \hat{i} + 2\sqrt{3} \hat{j} \, \text{m/s} \]

### Step 4: Find the magnitude of the final velocity
The magnitude of the velocity \( |\mathbf{v}| \) is given by:
\[ |\mathbf{v}| = \sqrt{(4)^2 + (2\sqrt{3})^2} \]
Calculating this:
\[ |\mathbf{v}| = \sqrt{16 + 12} = \sqrt{28} = 2\sqrt{7} \, \text{m/s} \]

### Step 5: Calculate the displacement after 2 seconds
Using the equation of motion for displacement:
\[ \mathbf{s} = \mathbf{u} t + \frac{1}{2} \mathbf{a} t^2 \]
Substituting the values:
\[ \mathbf{s} = (2 \hat{i} + 0 \hat{j}) \cdot 2 + \frac{1}{2} (1 \hat{i} + \sqrt{3} \hat{j}) \cdot (2^2) \]
This simplifies to:
\[ \mathbf{s} = (4 \hat{i} + 0 \hat{j}) + \frac{1}{2} (1 \hat{i} + \sqrt{3} \hat{j}) \cdot 4 \]
\[ \mathbf{s} = 4 \hat{i} + 0 \hat{j} + 2 \hat{i} + 2\sqrt{3} \hat{j} \]
\[ \mathbf{s} = (4 + 2) \hat{i} + (0 + 2\sqrt{3}) \hat{j} = 6 \hat{i} + 2\sqrt{3} \hat{j} \, \text{m} \]

### Step 6: Find the magnitude of the displacement
The magnitude of the displacement \( |\mathbf{s}| \) is given by:
\[ |\mathbf{s}| = \sqrt{(6)^2 + (2\sqrt{3})^2} \]
Calculating this:
\[ |\mathbf{s}| = \sqrt{36 + 12} = \sqrt{48} = 4\sqrt{3} \, \text{m} \]

### Final Results
- The magnitude of the final velocity is \( 2\sqrt{7} \, \text{m/s} \).
- The magnitude of the displacement is \( 4\sqrt{3} \, \text{m} \).