Velocity and acceleration of a particle at time `t=0`are `u=(2 hati+3 hatj) m//s and a=(4 hati+3 hatj) m//s^2` respectively. Find the velocity and displacement if particle at `t=2s.`

Explanation

To solve the problem, we need to find the velocity and displacement of a particle at time \( t = 2 \) seconds, given its initial velocity and acceleration.

### Step 1: Identify the given values
- Initial velocity \( \mathbf{u} = 2 \hat{i} + 3 \hat{j} \) m/s
- Acceleration \( \mathbf{a} = 4 \hat{i} + 3 \hat{j} \) m/s²
- Time \( t = 2 \) s

### Step 2: Calculate the velocity at \( t = 2 \) seconds
Using the formula for velocity under constant acceleration:
\[ \mathbf{v} = \mathbf{u} + \mathbf{a}t \]

Substituting the values:
\[ \mathbf{v} = (2 \hat{i} + 3 \hat{j}) + (4 \hat{i} + 3 \hat{j}) \cdot 2 \]
Calculating each component:
- For the \( \hat{i} \) component:
\[ v_x = 2 + 4 \cdot 2 = 2 + 8 = 10 \text{ m/s} \]
- For the \( \hat{j} \) component:
\[ v_y = 3 + 3 \cdot 2 = 3 + 6 = 9 \text{ m/s} \]

Thus, the velocity at \( t = 2 \) seconds is:
\[ \mathbf{v} = 10 \hat{i} + 9 \hat{j} \text{ m/s} \]

### Step 3: Calculate the magnitude of the velocity
The magnitude of the velocity \( V \) is given by:
\[ V = \sqrt{v_x^2 + v_y^2} \]
Substituting the values:
\[ V = \sqrt{10^2 + 9^2} = \sqrt{100 + 81} = \sqrt{181} \text{ m/s} \]

### Step 4: Calculate the displacement at \( t = 2 \) seconds
Using the formula for displacement under constant acceleration:
\[ \mathbf{s} = \mathbf{u}t + \frac{1}{2} \mathbf{a}t^2 \]

Calculating each component:
- For the \( \hat{i} \) component:
\[ s_x = (2 \hat{i}) \cdot 2 + \frac{1}{2} (4 \hat{i}) \cdot (2^2) \]
\[ s_x = 4 \hat{i} + \frac{1}{2} \cdot 4 \cdot 4 \hat{i} = 4 \hat{i} + 8 \hat{i} = 12 \hat{i} \text{ m} \]

- For the \( \hat{j} \) component:
\[ s_y = (3 \hat{j}) \cdot 2 + \frac{1}{2} (3 \hat{j}) \cdot (2^2) \]
\[ s_y = 6 \hat{j} + \frac{1}{2} \cdot 3 \cdot 4 \hat{j} = 6 \hat{j} + 6 \hat{j} = 12 \hat{j} \text{ m} \]

Thus, the displacement at \( t = 2 \) seconds is:
\[ \mathbf{s} = 12 \hat{i} + 12 \hat{j} \text{ m} \]

### Step 5: Calculate the magnitude of the displacement
The magnitude of the displacement \( S \) is given by:
\[ S = \sqrt{s_x^2 + s_y^2} \]
Substituting the values:
\[ S = \sqrt{12^2 + 12^2} = \sqrt{144 + 144} = \sqrt{288} = 12\sqrt{2} \text{ m} \]

### Final Results
- Velocity at \( t = 2 \) seconds: \( \sqrt{181} \) m/s
- Displacement at \( t = 2 \) seconds: \( 12\sqrt{2} \) m