At time `t= 0,` a particle is at `(2m, 4m).` It starts moving towards positive x-axis with constant acceleration `2 m//s^2` (initial velocity=0). After 2 s, an additional acceleration of `4 m//s^2` starts acting on the particle in negative y-direction also. Find after next 2 s.
(a) velocity and
(b) coordinates of particle.

Explanation

To solve the problem step by step, we will break it down into two main parts: the first 2 seconds of motion and the next 2 seconds after the additional acceleration starts acting.

### Step 1: Calculate the position and velocity after the first 2 seconds

1. **Initial Conditions**:
- Initial position \( r_0 = (2 \, \text{m}, 4 \, \text{m}) \)
- Initial velocity \( u = 0 \)
- Acceleration in the x-direction \( a_1 = 2 \, \text{m/s}^2 \)
- Time \( t_1 = 2 \, \text{s} \)

2. **Calculate the velocity in the x-direction after 2 seconds**:
\[ v_1 = u + a_1 t_1 = 0 + 2 \times 2 = 4 \, \text{m/s} \, \text{(in the x-direction)} \]

3. **Calculate the position in the x-direction after 2 seconds**:
\[ r_1 = r_0 + ut + \frac{1}{2} a_1 t_1^2 \]
\[ r_1 = (2 \, \text{m}, 4 \, \text{m}) + (0) + \frac{1}{2} \times 2 \times (2^2) = (2 + 4, 4) = (6 \, \text{m}, 4 \, \text{m}) \]

### Step 2: Calculate the position and velocity after the next 2 seconds with additional acceleration

1. **New Conditions**:
- After 2 seconds, an additional acceleration \( a_2 = -4 \, \text{m/s}^2 \) (in the negative y-direction) starts acting.
- Time for this phase \( t_2 = 2 \, \text{s} \)

2. **Calculate the velocity after the additional acceleration starts acting**:
\[ v_2 = v_1 + a_2 t_2 \]
Here, \( a_2 \) has components: \( 2 \, \text{m/s}^2 \) in the x-direction and \( -4 \, \text{m/s}^2 \) in the y-direction.
\[ v_2 = (4 \, \text{i}) + (2 \, \text{i} - 4 \, \text{j}) \times 2 \]
\[ v_2 = (4 + 4) \, \text{i} + (-8) \, \text{j} = 8 \, \text{i} - 8 \, \text{j} \, \text{m/s} \]

3. **Calculate the new position after the additional acceleration**:
\[ r_2 = r_1 + v_1 t_2 + \frac{1}{2} a_2 t_2^2 \]
\[ r_2 = (6 \, \text{m}, 4 \, \text{m}) + (4 \, \text{m/s} \times 2 \, \text{s}) + \frac{1}{2} (2 \, \text{i} - 4 \, \text{j}) \times (2^2) \]
\[ r_2 = (6, 4) + (8, 0) + (2 \times 2, -4 \times 2) = (6 + 8 + 8, 4 - 8) = (22 \, \text{m}, -4 \, \text{m}) \]

### Final Results:
- **Velocity after 4 seconds**: \( \mathbf{v_2} = 8 \, \text{i} - 8 \, \text{j} \, \text{m/s} \)
- **Coordinates after 4 seconds**: \( (22 \, \text{m}, -4 \, \text{m}) \)