Two particles are moving along x-axis. Particle-1 starts from `x=-10 m` with velocity `4 m//s` along negative x-direction and acceleration `2 m//s^2` along positive x-direction. Particle-2 starts from `x=+2 m` with velocity `6 m//s` along positive x-direction and acceleration `2 m//s^2` along negative x-direction.
(a) Find the time when they collide.
(b) Find the x-coordinates where they collide. Both start simultaneously.

Explanation

To solve the problem, we will analyze the motion of both particles and find the time and position at which they collide.

### Given Data:
- **Particle 1:**
- Initial position, \( x_1(0) = -10 \, \text{m} \)
- Initial velocity, \( u_1 = -4 \, \text{m/s} \) (moving in the negative x-direction)
- Acceleration, \( a_1 = 2 \, \text{m/s}^2 \) (in the positive x-direction)

- **Particle 2:**
- Initial position, \( x_2(0) = 2 \, \text{m} \)
- Initial velocity, \( u_2 = 6 \, \text{m/s} \) (moving in the positive x-direction)
- Acceleration, \( a_2 = -2 \, \text{m/s}^2 \) (decelerating)

### (a) Finding the time when they collide:

1. **Position equations for both particles:**
- For Particle 1:
\[ x_1(t) = x_1(0) + u_1 t + \frac{1}{2} a_1 t^2 \]
Substituting the values:
\[ x_1(t) = -10 - 4t + \frac{1}{2} \cdot 2 t^2 = -10 - 4t + t^2 \]
Thus,
\[ x_1(t) = t^2 - 4t - 10 \]

- For Particle 2:
\[ x_2(t) = x_2(0) + u_2 t + \frac{1}{2} a_2 t^2 \]
Substituting the values:
\[ x_2(t) = 2 + 6t - \frac{1}{2} \cdot 2 t^2 = 2 + 6t - t^2 \]
Thus,
\[ x_2(t) = -t^2 + 6t + 2 \]

2. **Setting the position equations equal to find the collision time:**
\[ t^2 - 4t - 10 = -t^2 + 6t + 2 \]
Rearranging gives:
\[ 2t^2 - 10t - 12 = 0 \]
Dividing the entire equation by 2:
\[ t^2 - 5t - 6 = 0 \]

3. **Factoring the quadratic equation:**
\[ (t - 6)(t + 1) = 0 \]
Thus, \( t = 6 \, \text{s} \) (we discard \( t = -1 \) as time cannot be negative).

### (b) Finding the x-coordinates where they collide:

1. **Substituting \( t = 6 \) into the position equation of either particle:**
Using Particle 1's position equation:
\[ x_1(6) = 6^2 - 4 \cdot 6 - 10 \]
Calculating:
\[ x_1(6) = 36 - 24 - 10 = 2 \, \text{m} \]

2. **Verifying with Particle 2's position equation:**
\[ x_2(6) = -6^2 + 6 \cdot 6 + 2 \]
Calculating:
\[ x_2(6) = -36 + 36 + 2 = 2 \, \text{m} \]

### Final Results:
- **Time of collision:** \( t = 6 \, \text{s} \)
- **x-coordinate of collision:** \( x = 2 \, \text{m} \)