Two particles are moving along x-axis. Particle-1 is `40 m` behind Particle-2. Particle-1 starts with velocity`12 m//s` and acceleration `4 m//s^2` both in positive x-direction. Particle-2 starts with velocity `4 m//s` and acceleration `12 m//s^2` also in positive x-direction. Find
(a) the time when distance between them is minimum.
(b) the minimum distacne between them.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Set Up the Equations for Each Particle
We need to find the equations of motion for both particles.

- For Particle 1:
- Initial position \( s_1(0) = -40 \, \text{m} \) (since it is 40 m behind Particle 2)
- Initial velocity \( u_1 = 12 \, \text{m/s} \)
- Acceleration \( a_1 = 4 \, \text{m/s}^2 \)

The position of Particle 1 at time \( t \) is given by:
\[ s_1(t) = s_1(0) + u_1 t + \frac{1}{2} a_1 t^2 \]
Substituting the values:
\[ s_1(t) = -40 + 12t + \frac{1}{2} \cdot 4 \cdot t^2 = -40 + 12t + 2t^2 \]

- For Particle 2:
- Initial position \( s_2(0) = 0 \) (starting at the origin)
- Initial velocity \( u_2 = 4 \, \text{m/s} \)
- Acceleration \( a_2 = 12 \, \text{m/s}^2 \)

The position of Particle 2 at time \( t \) is given by:
\[ s_2(t) = s_2(0) + u_2 t + \frac{1}{2} a_2 t^2 \]
Substituting the values:
\[ s_2(t) = 0 + 4t + \frac{1}{2} \cdot 12 \cdot t^2 = 4t + 6t^2 \]

### Step 2: Find the Time When the Velocities are Equal
To find the time when the distance between the two particles is minimum, we need to set their velocities equal to each other.

The velocity of Particle 1 at time \( t \) is:
\[ v_1 = u_1 + a_1 t = 12 + 4t \]
The velocity of Particle 2 at time \( t \) is:
\[ v_2 = u_2 + a_2 t = 4 + 12t \]

Setting \( v_1 = v_2 \):
\[ 12 + 4t = 4 + 12t \]
Rearranging gives:
\[ 12 - 4 = 12t - 4t \implies 8 = 8t \implies t = 1 \, \text{s} \]

### Step 3: Calculate the Distances Travelled by Each Particle at \( t = 1 \, \text{s} \)
Now, we will calculate the distance travelled by each particle at \( t = 1 \, \text{s} \).

- Distance travelled by Particle 1:
\[ s_1(1) = -40 + 12(1) + 2(1^2) = -40 + 12 + 2 = -26 \, \text{m} \]

- Distance travelled by Particle 2:
\[ s_2(1) = 4(1) + 6(1^2) = 4 + 6 = 10 \, \text{m} \]

### Step 4: Find the Minimum Distance Between the Particles
The minimum distance between the two particles at \( t = 1 \, \text{s} \) is given by:
\[ \text{Minimum Distance} = s_2(1) - s_1(1) = 10 - (-26) = 10 + 26 = 36 \, \text{m} \]

### Final Answers
(a) The time when the distance between them is minimum is \( t = 1 \, \text{s} \).

(b) The minimum distance between them is \( 36 \, \text{m} \).