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.

Options

A. `1 s` `36 m`
B. `12 s` `36 m`
C. `1 s` `360 m`
D. `1 s` `136 m`

Explanation

To solve the problem step by step, we will analyze the motion of both particles and determine the time when the distance between them is minimum, as well as the minimum distance itself.

### Given Data:
- Particle 1:
- Initial position: \( x_1(0) = -40 \, \text{m} \) (40 m behind Particle 2)
- Initial velocity: \( u_1 = 12 \, \text{m/s} \)
- Acceleration: \( a_1 = 4 \, \text{m/s}^2 \)

- Particle 2:
- Initial position: \( x_2(0) = 0 \, \text{m} \)
- Initial velocity: \( u_2 = 4 \, \text{m/s} \)
- Acceleration: \( a_2 = 12 \, \text{m/s}^2 \)

### Step 1: Determine the velocities of both particles as functions of time
The velocity of each particle at time \( t \) can be expressed as:
- For Particle 1:
\[ v_1(t) = u_1 + a_1 t = 12 + 4t \]
- For Particle 2:
\[ v_2(t) = u_2 + a_2 t = 4 + 12t \]

### Step 2: Set the velocities equal to find the time when they are the same
To find the time when the distance between them is minimum, we set \( v_1(t) = v_2(t) \):
\[ 12 + 4t = 4 + 12t \]
Rearranging gives:
\[ 12 - 4 = 12t - 4t \]
\[ 8 = 8t \]
\[ t = 1 \, \text{s} \]

### Step 3: Calculate the positions of both particles at \( t = 1 \, \text{s} \)
Using the equation of motion \( s = ut + \frac{1}{2}at^2 \):
- For Particle 1:
\[ s_1 = u_1 t + \frac{1}{2} a_1 t^2 = 12(1) + \frac{1}{2}(4)(1^2) = 12 + 2 = 14 \, \text{m} \]
Therefore, the position of Particle 1 at \( t = 1 \, \text{s} \):
\[ x_1(1) = -40 + 14 = -26 \, \text{m} \]

- For Particle 2:
\[ s_2 = u_2 t + \frac{1}{2} a_2 t^2 = 4(1) + \frac{1}{2}(12)(1^2) = 4 + 6 = 10 \, \text{m} \]
Therefore, the position of Particle 2 at \( t = 1 \, \text{s} \):
\[ x_2(1) = 0 + 10 = 10 \, \text{m} \]

### Step 4: Calculate the minimum distance between the two particles
The distance between the two particles at \( t = 1 \, \text{s} \) is:
\[ \text{Distance} = x_2(1) - x_1(1) = 10 - (-26) = 10 + 26 = 36 \, \text{m} \]
However, we need to consider the initial distance of 40 m:
\[ \text{Minimum Distance} = \text{Initial Distance} - \text{Distance} = 40 - 4 = 36 \, \text{m} \]

### Final Answers:
(a) The time when the distance between them is minimum: **1 second**
(b) The minimum distance between them: **36 meters**