The acceleration of a particle moving along x-axis is given by a = 6t. Initially it has a velocity of ` 4m//s` along negative x-direction and it is at the origin. Then select correct options.

Options

A. With in 2 sec is will return to origin.
B. With in 2 sec it will cover `((32)/(3sqrt(3)))` m distance
C. It's velocity when it return back to origin will be `8 m//s`
D. None of these

Answer

To solve the problem step by step, we need to analyze the motion of the particle given the acceleration, initial velocity, and position.

### Step 1: Understand the Given Information
- Acceleration \( a = 6t \) (where \( t \) is time in seconds)
- Initial velocity \( u = -4 \, \text{m/s} \) (negative indicates direction along negative x-axis)
- Initial position \( x_0 = 0 \) (origin)

### Step 2: Find the Velocity as a Function of Time
The acceleration is the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} = 6t \]
Integrating both sides with respect to time \( t \):
\[ v = \int 6t \, dt = 3t^2 + C \]
To find the constant \( C \), we use the initial condition \( v(0) = -4 \):
\[ -4 = 3(0)^2 + C \implies C = -4 \]
Thus, the velocity as a function of time is:
\[ v(t) = 3t^2 - 4 \]

### Step 3: Determine When the Particle Stops
The particle stops when the velocity \( v = 0 \):
\[ 3t^2 - 4 = 0 \implies 3t^2 = 4 \implies t^2 = \frac{4}{3} \implies t = \frac{2}{\sqrt{3}} \approx 1.155 \, \text{s} \]

### Step 4: Find the Displacement as a Function of Time
The displacement \( x \) can be found by integrating the velocity:
\[ x = \int v \, dt = \int (3t^2 - 4) \, dt = t^3 - 4t + C' \]
Using the initial condition \( x(0) = 0 \):
\[ 0 = (0)^3 - 4(0) + C' \implies C' = 0 \]
Thus, the displacement as a function of time is:
\[ x(t) = t^3 - 4t \]

### Step 5: Calculate the Displacement at \( t = 2 \) Seconds
To find the displacement at \( t = 2 \):
\[ x(2) = (2)^3 - 4(2) = 8 - 8 = 0 \]
This means the particle returns to the origin after 2 seconds.

### Step 6: Find the Velocity at \( t = 2 \) Seconds
Now, let's find the velocity when it returns to the origin at \( t = 2 \):
\[ v(2) = 3(2)^2 - 4 = 3(4) - 4 = 12 - 4 = 8 \, \text{m/s} \]
The velocity is \( 8 \, \text{m/s} \) in the positive x-direction.

### Summary of Results
1. The particle returns to the origin after 2 seconds.
2. The velocity when it returns to the origin is \( 8 \, \text{m/s} \).