A time varying force `F=6t-2t^(2)N`, at t=0 starts acting on a body of mass 2kg initially at rest, where t is in second. The force is withdrawn just at the instant when the body comes to rest again. We can see that at `t=0` the force `F=0`. Now answer the following:
Mark the correct statement:

Options

A. Velocity of the body is maximum when force acting on the body is maximum for the first time.
B. The velocity of the body becomes maximum when force acting on the body becomes zero again.
C. When force becomes zero again, velocity of the body also becomes zero at that instant.
D. All of the above

Explanation

To solve the problem, we will analyze the time-varying force and its effect on the motion of the body. The force is given by:

\[ F(t) = 6t - 2t^2 \, \text{N} \]

The mass of the body is:

\[ m = 2 \, \text{kg} \]

### Step 1: Determine when the force becomes zero
To find out when the force becomes zero, we set the force equation to zero:

\[ 6t - 2t^2 = 0 \]

Factoring out \( t \):

\[ t(6 - 2t) = 0 \]

This gives us two solutions:

1. \( t = 0 \)
2. \( 6 - 2t = 0 \) which simplifies to \( t = 3 \)

### Step 2: Analyze the motion of the body
The force acts on the body from \( t = 0 \) to \( t = 3 \) seconds. We need to find the velocity of the body when the force is withdrawn (i.e., at \( t = 3 \) seconds).

### Step 3: Calculate acceleration
The acceleration \( a(t) \) can be calculated using Newton's second law:

\[ a(t) = \frac{F(t)}{m} = \frac{6t - 2t^2}{2} = 3t - t^2 \]

### Step 4: Integrate acceleration to find velocity
To find the velocity, we integrate the acceleration with respect to time:

\[ \frac{dv}{dt} = 3t - t^2 \]

Integrating both sides from \( t = 0 \) to \( t = T \) and \( v = 0 \) to \( v = V \):

\[ \int_0^V dv = \int_0^T (3t - t^2) dt \]

Calculating the right-hand side:

\[ V = \left[ \frac{3t^2}{2} - \frac{t^3}{3} \right]_0^T = \frac{3T^2}{2} - \frac{T^3}{3} \]

### Step 5: Substitute \( T = 3 \)
Now substituting \( T = 3 \):

\[ V = \frac{3(3^2)}{2} - \frac{(3^3)}{3} = \frac{3 \times 9}{2} - \frac{27}{3} \]

Calculating:

\[ V = \frac{27}{2} - 9 = \frac{27}{2} - \frac{18}{2} = \frac{9}{2} \, \text{m/s} \]

### Step 6: Conclusion
At \( t = 3 \) seconds, the velocity of the body is \( \frac{9}{2} \, \text{m/s} \), which is not zero. Therefore, the correct statement is:

- The velocity of the body will become maximum when the force acting on the body becomes zero again.