The position of a particle moving in the xy plane at any time t is given by `x=(3t^2-6t)` metres, `y=(t^2-2t)` metres. Select the correct statement about the moving particle from the following

Options

A. The acceleration of particle is zero at `t=0` second
B. the velocity of the particle is zero `t=0` second
C. the velocity of the particle is zero at `t=1` second
D. the velocity and acceleration of the particle are never zero

Explanation

To solve the problem, we need to analyze the motion of the particle given its position functions in the xy-plane. The position of the particle is defined by the equations:

- \( x(t) = 3t^2 - 6t \)
- \( y(t) = t^2 - 2t \)

We will find the velocity and acceleration of the particle and check the statements provided about it.

### Step 1: Find the velocity components
The velocity components in the x and y directions can be found by taking the derivatives of the position functions with respect to time \( t \).

1. **Velocity in the x-direction**:
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(3t^2 - 6t) = 6t - 6 \]

2. **Velocity in the y-direction**:
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(t^2 - 2t) = 2t - 2 \]

### Step 2: Find the acceleration components
The acceleration components can be found by taking the derivatives of the velocity components with respect to time \( t \).

1. **Acceleration in the x-direction**:
\[ a_x = \frac{dv_x}{dt} = \frac{d}{dt}(6t - 6) = 6 \]

2. **Acceleration in the y-direction**:
\[ a_y = \frac{dv_y}{dt} = \frac{d}{dt}(2t - 2) = 2 \]

### Step 3: Analyze the conditions at specific times
Now, we will check the conditions at \( t = 0 \) seconds and \( t = 1 \) second.

1. **At \( t = 0 \) seconds**:
- Velocity:
\[ v_x(0) = 6(0) - 6 = -6 \quad \text{and} \quad v_y(0) = 2(0) - 2 = -2 \]
The velocity vector is \( (-6, -2) \), which is not zero.
- Acceleration:
\[ a_x = 6 \quad \text{and} \quad a_y = 2 \]
The acceleration vector is \( (6, 2) \), which is also not zero.

2. **At \( t = 1 \) second**:
- Velocity:
\[ v_x(1) = 6(1) - 6 = 0 \quad \text{and} \quad v_y(1) = 2(1) - 2 = 0 \]
The velocity vector is \( (0, 0) \), which means the particle is momentarily at rest at \( t = 1 \) second.

### Conclusion
Based on the calculations:
- The acceleration of the particle is never zero; it is constant at \( (6, 2) \).
- The velocity of the particle is zero at \( t = 1 \) second.

### Final Statements
1. The acceleration of the particle is not zero at \( t = 0 \) seconds.
2. The velocity of the particle is not zero at \( t = 0 \) seconds.
3. The velocity of the particle is zero at \( t = 1 \) second.