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 the particle is zero at `t = 0s`
B. The velocity of the particle is zero at `t = 0s`
C. The velocity of the particle is zero at `t = 1s`
D. The velocity and acceleration of the particle are zero

Explanation

To solve the problem, we need to analyze the motion of the particle in the xy-plane based on the given equations for its position:

1. **Given Equations**:
- \( x(t) = 3t^2 - 6t \) (in meters)
- \( y(t) = t^2 - 2t \) (in meters)

2. **Finding Velocity in x-direction**:
- The velocity in the x-direction \( v_x \) is the derivative of \( x(t) \) with respect to \( t \):
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(3t^2 - 6t) = 6t - 6 \]

3. **Finding Acceleration in x-direction**:
- The acceleration in the x-direction \( a_x \) is the derivative of \( v_x \):
\[ a_x = \frac{dv_x}{dt} = \frac{d}{dt}(6t - 6) = 6 \, \text{m/s}^2 \]

4. **Finding Velocity in y-direction**:
- The velocity in the y-direction \( v_y \) is the derivative of \( y(t) \) with respect to \( t \):
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(t^2 - 2t) = 2t - 2 \]

5. **Finding Acceleration in y-direction**:
- The acceleration in the y-direction \( a_y \) is the derivative of \( v_y \):
\[ a_y = \frac{dv_y}{dt} = \frac{d}{dt}(2t - 2) = 2 \, \text{m/s}^2 \]

6. **Evaluating Acceleration at \( t = 0 \)**:
- At \( t = 0 \):
\[ a_x = 6 \, \text{m/s}^2, \quad a_y = 2 \, \text{m/s}^2 \]
- The total acceleration \( a \) is given by:
\[ a = \sqrt{a_x^2 + a_y^2} = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} \, \text{m/s}^2 \]
- Thus, the acceleration is not zero at \( t = 0 \).

7. **Evaluating Velocity at \( t = 0 \)**:
- At \( t = 0 \):
\[ v_x = 6(0) - 6 = -6 \, \text{m/s}, \quad v_y = 2(0) - 2 = -2 \, \text{m/s} \]
- The total velocity \( v \) is given by:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(-6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} \, \text{m/s} \]
- Thus, the velocity is not zero at \( t = 0 \).

8. **Evaluating Velocity at \( t = 1 \)**:
- At \( t = 1 \):
\[ v_x = 6(1) - 6 = 0 \, \text{m/s}, \quad v_y = 2(1) - 2 = 0 \, \text{m/s} \]
- Therefore, the total velocity at \( t = 1 \) is:
\[ v = \sqrt{0^2 + 0^2} = 0 \, \text{m/s} \]
- Hence, the velocity is zero at \( t = 1 \).

9. **Conclusion**:
- The correct statement about the moving particle is that the velocity of the particle is zero at \( t = 1 \).