The position of a particle moving in the XY plane at any time $t$ is given by $x = (3t^2 - 6t)$ metres. Select

Question

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

Accepted Answer

1.  **Velocity ($v$):** Instantaneous velocity is the rate of change of position with respect to time, $v = \frac{dx}{dt}$ .
    Given $x = 3t^2 - 6t$.
    Differentiating with respect to $t$: $v = \frac{d}{dt}(3t^2 - 6t) = 6t - 6$.
2.  **Acceleration ($a$):** Acceleration is the rate of change of velocity, $a = \frac{dv}{dt}$ .
    Differentiating $v$ with respect to $t$: $a = \frac{d}{dt}(6t - 6) = 6 	ext{ m/s}^2$.
3.  **Analyze Options:**
    *   **Acceleration:** $a = 6 	ext{ m/s}^2$ (constant). It is non-zero at all times, including $t=0$.
    *   **Velocity at $t=0$:** $v(0) = 6(0) - 6 = -6 	ext{ m/s}$. (Non-zero).
    *   **Velocity at $t=1$:** $v(1) = 6(1) - 6 = 0 	ext{ m/s}$. (Zero).
    *   **Conclusion:** The velocity becomes zero at $t=1$ s, while the acceleration is constant and never zero.

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