Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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.

The acceleration of the particle is zero at t = 0 second

The velocity of the particle is zero at t = 0 second

The velocity of the particle is zero at t = 1 second

The velocity and acceleration of the particle are never zero

Step-by-Step Solution

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 $t$ : $v = \frac{d}{dt}(3t^2 - 6t) = 6t - 6$ .
Acceleration ( $a$ ): Acceleration is the rate of change of velocity, $a = \frac{dv}{dt}$ . Differentiating $v$ with respect to $t$ : $a = \frac{d}{dt}(6t - 6) = 6 \text{ m/s}^2$ .
Analyze Options:

Acceleration: $a = 6 \text{ m/s}^2$ (constant). It is non-zero at all times, including $t=0$ .
Velocity at $t=0$ : $v(0) = 6(0) - 6 = -6 \text{ m/s}$ . (Non-zero).
Velocity at $t=1$ : $v(1) = 6(1) - 6 = 0 \text{ m/s}$ . (Zero).
Conclusion: The velocity becomes zero at $t=1$ s, while the acceleration is constant and never zero.

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