Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A particle moves along a straight line such that its displacement at any time $t$ is given by $S = t^3 - 6t^2 + 3t + 4$ metres. The velocity when the acceleration is zero is:

3 ms⁻¹

-12 ms⁻¹

42 ms⁻¹

-9 ms⁻¹

Step-by-Step Solution

Velocity ( $v$ ): Instantaneous velocity is the rate of change of displacement with respect to time, $v = \frac{dS}{dt}$ . Given $S = t^3 - 6t^2 + 3t + 4$ . Differentiating with respect to $t$ : $v = \frac{d}{dt}(t^3 - 6t^2 + 3t + 4) = 3t^2 - 12t + 3$ .
Acceleration ( $a$ ): Instantaneous acceleration is the rate of change of velocity, $a = \frac{dv}{dt}$ . Differentiating $v$ with respect to $t$ : $a = \frac{d}{dt}(3t^2 - 12t + 3) = 6t - 12$ .
Condition for Zero Acceleration: Set $a = 0$ . $6t - 12 = 0 \implies 6t = 12 \implies t = 2 \text{ s}$ .
Calculate Velocity: Substitute $t = 2 \text{ s}$ into the velocity equation. $v = 3(2)^2 - 12(2) + 3$ $v = 3(4) - 24 + 3$ $v = 12 - 24 + 3 = -9 \text{ ms}^{-1}$ .

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