Solved: PHYSICS Question for NEET

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The position $x$ of a particle varies with time $t$ as $x = at^2 - bt^3$ . The acceleration of the particle will be zero at time $t$ equal to:

$\frac{a}{b}$

$\frac{2a}{3b}$

$\frac{a}{3b}$

zero

Velocity ( $v$ ): Instantaneous velocity is defined as the rate of change of position with respect to time, $v = \frac{dx}{dt}$ . Given $x = at^2 - bt^3$ . Differentiating with respect to $t$ : $v = \frac{d}{dt}(at^2 - bt^3) = 2at - 3bt^2$ .
Acceleration ( $a_{acc}$ ): Instantaneous acceleration is defined as the rate of change of velocity with respect to time, $a_{acc} = \frac{dv}{dt}$ . Differentiating $v$ with respect to $t$ : $a_{acc} = \frac{d}{dt}(2at - 3bt^2) = 2a - 6bt$ .
Condition: We need to find the time $t$ when the acceleration is zero. Set $a_{acc} = 0$ : $2a - 6bt = 0$ $6bt = 2a$ $t = \frac{2a}{6b} = \frac{a}{3b}$ .

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