A particle is moving in one dimension (along $x$ axis) under the action of a variable force. It's initial position was $16 \mathrm{~m}$ right of origin. The variation of its position $(x)$ with time $(t)$ is given as $x=-3 t^3+18 t^2+16 t$, where $x$ is in $\mathrm{m}$ and $\mathrm{t}$ is in $\mathrm{s}$.

The velocity of the particle when its acceleration becomes zero is _________ $\mathrm{m} / \mathrm{s}$.

<ul> <li><strong>Position (x):</strong> The particle's location on the x-axis at a given time.</li><br> <li><strong>Velocity (v):</strong> The rate of change of position with respect to time</li> </ul> <p>( $v = \frac{dx}{dt}$ ).</p> <ul> <li><strong>Acceleration (a):</strong> The rate of change of velocity with respect to time</li> </ul> <p>( $a = \frac{dv}{dt}$ ).</p> <p>The particle's position is given by:</p> <p>$x(t) = -3t^3 + 18t^2 + 16t$</p> <p>We need to find the velocity when the acceleration is zero.</p> <ol> <li><strong>Find the Velocity (v) and Acceleration (a) Functions:</strong></li> </ol> <ul> <br/><li>Velocity:</li> </ul> <p>$v(t) = \frac{dx}{dt} = -9t^2 + 36t + 16$</p> <ul> <li>Acceleration:</li> </ul> <p>$a(t) = \frac{dv}{dt} = -18t + 36$</p> <ol> <li><strong>Find the Time (t) When Acceleration is Zero:</strong></li> </ol> <p>$a(t) = 0$</p> <p>$-18t + 36 = 0$</p> <p>$t = 2 \text{ seconds}$</p> <ol> <li><strong>Calculate the Velocity at t = 2 seconds:</strong></li> </ol> <p>$v(2) = -9(2)^2 + 36(2) + 16$</p> <p>$v(2) = 52 \text{ m/s}$</p> <strong>Answer:</strong> <p>The velocity of the particle when its acceleration becomes zero is 52 m/s.</p> <p></p>