If the displacement ($s$ in metre) of a moving particle in terms of time ($t$ in second) is $s = t^3 - 6t^2 + 18t + 9$,then the minimum velocity attained by the particle is (in $m \ s^{-1}$)

If velocity and acceleration are in opposite directions in one-dimensional motion,what happens to the magnitude of velocity?

Acceleration of a particle changes when:

$A$ particle is moving in one dimension (along $x$-axis) under the action of a variable force. Its initial position was $16 \,m$ right of the origin. The variation of its position $(x)$ with time $(t)$ is given as $x = -3t^3 + 18t^2 + 16t$,where $x$ is in $m$ and $t$ is in $s$. The velocity of the particle when its acceleration becomes zero is . . . . . . $m/s$.

The correct position $(x)$ - time $(t)$ graph for a particle moving with negative acceleration is

The velocity of a particle is given as $\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k} \text{ m/s}$. The magnitude of acceleration at point $(1, 2, 4)$ is . . . . . . $\text{m/s}^2$.