The position vector of an object at any time $t$ is given by $\vec{r} = 3t^2 \hat{i} + 6t \hat{j} + \hat{k}$. Its velocity along the $y$-axis has the magnitude:

$A$ body moves $6\,m$ north,$8\,m$ east,and $10\,m$ vertically upwards. What is its resultant displacement from the initial position?

The magnitude of acceleration and velocity of a particle moving in a plane, whose position vector is $\vec{r} = 3t^2 \hat{i} + 2t \hat{j} + \hat{k}$ at $t = 2 \text{ s}$, are, respectively:

$A$ man moves in the $x-y$ plane along the path shown. At what point is his average velocity vector in the same direction as his instantaneous velocity vector? The man starts from point $P$.

Consider the motion of a particle described by $x = a \cos t$, $y = a \sin t$, and $z = t$. The trajectory traced by the particle as a function of time is

The trajectory of a particle moving in a plane is as shown in the figure. The coordinates of position $A$ are $(0, 2)$. The coordinates of another point at which the instantaneous velocity is the same as the average velocity between the points $(0, 2)$ and $(5, 3)$ are: