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:

$A$ point moves in the $x-y$ plane according to $x = kt$ and $y = kt(1 - \alpha t)$,where $k$ and $\alpha$ are positive constants. The equation of the trajectory is:

$A$ particle moves in the $x-y$ plane with velocity $\vec{v} = a\hat{i} + bx\hat{j}$,where $a$ and $b$ are constants. If the particle was initially at the origin,find the equation of its trajectory.

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:

$A$ 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 $P$.

The position vector of a moving body at any instant of time is given as $\overrightarrow{r} = (5t^2 \hat{i} - 5t \hat{j}) \text{ m}$. The magnitude and direction of velocity at $t = 2 \text{ s}$ is,