$A$ particle projected from the origin moves in the $x-y$ plane with a velocity $\vec{v} = 3 \hat{i} + 6x \hat{j}$,where $\hat{i}$ and $\hat{j}$ are the unit vectors along the $x$ and $y$ axes. Find the equation of the path followed by the particle.

The position of a particle is given by $\vec{r} = 3.0 t \hat{i} - 2.0 t^{2} \hat{j} + 4.0 \hat{k} \; m$,where $t$ is in seconds and the coefficients have the proper units for $\vec{r}$ to be in metres.
$(a)$ Find the velocity $\vec{v}$ and acceleration $\vec{a}$ of the particle.
$(b)$ What is the magnitude and direction of velocity of the particle at $t = 2.0 \; s$?

$A$ particle moves in space along the path $z = ax^3 + by^2$ in such a way that $\frac{dx}{dt} = c = \frac{dy}{dt}$,where $a, b,$ and $c$ are constants. The acceleration of the particle is:

$A$ particle initially at the origin starts moving in the $xy$-plane with a velocity component $\vec{V} = (6 + 2t) \hat{i} + (4 + 2\sqrt{3}t) \hat{j} \text{ m/s}$. The acceleration of the particle in $\text{m/s}^2$ is ($x, y$ are measured in meters, $t$ in seconds, respectively).

$A$ 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.

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