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

$A$ 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 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).

Derive the equations of motion for a body moving in two dimensions: $\vec{v} = \vec{v_0} + \vec{a}t$ and $\vec{r} = \vec{r_0} + \vec{v_0}t + \frac{1}{2}\vec{a}t^2$.

$A$ 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 moving in the $xy$-plane at any time $t$ is given by $x = (3t^2 - 6t) \text{ m}$ and $y = (t^2 - 2t) \text{ m}$. Select the correct statement about the moving particle from the following: