The coordinates of a moving particle at any time $t$ are given by $x = \alpha t^3$ and $y = \beta t^3$. The speed of the particle at time $t$ is given by

$A$ 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:

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 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).

$A$ body starts from rest from the origin with an acceleration of $6 \; m/s^2$ along the $x$-axis and $8 \; m/s^2$ along the $y$-axis. Its distance from the origin after $4 \; s$ will be (in $; m$)

Write the equations of motion for uniformly accelerated motion in a plane.