If the position vector of a particle is given by $\vec{r}(t) = (3t)\hat{i} + (4t^2)\hat{j}$,then obtain its velocity vector at $t = 2 \ s$.

Which two motions are considered to be combined for motion in a plane?

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 $t = 2.0 \; s$?

The coordinates of a moving particle at any time $t$ are given by $x = at^2$ and $y = bt^2$. The speed of the particle at any moment is

$A$ 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 $; m$)

$A$ particle starts from the origin at $t=0$ with a velocity of $10 \hat{j} \text{ ms}^{-1}$ and moves in the $x-y$ plane with a constant acceleration of $(8 \hat{i} + 2 \hat{j}) \text{ ms}^{-2}$. At an instant when the $x$-coordinate of the particle is $16 \text{ m}$, the $y$-coordinate of the particle is: (in $\text{ m}$)