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 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 $\text{ m}$)

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

Consider the motion of a particle described by $x = a \cos t$, $y = a \sin t$, and $z = t$. The trajectory traced by the particle as a function of time is

$A$ particle starts from the origin at $t=0$ with a velocity $5.0 \hat{i} \; m/s$ and moves in the $x-y$ plane under the action of a force which produces a constant acceleration of $(3.0 \hat{i} + 2.0 \hat{j}) \; m/s^2$.
$(a)$ What is the $y$-coordinate of the particle at the instant its $x$-coordinate is $84 \; m$?
$(b)$ What is the speed of the particle at this time?

The $x$ and $y$ coordinates of a particle at any time $t$ are given by $x = 5t - 2t^2$ and $y = 10t$ respectively,where $x$ and $y$ are in meters and $t$ is in seconds. The acceleration of the particle at $t = 2 \, s$ is . . . . . . $m/s^2$.