$A$ particle moves in the $xy$ plane according to the law $x = a \sin \omega t$ and $y = a(1 - \cos \omega t)$,where $a$ and $\omega$ are constants. The particle traces:

Three point particles $P, Q, R$ move in a circle of radius $r$ with different but constant speeds. They start moving at $t = 0$ from their initial positions as shown in the figure. The angular velocities (in $rad/s$) of $P, Q,$ and $R$ are $5\pi, 2\pi,$ and $3\pi$ respectively, in the same sense. Find the time interval after which they are at the same angular position.

$A$ particle starts from the origin at $t=0$ with a velocity $5 \hat{i} \text{ m/s}$ and moves in the $x-y$ plane under the action of a force which produces a constant acceleration of $(3 \hat{i} + 2 \hat{j}) \text{ m/s}^2$. If the $x$-coordinate of the particle at that instant is $84 \text{ m}$,then the speed of the particle at this time is $\sqrt{\alpha} \text{ m/s}$. The value of $\alpha$ is . . . . . . .

$A$ particle leaves the origin with an initial velocity $\vec{v} = (3 \hat{i}) \text{ m s}^{-1}$ and a constant acceleration $\vec{a} = (-1 \hat{i} - 0.5 \hat{j}) \text{ m s}^{-2}$. The position vector of the particle, when it reaches its maximum $x$-coordinate, is:

$A$ particle of mass $m$ is projected at a velocity $u$, making an angle $\theta$ with the horizontal ($x$-axis). If the angle of projection $\theta$ is varied keeping all other parameters same, then the magnitude of angular momentum $(L)$ at its maximum height about the point of projection varies with $\theta$ as,

$A$ truck starts from rest and accelerates uniformly at $2.0 \; m s^{-2}$. At $t = 10 \; s$,a stone is dropped by a person standing on the top of the truck ($6 \; m$ high from the ground). What are the $(a)$ velocity,and $(b)$ acceleration of the stone at $t = 11 \; s$? (Neglect air resistance.)