Two particles $A$ and $B$ start at the origin $O$ and travel in opposite directions along a circular path of radius $0.5\,m$ at constant speeds $0.5\,m/s$ and $1.5\,m/s$,respectively. The time when they collide with each other is ........ $\text{s}$.

The position vector of a particle moving in a plane is given by $r = a \cos \omega t \hat{i} + b \sin \omega t \hat{j}$, where $\hat{i}$ and $\hat{j}$ are the unit vectors along the rectangular axes $X$ and $Y$; $a$, $b$, and $\omega$ are constants, and $t$ is time. The acceleration of the particle is directed along the vector:

The real force $F$ acting on a particle of mass $m$ performing circular motion acts along the radius of a circle of radius $r$ and is directed towards the center of the circle. If the square root of the magnitude of such force is given by $\sqrt{F} = \frac{2 \pi}{T} \sqrt{m r}$,where $T$ is the periodic time,find the expression for the force $F$.

$A$ particle is moving such that its position coordinates $(x, y)$ are $(2 \, m, 3 \, m)$ at time $t = 0$,$(6 \, m, 7 \, m)$ at time $t = 2 \, s$,and $(13 \, m, 14 \, m)$ at $t = 5 \, s$. The average velocity vector $\vec{v}_{av}$ from $t = 0$ to $t = 5 \, s$ is:

$A$ body performing uniform circular motion of radius '$R$' has frequency '$n$'. Its centripetal acceleration per unit radius is proportional to $(n)^x$. The value of $x$ is

$A$ man wants to reach from $A$ to the opposite corner of the square $C$. The sides of the square are $100\, m$. $A$ central square of $50\, m \times 50\, m$ is filled with sand. Outside this square,he can walk at a speed of $1\, m/s$. In the central square,he can walk only at a speed of $v\, m/s$ $(v < 1)$. What is the smallest value of $v$ for which he can reach faster via a straight path through the sand than any path in the square outside the sand?