Two moving particles P and Q are $10\text{m}$ apart at a certain instant. The velocity of P is $8\text{m/s}$ making ${30}^{\circ }$ with the line joining P and Q and that of Q is $6\text{m/s}$ making ${30}^{\circ }$ with PQ in the figure. Then the angular velocity of Q with respect to P in rad/s at that instant is

Q. Two moving particles $P$ and $Q$ are $10\text{m}$ apart at a certain instant. The velocity of $P$ is $8\text{m/s}$ making an angle of ${30}^{\circ }$ with the line joining $P$ and $Q$. The velocity of $Q$ is $6\text{m/s}$ making an angle of ${30}^{\circ }$ with the line joining $P$ and $Q$ as shown in figure. The angular velocity of $P$ with respect to $Q$ is

Q. Two moving particles P & Q are 10m apart at any instant. Velocity of P is 8 m/s and that of Q is 6 m/s at ${30}^{\circ }$ angle with the line joining P & Q.
Calculate the angular velocity P w.r.t. Q

Q. Two particle $P$ and $Q$ are at a distance of $10m$ at any moment. Velocities of $P$ and $Q$ are $8m/s$ respectively. The make an angle of ${30}^o$ with line $PQ$ as shown in fig. Angular velocity of $P$ with respect to $Q$ will be:

Q. Two particles $P$ and $Q$ are moving on a circle. At a certain instant of time both the particles are diametrically opposite and $P$ has tangential acceleration $8m/s^2$ and centripetal centripetal $5m/s^2$ whereas $Q$ has only centripetal acceleration of $1m/s^2$. At that instant acceleration (in $m/s^2)$ of $P$ with respect to $Q$ is:

Q. A particle started moving from $P$ towards $S$ with uniform acceleration along a straight line. The average velocity of the particle from $P$ to intermediate point $Q$ is $8m/s$ and that $Q$ to $S$ is $12m/s$. If $QS=PQ$, then the average velocity from $P$ to $S$ is: