$A$ particle moves with a constant speed $v$ along a circle of radius $R$. What is the change in velocity of the particle as it moves from point $P_1$ to $P_2$?

The position vector of a particle is $\vec{r} = a \cos \omega t \hat{i} + a \sin \omega t \hat{j}$. Calculate the angle between its position vector and the velocity vector. $(\cos 0 = 1, \cos 90 = 0, \cos 60 = 0.5, \sin 30 = 0.5, \sin 0 = 0)$ (in $^\circ$)

As shown in the figure,a particle is moving with a constant speed $v = \pi \, m/s$ in a circular path. Considering its motion from point $A$ to point $B$,where the angle subtended at the center is $120^{\circ}$,the magnitude of the average velocity is:

Is the statement "Angular position $\theta$ is a scalar,whereas angular displacement is a vector" correct?

If ${a_r}$ and ${a_t}$ represent radial and tangential accelerations,the motion of a particle will be uniformly circular if

$A$ particle is performing uniform circular motion. The velocity of the particle at an instant of time is $(6 \hat{i} - 2 \hat{j}) \ m/s$ and the angular velocity is $\vec{\omega} = (a \hat{i} + b \hat{j}) \ rad/s$. Then the value of $\frac{b}{a}$ is: