For a particle moving with speed $v$ in a uniform circular motion, the acceleration at a point $P$$\left(R, \mathrm{\theta }\right)$ in the first quadrant on the circle of radius $R$ is $($here, $\mathrm{\theta }$ is measured from the $x$-axis$)$

Three coplanar vectors $\overrightarrow{A}$, $\overrightarrow{B}$ and $\overrightarrow{\mathrm{C}}$ have magnitudes $4$, $3$ and $2$, respectively. If the angle between any two vectors is $120°$, then which of the following vectors may be equal to $\frac{3\overrightarrow{A}}{4}+\frac{\overrightarrow{B}}{3}+\frac{\overrightarrow{C}}{2}$?

A particle $A$ moves along a circle of radius, $R=50 \mathrm{cm}$so that its radius vector $r$, relative to the point $O$ (figure), rotates with the constant angular velocity, $\omega =0.40 \mathrm{rad} {\mathrm{s}}^{-1}$. Then, the modulus of the velocity $\left(v\right)$ of the particle and the modulus of its total acceleration $\left(a\right)$ will be,

If $\left|\overrightarrow{a}\right|=3\sqrt{2}$ and $\left|\overrightarrow{b}\right|=6\sqrt{2}$, then $x-$component of $\left(\overrightarrow{a}+\overrightarrow{b}\right)$ is

The figure shows three vectors $\overrightarrow{a},\overrightarrow{ b}$ and $\overrightarrow{c}$. Here, $R$ is the midpoint of $PQ$. Then which of the following relations is correct?