The coordinates of a moving particle at any time $t$ are given by $x = a\, t^2$ and $y = b\, t^2$. The speed of the particle is

The sum of two forces $\overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ is $\overrightarrow{\mathrm{R}}$ such that $|\overrightarrow{\mathrm{R}}|=|\overrightarrow{\mathrm{P}}| .$ The angle $\theta$ (in degrees) that the resultant of $2 \overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ will make with $\overrightarrow{\mathrm{Q}}$ is

The vectors $\vec{A}$ and $\vec{B}$ are such that

$|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$

The angle between the two vectors is

Two vectors $\dot{A}$ and $\dot{B}$ are defined as $\dot{A}=a \hat{i}$ and $\overrightarrow{\mathrm{B}}=\mathrm{a}(\cos \omega t \hat{\mathrm{i}}+\sin \omega t \hat{j}$ ), where a is a constant and $\omega=\pi / 6 \mathrm{rad} \mathrm{s}^{-1}$. If $|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=\sqrt{3}|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|$ at time $t=\tau$ for the first time, the value of $\tau$, in, seconds, is. . . . . .

The vector sum of two forces is perpendicular to their vector differences. In that case, the forces

In an octagon $ABCDEFGH$ of equal side, what is the sum of $\overrightarrow{ AB }+\overrightarrow{ AC }+\overrightarrow{ AD }+\overrightarrow{ AE }+\overrightarrow{ AF }+\overrightarrow{ AG }+\overrightarrow{ AH }$ if, $\overrightarrow{ AO }=2 \hat{ i }+3 \hat{ j }-4 \hat{ k }$