Find the value of Relative velocity of any two particles moving in a frame of reference.
Consider there are two particles A and B in a frame of reference and having velocities $\overrightarrow{\mathrm{V}}_{\mathrm{A}}$ and $\vec{V}_{B}$
The velocity of particle$ A$ with respect to $B$ is given by
$\vec{V}_{A B}=\vec{V}_{A}-\vec{V}_{B}$
The velocity of particle B w.r.t. A is given by
$\vec{V}_{B A}=\vec{V}_{B}-\vec{V}_{A}$
Thus, we can write,
$\vec{V}_{A B}=-\vec{V}_{B A} \text { and }\left|\vec{V}_{A B}\right|=\left|\vec{V}_{B A}\right|$
In general If $\mathrm{P}$ and ' $\mathrm{Q}^{\prime}$ are moving along with $\mathrm{X}$
then $\vec{V}_{P Q}=\vec{V} \mathrm{PX}+\vec{V} \times Q$
$\overrightarrow{\mathrm{V}} \mathrm{PQ}=\overrightarrow{\mathrm{V}}_{\mathrm{PX}}-\overrightarrow{\mathrm{V}} \mathrm{QX} \quad \ldots$ (3) $[\because \overrightarrow{\mathrm{V}} \mathrm{XQ}=-\overrightarrow{\mathrm{V}} \mathrm{QX}]$
A rigid rod is sliding. At some instant position of the rod is as shown in the figure. End $A$ has constant velocity $v_0$. At $t = 0, y = l$ .
A particle initially at rest is subjected to two forces. One is constant, the other is a retarding force proportional to the particle velocity. In the subsequent motion of the particle :
Two balls are thrown horizontally from the top of a tower with velocities $v_1$ and $v_2$ in opposite directions at the same time. After how much time the angle between velocities of balls becomes $90^o$ ?
A particle has initial velocity $\left( {2\hat i + 3\hat j} \right)$ and acceleration $\left( {0.3\hat i + 0.2\hat j} \right)$. The magnitude of velocity after $10\, seconds$ will be
$List I$ describes four systems, each with two particles $A$ and $B$ in relative motion as shown in figure. $List II$ gives possible magnitudes of then relative velocities (in $ms ^{-1}$ ) at time $t=\frac{\pi}{3} s$.
Which one of the following options is correct?