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(A) Let the centre of the circle be $O$. The particle at $O$ moves towards $P$ with velocity $\vec{V}_1$ along the radius $OP$. The distance covered i

Vedclass · Accepted Answer

$	an \frac{\phi}{2} = \cot 	heta$

Vedclass · Answer

(A) Let the centre of the circle be $O$. The particle at $O$ moves towards $P$ with velocity $\vec{V}_1$ along the radius $OP$. The distance covered is $R$,so $t = \frac{R}{V_1}$.The particle at $Q$ moves towards $P$ with velocity $\vec{V}_2$. The distance $QP$ can be found using the triangle $OQP$. Since $OQ = OP = R$ and $\angle QOP = \phi$,the triangle $OQP$ is isosceles. The length $QP = 2R \sin(\frac{\phi}{2})$.Since both reach $P$ at the same time $t$,$t = \frac{QP}{V_2} = \frac{2R \sin(\frac{\phi}{2})}{V_2}$.Equating the times: $\frac{R}{V_1} = \frac{2R \sin(\frac{\phi}{2})}{V_2} \implies \frac{V_2}{V_1} = 2 \sin(\frac{\phi}{2})$.From the geometry of the figure,the angle between $\vec{V}_1$ and $\vec{V}_2$ is $	heta$. The velocity $\vec{V}_2$ makes an angle with the chord $QP$. By analyzing the triangle formed by the velocity vectors,we find that the angle between the direction of $QP$ and $OP$ is $\frac{\pi - \phi}{2} = 90^{\circ} - \frac{\phi}{2}$.Given the geometry,the angle $	heta$ between $\vec{V}_1$ and $\vec{V}_2$ relates to the isosceles triangle properties such that $	heta = 90^{\circ} - \frac{\phi}{2}$.Thus,$\frac{\phi}{2} = 90^{\circ} - 	heta$.Taking the tangent on both sides: $	an(\frac{\phi}{2}) = 	an(90^{\circ} - 	heta) = \cot 	heta$.

Column $I$	Column $II$
$(A)$ Displacement	$(p)$ $8 \sin 2$
$(B)$ Distance	$(q)$ $4$
$(C)$ Average velocity	$(r)$ $2 \sin 2$
$(D)$ Average acceleration	$(s)$ $4 \sin 2$

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