Four persons P, Q, R and S are initially at t | vedclass

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(B) Consider the motion of person $P$ towards $Q$. The velocity of $P$ is $v$ directed towards $Q$. The velocity of $Q$ is $v$ directed towards $R$. S

Vedclass · Accepted Answer

$\frac{d}{v}$

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(B) Consider the motion of person $P$ towards $Q$. The velocity of $P$ is $v$ directed towards $Q$. The velocity of $Q$ is $v$ directed towards $R$. Since $PQ$ and $QR$ are perpendicular,the angle between the velocity of $P$ and the velocity of $Q$ is $90^{\circ}$. The relative velocity of $P$ with respect to $Q$ along the line $PQ$ is given by $v_{rel} = v - v \cos(90^{\circ})$. Since $\cos(90^{\circ}) = 0$,we have $v_{rel} = v - 0 = v$. The initial separation between $P$ and $Q$ is $d$. Therefore,the time taken for them to meet is $T = \frac{	ext{distance}}{	ext{relative velocity}} = \frac{d}{v}$.

Four persons $P, Q, R$ and $S$ are initially at the four corners of a square of side $d$. Each person now moves with a constant speed $v$ in such a way that $P$ always moves directly towards $Q$,$Q$ towards $R$,$R$ towards $S$,and $S$ towards $P$. The four persons will meet after time:

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