Three point particles P, Q, R move in a circl | vedclass

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(C) Let the angular positions at $t=0$ be $	heta_P(0) = \pi$, $	heta_Q(0) = \pi/2$, and $	heta_R(0) = 0$.The angular positions at time $t$ are:$	h

Vedclass · Accepted Answer

$1/2 \, s$

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(C) Let the angular positions at $t=0$ be $	heta_P(0) = \pi$, $	heta_Q(0) = \pi/2$, and $	heta_R(0) = 0$.The angular positions at time $t$ are:$	heta_P(t) = 5\pi t + \pi$$	heta_Q(t) = 2\pi t + \pi/2$$	heta_R(t) = 3\pi t + 0$For particles $P$ and $Q$ to be at the same position: $	heta_P(t) - 	heta_Q(t) = 2n\pi$ (where $n$ is an integer).$3\pi t + \pi/2 = 2n\pi \implies 3t = 2n - 1/2 \implies t = (4n-1)/6$.For $n=1, t = 3/6 = 1/2 \, s$. For $n=2, t = 7/6 \, s$. For $n=3, t = 11/6 \, s$.For particles $Q$ and $R$ to be at the same position: $	heta_Q(t) - 	heta_R(t) = 2m\pi$.$-\pi t + \pi/2 = 2m\pi \implies t = 1/2 - 2m$.For $m=0, t = 1/2 \, s$. For $m=-1, t = 2.5 \, s$.Comparing the times, they all meet at $t = 1/2 \, s$.

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