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Three particles are arranged at the vertices of a equilateral triangle as shown in figure. They move with constant speed V in such a way that A always follows B, B always follow C, C always follows A. Find the distance travelled by A in one complete rotation around centroid of triangle

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a

2a31−e−23π

b

2a3e23π−1

c

2a3e−23π

d

2a3e23π

answer is A.

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Detailed Solution

drdt=−vcos⁡30---(1) rdθdt=vsin⁡30---(2)(1)÷(2)drrdθ=-vcos⁡30vsin⁡30drr=-3dθ∫r0rdrr=-3∫0θdθon solving r=r0e-3θFor one revolution, θ=2π r=r0e-3(2π)From(1)∫r0rdr=−∫0tvcos⁡30dtr-r0=-32vtt=2(r0-r)3v=t=2(r0-r0e-3(2π))3vd=vt=2r0(1-e-3(2π))3d=23×a3(1-e-3(2π))d=2a3(1-e-3(2π))

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Three particles are arranged at the vertices of a equilateral triangle as shown in figure. They move with constant speed V in such a way that A always follows B, B always follow C, C always follows A. Find the distance travelled by A in one complete rotation around centroid of triangle