Three particles `A, B` and `C` situated at vertices of an equilateral triangle, all moving with same constant speed such that A always move towards `B, B` always towards `C` and `C` always towards `A`. Initial seperation between each of the particle is a. `O` is the centroid of the triangle. Distance covered by particle A when it completes one revolution around `O` is

Options

A. `2a(1-e^(-2sqrt(3pi)))`
B. `(2a)/(3)(1-e^(-2sqrt(3pi)))`
C. `a(1+e^(-2sqrt(3pi)))`
D. `(2a)/(3)(1-e^(-sqrt(3pi)))`

Answer

B

Explanation

`(dr)/(dt)=-v cos 30^(@)=-(sqrt(3))/(2)V`
`r(dtheta)/(dt)=v sin 30^(@)=sqrt(2)`
`(1)/(r) (dr)/(dtheta)=-sqrt(3)`
`underset(r_(0))overset(r)int(dr)/(r)=-sqrt(3)underset(0)overset(theta)intrArr r=r_(0)e^(-sqrt(30))`
When A completes are revolution `theta=2pi`
Time taken `t=(r_(0)(1-e^(-2sqrt(3pi))))/(sqrt(3)v//2)`
Distance travelled `D=vt=(2r_(0))/(sqrt(3))(1-e^(-2sqrt(3pi)))`
`D=(2a)/(3)(1-e^(-2sqrt(3pi)))`