Question

Three particles A, B and C are situated at the vertices of an equilateral triangle ABC of side d at time `t=0.` Each of the particles moves with constant speed v. A always has its velocity along AB, B along BC and C along CA. At what time will the particles meet each other?

Answer 1

`1st` method. In 2 (a) 45, the velocity of (A) is along (AB) The velocity of (B) is along (BC) The component velocity of (B) along ` BA= v cos 60^@ = v//2 `. Thus the separation of partivles at (A) and (B) dereases with the speed` = v+ v//2 = 3 v//2 `.
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Since, this speed is constant, the time taken in reducing the separation ` AB` from ` (A) to zero is, ltbRgt ` =a// ( 3 v//2) = 2 a//3 v`
` 2nd` method. The motion of three particles at ` A` , B` and ` C` is shown in Fig. 2 (a) 46` by solid curved lines . They will be meeting at the centroid (O) of the triangle ` ABC`. It means a particle at (A) will be covertin a distance ` AO` before meeting the other particle. The velocity of particele at (A) along
` AO = v cos 30^@ = v sqrt 3 //2`
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distance, ` AO( = 2/3 AD = 2/3 sqrt( a^2 -a^2 //4) = a/ (sqrt3)`
Therefore, the time taken by particle at (A) to go from (A) to (O)
` (AO)/ ( v cos 30^@) = ( a //sqrt 3)/( v sqrt 3 //2 ) = ( 2a)/ (3v)`.