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

Asked by sunarmik das | 26th Jun, 2011, 02:57: AM

Expert Answer:

                                                                                                   

Figure above shows the initial direction of velocities of A, B and C i.e. when A, B,C are at the ends of an equilateral triangle of side 'a' each.

The velocity of A is along AB. So, as B moves, then the direction of velocity of A changes continuously towards the direction of B.

Similarly, the velocity of B is along BC and velocity of C is along CA.

The component of velocity of B along BA = v cos60o =

So, the effective speed with which A and B approach each other 

 Time taken in reducing the separation AB from 'a' to zero is 

Since all three particles A, B and C move with same constant speed v, hence C will also meet A and B at the same time.

 

 

Answered by  | 8th Jul, 2011, 12:08: PM

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