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28 Mar 2007 13:45:46 IST

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HC VERMA KINEMATICS Q

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6 particles are located at the 6 corners of a regular hexagon of side a. If each moves with a constant velocity of v always towards the particle at the next corner then find the time taken for the particles to meet. (ANS= 2a/v)

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Priyadharshini Venkataraman

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Joined: 9 Mar 2007

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28 Mar 2007 13:56:13 IST

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if u see ex20 ull get the idea

nikhil handa

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29 Mar 2007 16:56:08 IST

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in this Q
first find the angle by using (n-2)180/n with this we shall get it to be 120.
now as it is a regular hexagon join any two adjacent points to the centre.the lines joining bisects the angles, thus making it an equilateral triangle.
now vcos is the component on the line i.e. v/2
length to be covered is a
therefore 2a/v.
plsss rate me

deepak_agarwal

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29 Mar 2007 17:52:59 IST

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due to symmetry we can conclude that the particle meets at centroid...this can be visualized easily...now along the hexaagon side velocity is v so make its components alon x and y axis as...v/2 and root(3)v/2 since angle hexagon forms with x axis is 60....bas now length of one side is a and relative velocity along it is v/2...hence...so is the aans

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