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three particles A,B and C are situated at the vertices of an equilateral triangle ABC of side d at 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 ?

three particles A,B and C are situated at the vertices of an equilateral triangle ABC of side d at 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 ?

Grade:11

7 Answers

pragya yadav
10 Points
8 years ago

if they are moving with same velocities then i think they will never meet 

Ayush Agrawal
39 Points
8 years ago

They will meet after 3v/2 seconds .... where "v" is the constant speed of all the three!

Prasenjit Dubey
17 Points
8 years ago

the particle will meet at the centroid of the equilateral triangle whose distace fro any of the vertices is equal to d=a/(root 3).(a+ side of triangle. hence time=d/v.

pracheesh singh
26 Points
8 years ago

velocity of A is v along AB 

the velocity of B is along BC . its component along BA is vcos60 = v/2. thus the separation AB decreases at the rate

 

v+v/2=3v/2

sice this rate is constt. the time taken in reducing the separation AB from d to 0 is 

 

t=d/(3v/2) =2d/3v

SOURAV MISHRA
37 Points
8 years ago

this question can be solved by two methods.

first note that the particles collide at the centroid of the triangle.

for this the particles have to cover a distance of d/√3.

and at any instant of time the speeds of the particles is same as v and they preserve the equilateral triangle shape till they collide.

the component of the speed of the particle along this distance is (√3/2)*v

so the time taken to collide is 2d/3v.

in the second method we look at one of the particles from another one. the relative speed is constant and equals √3v/2.

this is inclined at 30º to the line joining the particles always.

the component of this speed along this line is 3v/2.

in order to collide the distance covered by one particle relative to the other is d and this is along the line joining them.

so the time taken is 2d/3v. 

Vinayk Gupta
21 Points
5 years ago
 a particle moves over the sides of an equilateral triangle of sides l with constant speed v as shown in figure.Then 1.What is the magnitude of average velocity from A to C? 2.What is the magnitude of average acceleration from A to C?
 
Ishaan
14 Points
4 years ago
All the three particles will meet at centroid and value of centroid is a, of an equilateral triangle. The velocity of particle A will be divided into two components i.e. vcos60 and vsin60.V=v+v cos 60V=v+v/2V=3v/2Time =tS =Distance travelled by A= that by B= that by C = centroid=aV=s/tt=s/vt=a÷3v/2t= 2a/3v

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