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 ?

kirti nidhi, 10 years ago

Grade:11

7 Answers

pragya yadav

10 Points

10 years ago

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

Ayush Agrawal

39 Points

10 years ago

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

Disapproved

Prasenjit Dubey

17 Points

10 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.

Disapproved

pracheesh singh

26 Points

10 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

Approved

SOURAV MISHRA

37 Points

10 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

7 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

6 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