I just read an example of vectors in my book which is confusing me.

Three particles A,B and C are at the vertices of an equilateral trinagle ABC. Each of the particle moves with constant speed v. A always has its velocity along AB, B along BC and C along CA. They meet each other at the centroid. At any instant, the component of velocity of B along BA is $v\cos60^\circ$.

I don't understand how the particles meet at the centroid and why the component of velocity of B along BA is $v\cos 60^\circ$.

  • 1
    $\begingroup$ I think the book is trying to say that the particles themselves define the triangle, so as B moves along BC, the definition of AB and BC change. $\endgroup$ – Scott Lawrence Apr 25 '14 at 7:04

This question is exactly 11th class's question in Kinematics chapter. It is used in many books. The book in my hands is H.C Verma's Concept of physics and chapter-3 solved example-20.

Your answer: By symmetry points$A(t),B(t),C(t)$ will always make an equilateral triangle. Since the angle b/w $BC$ and $BA$ is always $60^0$ so the component of velocity of $B$ along $BA$ is always $v\cos60^0$.
All the triangles $A_1B_1C_1,A_2B_2C_3$ and $A_nB_nC_n$ are concentric. Hence at the end when $A$,$B$ and $C$ approach each other they form an infinitesmall triangle whose centroid is the same as that of the initial triangle $A_1B_1C_1$ and is forthcoming point of meeting of $A$,$B$ and $C$.

| cite | improve this answer | |
  • $\begingroup$ A variation often seen is the "bugs on a table" with 4 corners of a square. Along with "will they meet" is "If so, how far will they go"? In this case, since that are always normal to each other, the curve to the center is exactly the length of a side of the square. (The DEQ is much easier if you put them on the axis to start so it is a "diamond"). Give me the length of travel in this triangle case and I'll give your answer a vote :-) $\endgroup$ – C. Towne Springer Apr 25 '14 at 8:57
  • $\begingroup$ I was not commenting on the OPs question. You do that above, under the question. $\endgroup$ – C. Towne Springer Apr 26 '14 at 3:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.