Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE, JEE/Main and JEE/Advance, UG NEET, AIIMS or any other Engineering and Medical entrance examination. All the questions in these books are of high level, which requires all basics to applied concept of physics. We here makes your task very easy, we have presented complete solution with detailed explanation step by step. The solution of these books teaches students also teachers in a suitable manner and then tests you with some tricky questions. To answer these questions you need to have thorough understanding of the concepts and this is where most students falter.
Problem: 1.12
Three points are located at the vertices of an equilateral triangle whose side equals \(a\). They all atart moving simultaneously with velocity \(v\) constant in modulus, with the first point heading continually for the second, the second for the third, and the third for the first. How soon will the points converge?
Solution: 1.12
While a very mathematical and rigorous treatment of the problem is possible however, it can be solved by simply exploiting the inherent symmetry in the problem. The first step is to understand that because of the inherent symmetry in the positions and motions of the three particles, their positions at any given time always form equilateral triangle as shown in figure.
Thus, the motion of particles will depict a shrinking equilateral triangle that is also rotating about the centroid as it shrinks. This is depicted in the figure.
The problem becomes quite simple if one solves it from the point of view of an observer sitting on one of the particles. To an observer sitting on one the particles, the rotational motion of the triangle will not be present (since the triangle is always an equilateral triangle at any given point of time) and he only perceive a shrinking equilateral triangle as shown in the second figure.
To an observer located on one of the particles the components of velocities of its adjacent particle at any given point of time are depicted in the third figure. Thus, any given particles feel that while it is moving towards its adjacent particle at a speed \(v\), the adjacent particle is moving towards it at a speed \(\frac{v}{2}\). Thus the particles seem to approach each other at a constant rate of \(\left( {v + \frac{v}{2}} \right) = \frac{{3v}}{2}\). Since the initial distance was \(a\), the time takes to meet is given by \(t = \frac{a}{{\frac{{3v}}{2}}} = \frac{{2a}}{{3v}}\).
Problem: 1.12
Three points are located at the vertices of an equilateral triangle whose side equals \(a\). They all atart moving simultaneously with velocity \(v\) constant in modulus, with the first point heading continually for the second, the second for the third, and the third for the first. How soon will the points converge?
Solution: 1.12
While a very mathematical and rigorous treatment of the problem is possible however, it can be solved by simply exploiting the inherent symmetry in the problem. The first step is to understand that because of the inherent symmetry in the positions and motions of the three particles, their positions at any given time always form equilateral triangle as shown in figure.
Thus, the motion of particles will depict a shrinking equilateral triangle that is also rotating about the centroid as it shrinks. This is depicted in the figure.
The problem becomes quite simple if one solves it from the point of view of an observer sitting on one of the particles. To an observer sitting on one the particles, the rotational motion of the triangle will not be present (since the triangle is always an equilateral triangle at any given point of time) and he only perceive a shrinking equilateral triangle as shown in the second figure.
To an observer located on one of the particles the components of velocities of its adjacent particle at any given point of time are depicted in the third figure. Thus, any given particles feel that while it is moving towards its adjacent particle at a speed \(v\), the adjacent particle is moving towards it at a speed \(\frac{v}{2}\). Thus the particles seem to approach each other at a constant rate of \(\left( {v + \frac{v}{2}} \right) = \frac{{3v}}{2}\). Since the initial distance was \(a\), the time takes to meet is given by \(t = \frac{a}{{\frac{{3v}}{2}}} = \frac{{2a}}{{3v}}\).
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