I came across this problem in the book "Problems in General Physics by IE Irodov"-

Three points are located at the vertices of an equilateral triangle whose side equals s. They all start 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?

The problem is famously solved in the reference frame of a particle (say $A$). According to particle $A$, particle $B$ is approaching it with a constant relative speed of $(v\cos(\pi/3)+v)=3v/2,$ and since the initial side length of the triangle was "$s$", time taken by them to meet will be $t=2s/3v$.

But from Ground Frame, it is clear that the particles are accelerating (since particles follow the triangular spiral shown in the figure below).

enter image description here

And, since the whole system should be symmetric around the triangle, their acceleration vectors should be symmetric as well. So, how are we changing reference frames (from ground frame to particle $A$'s frame) without taking into account the acceleration vectors? For their acceleration vectors to cancel in Frame of Particle $A$, their magnitude and direction should be equal. But their acceleration vectors cannot possibly be unidirectional, since that will not be symmetric. This implies that particle $B$ should be accelerating with respect to particle $A$'s frame.

What I got so far: At any instant in time, the particles are at the vertices of an equilateral triangle, and instantaneously, any particle is in a circular motion about the centroid of the triangle with the radius of the circle decreasing with time. Thus, the acceleration vector of a particle at any instant in time will be directed towards the centroid of the triangle.

enter image description here

From the figure above, we get that the magnitude of acceleration vector at any time is $a=\frac{v_{tangential}^2}{r}=\frac{(v\sin(\pi/6))^2}{R-v\cos(\pi/6)t}$ where R is the initial distance of a vertex from the centroid of the triangle. Thus, at any instant in time, according to particle $A$, particle $B$ should have an acceleration of $|\overrightarrow{a_B}-\overrightarrow{a_A}|=2a\cos(\pi/6)=\sqrt3a$.

So, why according to particle $A$, particle $B$ is in uniform motion? What happens to their acceleration vectors while frames are changed? I might be missing a very obvious fact here, and if that is the case, please let me know.

  • $\begingroup$ I originally posted this on Physics SE here-physics.stackexchange.com/questions/582682/… $\endgroup$ Oct 2, 2020 at 11:48
  • $\begingroup$ I don't understand: if the velocities are constant, the acceleration is $O$, no ? $\endgroup$
    – Jean Marie
    Oct 2, 2020 at 12:00
  • $\begingroup$ That is the exactly the contradiction I'm having. If velocities are constant, acceleration should be zero but by approaching the problem another way, I get that the particles must be accelerating wrt each other. $\endgroup$ Oct 2, 2020 at 12:04
  • 1
    $\begingroup$ The velocity isn’t constant now is it? If it were it would be moving in a straight line! Remember, velocity is a vector. The direction matters. I believe your confusion stems from a misunderstanding of uniform circular motion $\endgroup$ Oct 2, 2020 at 13:11
  • $\begingroup$ The velocity $v$ is constant only in modulus and so the acceleration is obviously not zero. So I do not see the contradiction. For the hypothetical "independent" particle sitting at the centroid, I do not think we can look at A and B in isolation. So I am not sure your argument (1) holds. $\endgroup$
    – Math Lover
    Oct 2, 2020 at 13:11

1 Answer 1


The issue in this interesting problem is that the acceleration vector does not have a single component, but two components. In particular, as shown below, as a result of this double component the acceleration vector is perpendicular to the velocity vector. It is known that the perpendicular directions of the acceleration and velocity vectors represent the typical situation occurring whenever velocity is a constant. This explains the apparent contradiction described in the OP.

As correctly noted, the scenario of this problem resembles that of a uniform circular motion, i.e. a type of motion where an object moves among a circular path with a constant speed. Differently from one-dimensional problems, where objects with constant velocity have zero acceleration, in 2D or 3D problems an object can have acceleration if it motion follows a curved trajectory. This is the case of uniform circular motion, in which a particle with constant velocity $v$ moving on a circular trajectory with radius $R$ is subjected to a centripetal acceleration with magnitude $v^2/R$, directed along the radial director towards the center of the circle. The role of centripetal acceleration is to change the direction of the velocity vector, so that the motion remains tangential to the path.

However, in the problem described by the OP, we are not dealing with a true uniform circular motion. The similarity comes from the fact that, at any point in the path of the particle, we rotate and scale the equilateral triangle to reproduce the initial one. Actually the three points do not travel a circular path, but rather a spiral. Therefore, another acceleration component must necessarily exist that transforms the circular path into a spiral.

To better illustrate this, it is easier to use the polar coordinate system. The fundamental components of this system are the unit radial vector $\hat {\textbf{r}}$ and the unit tangential vector $\hat{\boldsymbol{\theta}}$. In our case of spiral path, we have that the velocity vector results from a radial component (forming an angle of $5\pi/6$ with $\textbf{v}$) and a tangential component (forming an angle of $\pi/3$ with $\textbf{v}$). Here is a picture:

enter image description here

Assuming that, for the radial vector, positive values are directed externally, the velocity vector $\textbf{v}=v$ is given by

$$\textbf{v}=v \cdot \cos\left(\frac{5\pi}{6}\right) \hat {\textbf{r}}+v \cdot \cos\left(\frac{\pi}{3}\right) \hat{\boldsymbol{\theta}}\\ = - \frac{v\,\sqrt{3}}{2} \hat {\textbf{r}}+ \frac{v}{2} \hat{\boldsymbol{\theta}} $$

Note that, using the standard dot notation for time derivatives, the coefficients of the last equation satisfy the relations $-v\sqrt{3}/2=\dot{r}$ and $v/2=r\,\dot{\theta}$. This last relation can also be written as $\dot{\theta}=v/(2r)$, and will be used in the next steps.

The acceleration vector is obtained by differentiating the velocity equation:

$$\textbf{a}= - \frac{v\, \sqrt{3}}{2}\, \dot {\hat{\textbf{r}}}+ \frac{v}{2} \dot{\hat{\boldsymbol{\theta}}} $$

Since it is known that the derivative of the radial and tangential vector can be expressed as $\dot {\hat{\textbf{r}}}=\dot{\theta} \hat{\boldsymbol{\theta}}$ and as $\dot {\hat{\boldsymbol{\theta}}}=-\dot{\theta} \hat{\boldsymbol{r}}$, respectively, we get

$$\textbf{a}= - \frac{v}{2} \dot{\theta} \hat{\boldsymbol{r}} - \frac{v\, \sqrt{3}}{2}\, \dot{\theta} \hat{\boldsymbol{\theta}} $$

and substituting $\dot{\theta}=v/(2r)$ we finally obtain

$$\textbf{a}= - \frac{v^2}{4r} \hat{\boldsymbol{r}} - \frac{v^2\, \sqrt{3}}{4r}\, \hat{\boldsymbol{\theta}} $$

The figure below shows the acceleration vector, whose magnitude is $v^2/(2r)$:

enter image description here

It is also clear that, if we consider two of the three particles and try to describe the motion of one of them from the others' point of view, the resulting vectors are still perpendicular.

In conclusion, the scenario of the OP is characterized by moving points whose acceleration vectors are perpendicular to the corresponding velocity vectors. As stated at the beginning of this answer, this is a classical situation where velocity is constant, and this explains why, according to any of the three particles and from its frame, the other two are in uniform motion. Just to provide a very intuitive and simplified scenario that illustrates well the situation of a constant speed with velocity vector perpendicular to the acceleration vector, we can think of an observer stationary at the center of an Archimedean spiral and a second observer who travels the spiral at a constant speed, so that the distance between them decreases linearly. From the point of view of the stationary observer, the running observer has a constant speed and the distance between them decreases uniformly, although the running observer has not zero acceleration.

  • $\begingroup$ So, in a nutshell, in reference frame of $A$, particle $B$ has acceleration perpendicular to the line joining $A$ and $B$ which does not change its relative velocity modulus ($3v/2$) but it changes its direction making it rotate around $A$? $\endgroup$ Oct 3, 2020 at 22:55
  • $\begingroup$ Yes, this is a good way to summarize what happens. $\endgroup$
    – Anatoly
    Oct 3, 2020 at 22:59
  • $\begingroup$ But in $A$'s frame, $B$ also have a constant perpendicular velocity ($v\sin(\pi/3$)). Why doesn't the acceleration vector (unidirectional to $v\sin(\pi/3$)) increase that velocity but instead, change the direction of the velocity parallel to line joining the two? I mean, since this is a particularly physical situation, we can tell that it should do what it is doing but how to pin point a reason for any general case? (I apologize for asking stupid questions) $\endgroup$ Oct 3, 2020 at 23:02
  • $\begingroup$ And also, I plotted the trajectories of the particles in the cartesian coordinates-desmos.com/calculator/cbvl6ynwet. But after double differentiating it, I got that the acceleration vector (of the particle initially at (1,0)) is always pointing towards the centroid, which is clearly wrong. What did I do wrong? $\endgroup$ Oct 3, 2020 at 23:14
  • $\begingroup$ Looking at single components is often misleading. We have to look at the whole system. In A's frame, the motion of B results from the combination of their velocities and accelerations, which finally gives perpendicular vectors. The normal directions of velocity and acceleration is the reason why acceleration changes the direction of the velocity (and not its modulus), just as typically occurs in uniform circular motion. $\endgroup$
    – Anatoly
    Oct 3, 2020 at 23:30

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