3 particles A,B and C are situated at the vertices..

Question

I had this question

"3 particles A,B and C are situated at the vertices of an equilateral triangle ABC of side d at t=0. Each particle moves with a constant speed v. A always has its velocity along AB, B along BC and C along CA. At what time will the particles meet ?"

In this my book says that the distance travelled by the particle let's say A (it will be the same for all three) will travel a distance AO ( O being the centroid where all three will meet) .

In my book value of AO is given to be

AO = $\frac{2}{3} \sqrt{d^2 - (\frac{d}{2})^2}$ = $\frac{d}{\sqrt{3}}$

The only thing I couldn't understand was that how did expression come AO = $\frac{2}{3} \sqrt{d^2 - (\frac{d}{2})^2}$

Can somebody please give me an explanation for this . Regards

The question isn't making much sense. Are you sure you have posted it correctly? — Vizag, May 24 '19 at 11:16
@Vizag If the particles are really moving such that the speed vector of A points towards B and so on, then they will travel a distance greater than the straight-line length of AO. — Michal Adamaszek, May 24 '19 at 11:20
Your question is how the distance traveled is AO, or how $AO = \frac{2}{3} \sqrt{d^2 - (\frac{d}{2})^2}$, if it is the second one, that is standard property of equilateral triangle. — novice_2, May 24 '19 at 11:51
Try searching for "pursuit curve" and "equilateral triangle". — Somos, May 24 '19 at 12:04

score 0 · Accepted Answer · answered May 24 '19 at 19:06

I'm not sure where that expression came from either. But if you look at:

You'll notice that $AOE$ is a right angled 30-60-90 triangle so $AO=\frac{2}{3}\sqrt{3}AE=\frac{1}{\sqrt{3}}AC$ which is what you wanted.

I guess I'll add this for completeness:

To finish off the question, we note that if the particles were travelling in straight lines, the distance to the center of the equilateral triangle becomes $$\sqrt{(d-v\Delta t)^2+\frac{1}{3}d^2}$$ so $\frac{\mathrm d}{\mathrm dt}(\text{distance to center})=-\frac{\sqrt{3}}2$. Since the distance to the center is $\frac{\sqrt{3}}{2}$ of the sidelength $d$, we know that the time taken is $\cfrac{3}{2}\cfrac{s}{v}$ for particles of speed $v$ starting at distance $s$ from each other.

3 particles A,B and C are situated at the vertices..

1 Answers1