The bugs are at the corners of a regular polygon with N sides of length L. Each bug walks directly towards the next bug and the bugs velocity is v. How long does it take for the bugs to meet?

Also solve for v=1.0 cm/s, L=30 cm and N = 3 (an equilateral triangle).

###### Solution.

Let’s consider the general case.

Please look on the picture below.

The formula of the angle α for the regular polygon is α = 2π/N.

The velocity of the bug at the left is v and the component of the velocity of the bug at the right towards the same direction is v * cos(α) .

Then the relative velocity will be v – v * cos(α) .

Now, we should use the following formula t = L/V , where L – the length, t – time and V – the relative velocity.

The final formula will be

t = L / ( v – v *cos (2π/N) )

Let’s calculate

t = 30 / ( 1 – 1 *cos (2π/3) ) = 20 s

###### Animation.

GIF Animation Credit: math.stackexchange.com

Without making any approximation about neglecting the effect of friction, I tend to disagree with the answer here…First of all, each bug in that diagram seems to be resisted by different values of frictional force as they struggle to overcome forces along their individual path. Secondly, the direction of air drag might also have effect on at least two of the bugs…. Until it can be assumed that friction is negligible, I think further analysis is still required 😀

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