Just Keep Moving

It’s always so beautiful when a problem which seems to be so arcane, and appears to require advanced techniques such as differentiation at the very least, breathes new life in a simplistic and sleek solution. This is one such problem.

Consider an equilateral triangle of side length 1m with 3 insects sitting at the vertices. These insects have a speed of 1/3 ms-1, and all are pointed in the direction of the insect clockwise to it, and begin moving with their constant speed of 1/3 ms-1, however they maintain their direction as directly pointing towards the next insect in the triangle. How long do they take to meet, and what distance do they travel? Do take your time and think and try this one through before scrolling below this picture, where you’ll see the solution. The picture should give you an idea of how the situation unfolds.

The first basic observation we can make is that whatever happens in symmetric, and so rotation by 120° will keep it the same at any time, and therefore we know they will all meet at the center, as that is the only point where this holds true for the endpoint.  Each insect is moving towards each other one with some vector component of its speed, and so let’s try and see what happens if we try and delve into that concept of relative velocities.

B is moving toward A with a vector speed of 1/3 ms-1  and the angle BAC is 60° clearly.  What is the vector component of A’s speed in the direction of B?  Let’s draw a triangle.

Screen Shot 2016-02-20 at 3.28.23 PM.png

PQ is the vector of A’s speed, in the direction of C, and therefore QR would be the vector of A’s speed in the direction of B, which would by simple trigonometry be PQ x cos 60. This gives the vector PQ/2, which is half the vector speed of A towards C, which gives a vector of (1/3)/2 ms-1 , 1/6 ms-1 . B is moving towards A at 1/3 ms-1  and A is moving at a relative velocity towards B at 1/6 ms-1 .  Essentially, since these two insects are moving towards each other, they are coming together at a rate 1/3 + 1/6, so 1/2 ms-1 . We know that distance AB is 1m, as the side length of the triangle is 1m, so the time for A and B to meet is distance/speed which is:

Time = D/S

Time = 1m / (1/2 ms-1 ) = 2 seconds

So the time we get is 2 seconds.  Now what about the distance.  Thinking about the arc and path that these insects follow, anyone can get a little disillusioned about this.  But we know these insects have constant speed.  And we know the time.  What is distance? Speed x time.

Distance = S x T

Distance = (1/3 ms-1 ) x 2 s = 2/3 m

I can’t imagine anyone ever saying that this problem looked easier than it actually is.  And this is why we like doing math.


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