Euler's Run, prove this?

Question

An elastic track on the ground is of length 1, fixed at the start position, the other end pulled by a tractor so that the track stretches uniformly.  At time = 0, a runner is at the start position, and begins running at constant speed v on the track.  The tractor also begins pulling at constant speed v.  Prove that when the runner reaches the tractor, the track is of length e, or Euler's number.

? · Accepted Answer

The runner is running with constant speed v with respect to the elastic track whose tip attached to the tractor is moving at a constant rate of v.

Therefore the absolute speed of the runner is the sum of its relative speed (with respect to the track) plus the speed of the point the runner has reached at time t.

i.e.  dx/dt = v + v1

Now v1 / v  = x / L  , where L is the length of the total track at time t.

Thus v1 = v x / (v t + 1)

Thus dx/dt = v + v x / (v t + 1)

In standard form, the differential equation becomes

dx/dt - (v / (v t + 1) ) x = v

Using integrating factor of exp( integral (- v / (v t + 1) ) ) = exp ( - ln (vt + 1) ) = 1 / ( v t + 1)

we get  d/dt ( (1/(vt+1)  x ) = v (1/ (vt + 1)

Therefore integrating both sides, we get

x / ( v t + 1) = ln ( v t + 1)

i.e. x = (vt + 1) ln (v t + 1)

since L = vt + 1 , then when x = L we must have ln (vt + 1) = 1 that is (v t +1 ) = e

but ( v t + 1) = L, so L = e.

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Euler's Run, prove this?

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