2

This question is inspired by the Dichotomy Paradox but with a twist: Let's say that Telemachus is running between two light posts, distance L length units apart. He starts at the first light post with an initial velocity of V0. When he reaches L/2, half the distance between the light posts, he doubles the speed (=2*V0). When the distance between Telemachus and the light post is L/4 – that is: when he has run L/2 + L/4 = 3/4 L length units – he doubles the speed once again. Telemachus's speed is now 22*V0.

In other words: whenever the remaining distance between Telemachus and the light post is L/2n the speed is doubled. Ignoring special relativity so that the speed tends towards infinity.

When does Telemachus reach his final destination?

/V.Vocor Undergraduate, Engineering Physics

V.Vocor
  • 23

2 Answers2

1

Let $t_0=\frac{L}{2V_0}$ the time need to reach the distance $\frac{L}{2}$. The next step the distance is divided by $2$ and the velocity multiplied by $2$, so the time is divided by $4$, and the same for each subsequent step.

So the time needed to reach $L$ is $$\sum_{i=0}^{\infty}\frac{t_0}{4^i}=\frac{4t_0}{3}=\frac{2L}{3V_0} $$

Xoff
  • 10,310
0

This is a weaker form of the paradox, because the original paradox holds when the speed is held constant. To resolve such a paradox, regardless of speed, one has to consider the infinite series (sum) of times to reach destinations, instead of focusing on one destination at a time on and on forever. Then you realize that the infinite series is convergent, so the time to reach the final destination is constant. In your case, the infinite series is equivalent to

$$ 1/2 + 1/8 + \ldots + 1/2^{2n+1} + \ldots$$

which reaches a finite value because

$$1/2 + 1/4 + 1/8 + \ldots + 1/2^n + \ldots$$

reaches a finite value (1), and the first series is strictly smaller.

user2566092
  • 26,142