This week the #puzzle is: Can You Race Against Zeno? #integration #summation #speed #distance #time (Link at the bottom.)

I’ve been experimenting with different strategies in 5000-meter races (known as “5K”s). If I run the distance at a steady pace, I’ll finish in precisely 23 minutes.

However, I tend to find a burst of energy as I near the finish line. Therefore, I’ve tried intentionally running what’s called a “negative split,” meaning I run the second half of the race faster than the first half—as opposed to a “positive split,” where I run a slower second half.

I want to take the concept of a negative split to the next level. My plan for an upcoming race—the “Zeno Paradox 5K”—is to start out with a 24-minute pace (i.e., running at a speed such that if I ran the whole distance at that speed, I’d finish in 24 minutes). Halfway through the race by distance (i.e., after 2500 meters), I’ll increase my speed (i.e., distance per unit time) by 10 percent. Three-quarters of the way through, I’ll increase by another 10 percent. If you’re keeping track, that’s now 21 percent faster than my speed at the start.

I continue in this fashion, upping my speed by 10% every time I’m half the distance to the finish line from my previous change in pace. (Let’s put aside the fact that my speed will have surpassed the speed of light somewhere near the finish line.)

Using this strategy, how long will it take me to complete the 5K? I’m really hoping it’s faster than my steady 23-minute pace, even though I start out slower (at a 24-minute pace).

And for extra credit:

I still want to run a negative split, but upping my tempo in discrete steps is such a slog. Instead, my next plan is to continuously increase my pace in the following way:

At the beginning of the race, I’ll start with a 24-minute pace. Then, wherever I am on the race course, I’m always running at a 10 percent faster speed than I was when I was twice as far from the finish line. Also, my speed should always be increasing in a continuous and smooth fashion.

(On the off chance you find more than one speed as a function of distance that satisfies all these conditions, I’m interested in seeing how. If only one such function exists, can you prove it? Regardless, my expectation is for the function to be as straightforward as possible.)

Using this strategy, how long will it take me to complete the 5K? Once again, I’m hoping it’s faster than my steady 23-minute pace, even though I started out slower.

Highlight to reveal (possibly incorrect) solution:

Desmos 1 – just to be confusing, the definition of t_Zeno1 here is slightly different.

Let’s look at what we know or define:

• d: the full distance (5 km – all we metric centric people thank you)
• t_init: the time spent running with the initial speed all the way (24 m)
• v₀: the initial speed (d/t_init) (also, t_init = d/v₀)
• a: the increase of speed as a factor (1.1)
• t_Zeno1: the time spent running under the new rule

With these definitions in place, let’s go running.

• Run at v₀ to cover the distance d/2^1. (Half of the distance is run at the initial speed.)
• Then run at v₀ * a to cover the distance d/2². (Half of the remaining distance is run after 1 speed increase.)
• Then run at v₀ * a² to cover the distance d/2³.
• Then run at v₀ * a³ to cover the distance d/2⁴.
• In general run at v₀ * aⁱ to cover the distance d/2ⁱ⁺¹.
• In general spend t_i to cover distance d/2ⁱ⁺¹.
• t_i = d/2ⁱ⁺¹ / (v₀ * aⁱ)
• t_i = (d/v₀) / (2ⁱ⁺¹ * aⁱ)
• t_i = t_init / (2ⁱ⁺¹ * aⁱ)
• t_Zeno1 = Σ_i=0 t_i
• t_Zeno1 = Σ_i=0 t_init / (2ⁱ⁺¹ * aⁱ)
• t_Zeno1 = t_init * Σ_i=0 1 / (2ⁱ⁺¹ * aⁱ)
• To get a good approximation, I can let the upper bound be finite, but growing.
• According to Desmos, t_Zeno1 = 22 m.

But seeing I am so close, maybe I can do this in a better way.

• The general formula for v can be written as
• v(t) = a^{(⌊-log₂(1-t)⌋)} * v₀
• Some examples:
  • v(0) = a^{(⌊-log₂(1-0)⌋)} = a^{(⌊-log₂(1)⌋)} = a^(⌊0⌋) = 1 * v₀
  • v(0.4) = a^{(⌊-log₂(1-0.4)⌋)} = a^{(⌊-log₂(0.6)⌋)} = a⁰ = 1 * v₀
  • v(0.5) = a^{(⌊-log₂(1-0.5)⌋)} = a^{(⌊-log₂(0.5)⌋)} = a^{(⌊-(-1)⌋)} = a * v₀
• Instead of the sum I can use an integral.
• t_Zeno1 = t_init * ∫ ₀¹ 1 / (v(t)/v₀) dd
• Because, for a small distance d_j the contribution to the time is:
• t_j = d_j / v(t_j)
• t_Zeno1 = Σ_{all j} t_j
• t_Zeno1 = Σ_{all j} d_j / v(t_j)
• t_Zeno1 = Σ_{all j} d * d_j/d / v(t_j)
• t_Zeno1 = d * Σ_{all j} d_j/d / v(t_j)
• t_Zeno1 = d * ∫ ₀¹ 1 / v(t) dd
• t_Zeno1 = d * ∫ ₀¹ 1 / (a^{(⌊-log₂(1-t)⌋)} * v₀) dd
• t_Zeno1 = d/v₀ * ∫ ₀¹ 1 / a^{(⌊-log₂(1-t)⌋)} dd
• t_Zeno1 = t_init * ∫ ₀¹ 1 / a^{(⌊-log₂(1-t)⌋)} dd
• qed
• Again, desmos calculates this as the integral being 0.916666666667, therefore t_Zeno1 = 24 m * 0.916666666667 = 22 m.

And for extra credit:

Desmos 2

Again, let’s look at what we know:

• v(t) is the speed at any given time.
• v(0) = v(1-1) = v₀ above.
• v(0.5) = v(1-1/2¹) = v₀ * a
• v(0.75) = v(1-1/2²) = v₀ * a²
• v(0.875) = v(1-1/2³) = v₀ * a³
• In general v(1-1/2ⁱ) = v₀ * aⁱ
• And v(t) is a smooth function.
• This calls for a double log system!
• Let’s try simply plugging in some numbers first in Desmos. Looks good.
• Then let’s try to describe this function.
• v(t) = v_opp(1-t) (running the opposite way)
• v_opp(1/2ⁿ) = v₀ * aⁿ
• v_opp(log₂(t)) = v₀ * a^-log₂(t)
• v(t) = v_opp(1-t) = v₀ * a^-log₂(1-t)
• Double log confirms, just writing it as v(t) = v₀ * (1-t)^-0.137504 instead.
• t_Zeno2 = t_init * ∫ ₀¹ 1 / a^-log₂(1-t) dd
• The value of the integral is 0.879118155787, therefore t_Zeno1 = 24 * 0.879118155787 = 21.0988.

***

Can You Race Against Zeno?