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Wizard
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May 10th, 2019 at 4:56:56 PM permalink
Nathan and Miplet have a unicycle race. Each has a uniform acceleration from a standing start.

Nathan covers the last 1/5 of the distance in 4 minutes. Miplet covers the last 1/4 of the distance in 5 minutes. Who won?

For the beer, I'd like to not just see a common sense answer but something showing the acceleration of both, for which you may assume a total distance of 1.

24-hour hold rule applies.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Gialmere
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May 10th, 2019 at 5:07:55 PM permalink
Round or square wheels?
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Wizard
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May 10th, 2019 at 5:11:16 PM permalink
Quote: Gialmere

Round or square wheels?



Good one!
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SkittleCar1
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May 10th, 2019 at 6:43:59 PM permalink
But can you pop a wheelie?
Wizard
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May 10th, 2019 at 10:55:51 PM permalink
Quote: SkittleCar1

But can you pop a wheelie?



Not on a unicycle, without breaking the law of gravity.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
DogHand
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May 11th, 2019 at 8:41:19 AM permalink
Wizard,
Here is my answer.



For constant acceleration a, if the initial velocity is 0, the distance traveled x is
x = a*t²
The time to reach a given distance x is given by t = √(x/a).
In the race, both N and M cover the same distance d (assumed to be 1 mile), so whoever requires less time is the winner.
Each contestant covers the final fraction f of the total distance in a time of ∆t.
Thus, ∆t = (time to reach d) - (time to reach (1-f)*d)
∆t = √(d/a) - √(d*(1-f)/a))
∆t = √(d/a)*(1 - √(1-f))
Solving for the constant acceleration a gives
a = d*(1 - √(1-f))²/(∆t)²
N covers the last f_N = 1/5 of the distance in ∆t_N = 4 minutes, so
a_N = (1 mile)*(1 - √(1-0.2))²/(4 min)² = 0.0006966... miles/min²
M covers the last f_M = 1/4 of the distance in ∆t_M = 5 minutes, so
a_M = (1 mile)*(1 - √(1-0.25))²/(5 min)² = 0.0007179... miles/min²
The time each requires to finish the distance d is
t_N = √(d/a_N) = √((1 mile)/(0.0006966... miles/min²)) = 37.88... min
t_M = √(d/a_M) = √((1 mile)/(0.0007179... miles/min²)) = 37.32... min
Therefore, M wins by just over a half minute.



Dog Hand

Edit: fixed a typo in spoiler.
Last edited by: DogHand on May 11, 2019
MichaelBluejay
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May 11th, 2019 at 10:38:29 AM permalink
Quote: Wizard

Nathan and Miplet have a unicycle race. Each has a uniform acceleration from a standing start. Nathan covers the last 1/5 of the distance in 4 minutes. Miplet covers the last 1/4 of the distance in 5 minutes.

How do you know all this?
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Wizard
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May 11th, 2019 at 11:42:21 AM permalink
Quote: MichaelBluejay

How do you know all this?



The Great and Powerful Wizard knows all!
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Nathan
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May 11th, 2019 at 12:07:18 PM permalink
Quote: Wizard

The Great and Powerful Wizard knows all!



Hey, a Wizard Of Oz reference! 😁
In both The Hunger Games and in gambling, may the odds be ever in your favor. :D "Man Babes" #AxelFabulous "Olive oil is processed but it only has one ingredient, olive oil."-Even Bob, March 27/28th. :D The 2 year war is over! Woo-hoo! :D I sometimes speak in metaphors. ;) Remember this. ;) Crack the code. :D 8.9.13.25.14.1.13.5.9.19.14.1.20.8.1.14! :D "For about the 4096th time, let me offer a radical idea to those of you who don't like Nathan -- block her and don't visit Nathan's Corner. What is so complicated about it?" Wizard, August 21st. :D
ThatDonGuy
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May 11th, 2019 at 8:31:47 PM permalink

This is easier to solve if the distance is 20

From a standing start with constant acceleration, distance =acceleration x time squared

Let An and Am be Nathan and Miller's accelerations, and Tn and Tm their total times in minutes.

16 = An (Tn - 4)2 / 2
20 = An Tn2 / 2
An = 1 / (Tn - 2)
40 = Tn2 / (Tn - 2)
Tn = 20 + sqrt(320)

15 = Am (Tm - 5)2 / 2
20 = Am Tm2 / 2
Am = 1 / (Tm - 5/2)
40 = Tm2 / (Tm - 5/2)
Tm = 20 + sqrt(300)

Miplet wins

DogHand
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May 12th, 2019 at 5:07:56 AM permalink
Quote: DogHand

Wizard,
Here is my answer.



For constant acceleration a, if the initial velocity is 0, the distance traveled x is
x = a*t²
The time to reach a given distance x is given by t = √(x/a).
In the race, both N and M cover the same distance d (assumed to be 1 mile), so whoever requires less time is the winner.
Each contestant covers the final fraction f of the total distance in a time of ∆t.
Thus, ∆t = (time to reach d) - (time to reach (1-f)*d)
∆t = √(d/a) - √(d*(1-f)/a))
∆t = √(d/a)*(1 - √(1-f))
Solving for the constant acceleration a gives
a = d*(1 - √(1-f))²/(∆t)²
N covers the last f_N = 1/5 of the distance in ∆t_N = 4 minutes, so
a_N = (1 mile)*(1 - √(1-0.2))²/(4 min)² = 0.0006966... miles/min²
M covers the last f_M = 1/4 of the distance in ∆t_M = 5 minutes, so
a_M = (1 mile)*(1 - √(1-0.25))²/(5 min)² = 0.0007179... miles/min²
The time each requires to finish the distance d is
t_N = √(d/a_N) = √((1 mile)/(0.0006966... miles/min²)) = 37.88... min
t_M = √(d/a_M) = √((1 mile)/(0.0007179... miles/min²)) = 37.32... min
Therefore, M wins by just over a half minute.



Dog Hand

Edit: fixed a typo in spoiler.



I found a mistake in my solution.

In my original solution, I forgot the fact of ½ in the first equation. On the plus side, my t's were correct, as the error cancelled out.

Here is the corrected solution:

For constant acceleration a, if the initial velocity is 0, the distance traveled x is

x = ½*a*t²

The time to reach a given distance x is given by

t = √(2x/a)

In the race, both N and M cover the same distance d (assumed to be 1 mile), so whoever requires less time is the winner.

Each contestant covers the final fraction f of the total distance in a time of ∆t. Thus,

∆t = (time to reach d) - (time to reach (1-f)*d)

∆t = √(2d/a) - √(2d*(1-f)/a))

∆t = √(2d/a)*(1 - √(1-f))

Solving for the constant acceleration a gives

a = 2d*(1 - √(1-f))²/(∆t)²

N covers the last f_N = 1/5 of the distance in ∆t_N = 4 minutes, so

a_N = 2(1 mile)*(1 - √(1-0.2))²/(4 min)² = 0.0013932... miles/min²

M covers the last f_M = 1/4 of the distance in ∆t_M = 5 minutes, so

a_N = 2(1 mile)*(1 - √(1-0.25))²/(5 min)² = 0.0014358... miles/min²

The time each requires to finish the distance d is

t_N = √(2d/a_N) = √((2*1 mile)/(0.0013932... miles/min²)) = 37.88... min

t_M = √(2d/a_M) = √((2*1 mile)/(0.0014358... miles/min²)) = 37.32... min

Therefore, M wins by just over a half minute.


Dog Hand
Wizard
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May 12th, 2019 at 6:24:31 AM permalink
Dog, I agree. Don, I haven't checked your work for a distance of 20 miles, but I have little doubt you're correct.

If forced, I think the beer should go to Dog.

Don, great meeting you yesterday and enjoy the rest of your Vegas vacation!
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IndyJeffrey
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May 12th, 2019 at 10:46:53 AM permalink
I'd add, "I wish I could remember these formulas from my school days" to the poll responses. On the plus side, I have learned how to search on Google.
Wizard
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May 12th, 2019 at 3:23:52 PM permalink
Quote: IndyJeffrey

I'd add, "I wish I could remember these formulas from my school days" to the poll responses.



You just need to remember that velocity is the derivative of distance and acceleration is the derivative of velocity.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
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