Unicycle race
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8 members have voted
May 10th, 2019 at 4:56:56 PM
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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.
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)
May 10th, 2019 at 5:07:55 PM
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Round or square wheels?
Have you tried 22 tonight? I said 22.
May 10th, 2019 at 5:11:16 PM
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Quote: GialmereRound or square wheels?
Good one!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 10th, 2019 at 6:43:59 PM
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But can you pop a wheelie?
May 10th, 2019 at 10:55:51 PM
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Quote: SkittleCar1But 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)
May 11th, 2019 at 8:41:19 AM
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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.
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
May 11th, 2019 at 10:38:29 AM
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How do you know all this?Quote: WizardNathan 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.
I run Easy Vegas ( https://easy.vegas )
May 11th, 2019 at 11:42:21 AM
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Quote: MichaelBluejayHow do you know all this?
The Great and Powerful Wizard knows all!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 11th, 2019 at 12:07:18 PM
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Quote: WizardThe 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
May 11th, 2019 at 8:31:47 PM
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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
May 12th, 2019 at 5:07:56 AM
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Quote: DogHandWizard,
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.
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
May 12th, 2019 at 6:24:31 AM
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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!
If forced, I think the beer should go to Dog.
Don, great meeting you yesterday and enjoy the rest of your Vegas vacation!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 12th, 2019 at 10:46:53 AM
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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.
May 12th, 2019 at 3:23:52 PM
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Quote: IndyJeffreyI'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)
