Lev Okun gave this problem to Andrei Sakharov to pass the time while they were being

[Crunchy Black]

Lev Okun gave this problem to Andrei Sakharov to pass the time while they were being driven from Moscow to JINR in Dubna. However, it didn't work (to pass the time) because immediately after being told the problem, Sakharov pulled out a pen, took Lev's magazine and wrote down the solution, without any hesitation whatsoever.

how long does the problem take you? problem is in following post

[Crunchy Black]

An infinitely stretchable rubber band has one end nailed to a wall, while the other end is pulled away from the wall at the rate of 1 m/s; initially the band is 1 meter long. A bug on the rubber band, initially near the wall end, is crawling toward the other end at the rate of 0.001 cm/s. Will the bug ever reach the other end? If so, when?

[yo soy el mejor]

uhm no?

[Ol' Couch Ass]

What is circumference of the earth again?

[bardy]

I just did this little experiment and the object on the rubber band was only moved half of the distance I pulled the rubber band

so like I move rubber band 4 inches and the thing on the rubber band is only moved 2 inches

so the bug will never reach the end of the rubber band, right?

I mean the bug will always be moving at a slower velocity than the band is increasing, he would be like 0.500000000000000000000000000001 (exaggreated) m/s and the thing is increasing at 1 m/s

he should be given a 27.77777777777777 hour head start

[Crunchy Black]

youve solved the wrong problem, bardy. thats a good start, though, eh?

[bardy]

I modified?

[BlissedandGone2]

yes if the rubberband is stretched evenly. some huge number in time though.

[???]

um theoretically i guess the band would always be pulling the bug to the halfway point, which would mean the bug was always proportionately getting closer to the end of the band. i don't think this would hold up to scientific scrutiny though, would it?

[Crunchy Black]

at the wall, the bug is moving bug crawl speed. at the halfway point, the bug is moving half of the band stretching speed plus its crawl speed. if it ever makes it to the end, it will be moving as fast as the band expands plus its crawl speed

[Crunchy Black]

well andrei sakharov has clearly put me to shame so far, but i will not giv eup hope

[bardy]

yeah the further you pull the rubber band the less the bug is going to be pulled by it, upon futher inspection... so I dunno????????????????????????????????????????????? ?? I am bored at work and playing with rubber bands I need to go home


edit: nm

[yo soy el mejor]

walking on sunshine

[Crunchy Black]

edit - nvm read that wrong

[bardy]

it would probably pull the nail out after a while

[KrazeeStacee]

Not if it's infinitely stretchable...

[bardy]

nm

[Nimrod's Son]

eventually the bug will get tired and fall off the rubber band

[Crunchy Black]

i wonder if i should just give up and look at the answer... who knew this problem would be so hard!

[Crunchy Black]

i guess i might as well have something to do during analytical chemistry

[yo soy el mejor]

something to pass the time.

[bardy]

someone fix my errors

I gave the band an original length of 1m which is probably the wrong way to look at this, but whatever

Length of Band = L
Bug Velocity B(t)
Pull Velocity P(t)
distance bug travled Db

L = 1m + 1m/s * t

del for a second

[bardy]

someone figure out the acceleration of the bug I dont want to

[Future Boy]

Assume that the bug's distance from the wall when the band is not stretched is f(t) and the length of the rubber band (when it is stretched) is r(t). Let the distance from the wall to the bug (crawling along the band as it is stretched) be b(t). Then for an infinitesimal period dt, we have

b(t+dt) = b(t)r(t+dt)/r(t) + df(t).

Setting r(t+dt) = r(t) + r'(t)dt and df(t) = f'(t)dt, rearranging,

(b(t+dt)-b(t))/dt = b(t)r'(t)/r(t) + f'(t).

Taking the limit as dt approaches 0,

b'(t) = b(t)r'(t)/r(t) + f'(t).

The solution to this differential equation with boundary condition b(0)=0, is

b(T) = r(T)(Integral t=0->t=T f'(t)/r(t) dt),

If the bug reaches the end of the band at time T, so that b(T)=r(T), then we have

Integral t=0->t=T f'(t)/r(t) dt = 1.

The solution in T of this equation (when one exists) gives the answer for the general case.

Let f(t) = 0.00001t, and r(t) = t***. Then we have

Integral t=0->t=T 0.00001/(t***) dt = 1,

or, carrying out the integral,

0.00001 ln(T***) = 1.

So, T = e100000 - 1 seconds.

[bardy]

I will let everyone know that future boy obtained his solution from this page:

http://www.feynmanlectures.info/solu...and_sol_3.html

[Future Boy]

no way, i did that just now between gears of war matches.

[bardy]

just giving credit where credit is due

[BumbleBeeMouth]

Originally Posted by Future Boy

Assume that the bug's distance from the wall when the band is not stretched is f(t) and the length of the rubber band (when it is stretched) is r(t). Let the distance from the wall to the bug (crawling along the band as it is stretched) be b(t). Then for an infinitesimal period dt, we have

b(t+dt) = b(t)r(t+dt)/r(t) + df(t).

Setting r(t+dt) = r(t) + r'(t)dt and df(t) = f'(t)dt, rearranging,

(b(t+dt)-b(t))/dt = b(t)r'(t)/r(t) + f'(t).

Taking the limit as dt approaches 0,

b'(t) = b(t)r'(t)/r(t) + f'(t).

The solution to this differential equation with boundary condition b(0)=0, is

b(T) = r(T)(Integral t=0->t=T f'(t)/r(t) dt),

If the bug reaches the end of the band at time T, so that b(T)=r(T), then we have

Integral t=0->t=T f'(t)/r(t) dt = 1.

The solution in T of this equation (when one exists) gives the answer for the general case.

Let f(t) = 0.00001t, and r(t) = t***. Then we have

Integral t=0->t=T 0.00001/(t***) dt = 1,

or, carrying out the integral,

0.00001 ln(T***) = 1.

So, T = e100000 - 1 seconds.

No, sorry, what?

Is it basically one of those mathematical pieces of reasoning that takes advantage of the adhoc nature of physics? like these things?

Given that a and b are integers such that a = b + 1, Prove: 1 = 0

1. a = b + 1 1. Given
2. (a-b)a = (a-b)(b***) 2. Multiplication Prop. of =
3. a2 - ab = ab + a - b2 - b 3. Distributive Propoerty
4. a2 - ab -a = ab + a -a - b2 - b 4. Subtraction Prop. of =
5. a(a - b - 1) = b(a - b - 1) 5. Distributive Propoerty
6. a = b 6. Division Property of =
7. b + 1 = b 7. Transitive Property of = (Steps 1, 7)
8. Therefore, 1 = 0 8. Subtraction Prop. of =

[bardy]

uh no that is the solution

there are a few ways to get to it but that is the right answer

[Mayfuck]

lev okun and andrei sakharov sound like some shitty dudes to hang out with