Trolley problem. Just that.

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Racosharko

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Piisfun

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I sense a 2nd order differential equation...

2.6 seconds + how ever much time it is in contact with the spring.
18.308 seconds, assuming the spring can be compressed by at least 25 meters. Otherwise it crashes.

Have fun; it's a second order differential equation that requires complex numbers and hyperbolic trigonometry to solve.
 
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owotrucked

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I sense a 2nd order differential equation...

2.6 seconds + how ever much time it is in contact with the spring.
18.308 seconds, assuming the spring can be compressed by at least 25 meters. Otherwise it crashes.

Have fun; it's a second order differential equation that requires complex numbers and hyperbolic trigonometry to solve.
I don't understand how you got 2.6s
I assume the trolley moves out from the spring at the same velocity as it moved in because of energy conservation. So the total time going in and out should be T=L/V=(8.5*2) / 5 = 3.4s

For the second order differential equation, the analytical solution needs trigonometry and the problem poses pi=3.

But if you consider a computational simulation, pi doesn't intervene in the numerical solution so it's absolute bullshit lol
 

laccoff_mawning

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Have fun; it's a second order differential equation that requires complex numbers and hyperbolic trigonometry to solve.
To find the time taken for contact with the spring:

m a = - k e

4 x 10^6 x" = -100,000 x

x " = - x / 40

x = A * sin( wt) + A * cos(wt) , w = sqrt(1/40)

x(0) = 0, so x = A sin(wt)

The time in contact with the spring is half an oscillation, so wt = pi (velocity here doesn't actually matter. At least under the various assumptions a model spring makes.)

so t = pi/w = pi sqrt(40) = 2sqrt(10) pi

so t = 19.869... (usually)

However, we've been explicitly told pi = 3, so t = 18.97

so the final answer should be 3.4 + 18.97 = 22.4s (3 s.fs), if I've done my maths right.
 

owotrucked

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x = A * sin( wt) + A * cos(wt) , w = sqrt(1/40)

x(0) = 0, so x = A sin(wt)

The time in contact with the spring is half an oscillation, so wt = pi (velocity here doesn't actually matter. At least under the various assumptions a model spring makes.)

You sure its not
x = A * sin( wt) + B * cos(wt)
x' = something something

x(0) = 0
x'(0) = 5
 

laccoff_mawning

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You sure its not
x = A * sin( wt) + B * cos(wt)
x' = something something

x(0) = 0
x'(0) = 5
Thats correct,

x' = A w cos(wt), so you can substitute that in to get 5 = A w, and solve for the amplitude of the motion. That would give you how far the spring compresses, but it's not necessary if you just want to work out the time taken for the motion.
 

owotrucked

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Thats correct,

x' = A w cos(wt), so you can substitute that in to get 5 = A w, and solve for the amplitude of the motion. That would give you how far the spring compresses, but it's not necessary if you just want to work out the time taken for the motion.
Right

I think I understood why its a trolley problem. It gives the choice to kill maths or physical reality

Because pi is built in so many physical phenomenon like probability that you have to completely wipe the very concept of existence as we know it to imagine a world where pi=3. So if you save maths, we all get deleted


For instance:
What kind of sick world do you have to build to make pi converge to 3
 
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Piisfun

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I don't understand how you got 2.6s
I assume the trolley moves out from the spring at the same velocity as it moved in because of energy conservation. So the total time going in and out should be T=L/V=(8.5*2) / 5 = 3.4s

For the second order differential equation, the analytical solution needs trigonometry and the problem poses pi=3.

But if you consider a computational simulation, pi doesn't intervene in the numerical solution so it's absolute bullshit lol
The 2.6 seconds:
8.5 meters / 5 meters per second * 2
Double since it is a round trip: this is the total time that it is not in contact with the spring.

Ah, entry error. Should be 3.4 seconds
 
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