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Topic: Particle Time (Read 5063 times) |
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tim
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Particle Time
« on: Aug 15th, 2002, 5:56am » |
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I can't find any upper bound on how long the particle takes. Perhaps I'm misinterpreting the rather odd condition "never increases its acceleration along its journey".
It seems to me that a particle may follow a circular path from A to B at constant speed V, with as large a radius as you wish. The particle never increases its acceleration along its journey, in fact the magnitude of the acceleration is constant throughout.
Is there a reason why such a path is disallowed?
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AlexH
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Re: Particle Time
« Reply #1 on: Aug 15th, 2002, 7:45am » |
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Well you could interpret the acceleration as a vector so that it would be unclear whether a circular path "increases" the acceleration. For example (a \dot (B-A)) will be going up, so your "acceleration toward B" will be increasing . Even that rather strained interpretation doesn't help though if you're allowed an initial velocity. Just pick some velocity directed opposite from B and you can take as long as you like. I'm guessing its a trick question.
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S. Owen
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Re: Particle Time
« Reply #2 on: Aug 15th, 2002, 8:31am » |
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Yeah, I am assuming that this is a 1-dimensional problem - A and B are on the number line or something. Agreed, I think you also have to assume a nonnegative initial velocity. Then it's interesting... I still haven't come up with an answer though.
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AlexH
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Re: Particle Time
« Reply #3 on: Aug 15th, 2002, 8:55am » |
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If you assume non-negative velocity toward B in one dimension then with fixed final velocity V the longest time you can take is with full acceleration throughout starting from a standstill, otherwise you'd have a higher velocity at all earlier times. If X is the distance and T the duration of the trip we have X=.5aT^2 and a=V/T --> T= 2X/V
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S. Owen
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Re: Particle Time
« Reply #4 on: Aug 15th, 2002, 11:00am » |
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Yeah I think you're right, it's as simple as that, come to think of it. You want to defer the acceleration as much as possible, but since it can't increase over time, the best way must be to keep steady acceleration throughout at whatever minimal acceleration will get you to X with velocity V.
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Chronos
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Re: Particle Time
« Reply #5 on: Aug 15th, 2002, 11:27am » |
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Should we assume, based on the lack of a red M under the problem, that we're not considering special relativity?
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tim
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Re: Particle Time
« Reply #6 on: Aug 15th, 2002, 5:49pm » |
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Yes, a one-dimensional space would make my circular path illegal.
I don't see why you can't start with a particle at A moving at an arbitrary speed away from B, with a constant acceleration toward B.
I suppose the problem is meant to be one dimensional, with the particle starting at A and never again passing through A. In which case the solution is trivial as Alex has shown.
I'm afraid I must still be missing something. Or the problem is. 
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tim
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Re: Particle Time
« Reply #7 on: Aug 15th, 2002, 7:08pm » |
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Chronos: (btw, apt name given the problem under consideration  )
Special relativity doesn't involve a lot more math, but it does require that you define how you're measuring the elapsed time
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tim
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Re: Particle Time
« Reply #8 on: Aug 15th, 2002, 10:46pm » |
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Oops, I forgot to give the relativistic formula 
We once again have constant acceleration starting from rest as the optimum path. We have a.X = (gamma-1) c^2, where gamma is the usual c/sqrt(c^2-v^2).
So that gives us the acceleration a, from which we use
v = c tanh (a t/c) to find t:
t = (c/a) tanh^-1 (v/c).
Note that this is the proper time along the world-line of the particle. The elapsed time from B's point of view is
t' = (c/a) sqrt((a X/c^2 + 1)^2 - 1)
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James Fingas
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Re: Particle Time
« Reply #9 on: Aug 28th, 2002, 1:40pm » |
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William has revised the question to include a zero-velocity initial condition.
I would also assume that he's talking about the particle moving in a straight line. This makes the solution just as simple as Alex has said.
The other thing is that he would have to give us a vector-valued final velocity if this were a 3d problem. But the value he gave us was a scalar (or was it?)
I think this is just another one of those problems where a seemingly meaningless constraint leads to a single definite answer (see also the Funkytown riddle). If so, it shouldn't be in Hard!
In your relativistic formula, from what reference point are you measuring acceleration?
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| « Last Edit: Aug 28th, 2002, 1:41pm by James Fingas » |
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AlexH
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Re: Particle Time
« Reply #10 on: Aug 28th, 2002, 2:47pm » |
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He is measuring constant a from the perspective of the traveller. If you chose the rest frame of A and B then the relativistic answer would look exactly like the newtonian one (though obviously the traveller couldn't maintain constant acceleration according to the rest frame if it would push him past c).
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LuisFigo
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Hi all, im kinda confused
isnt the answer just T = sqr(3D/V) ?
dv/dt = k1 so v(t)=(k1)t + k2 with initial and terminal conditions
v(0)=0 v(T) = V
v(t) = (Vt)/T
D=(integral t= 0 to t=T)v(t)tdt
so D = (V(T^2))/3
no friction, straight line, no nothing....?
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Icarus
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Re: Particle Time
« Reply #12 on: Dec 3rd, 2002, 9:49pm » |
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On the off-chance Luis is still around to read this after 2.5 months:
Quote:D=(integral t= 0 to t=T)v(t)tdt
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No, D=(integral t=0 to t=T)v(t)dt = (V/T)(T^2)/2 = VT/2.
However, no one however has really shown yet that 2D/V is the maximum time (classically). Everyone is assuming that the acceleration is constant when the problem only says that it is not increasing at any time. Is a constant acceleration rate the slowest route? I would think this out more now, but its nearly midnight.
[e]Eleven months later I reread this thread and see that the concern I brought up had in fact been addressed early on.[/e]
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| « Last Edit: Nov 15th, 2003, 7:27am by Icarus » |
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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titan
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Re: Particle Time
« Reply #13 on: Oct 15th, 2013, 9:16pm » |
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on Oct 15th, 2013, 9:15pm, wrote:| with fixed final velocity V the longest time you can take is with full acceleration throughout starting from a standstill, otherwise you'd have a higher velocity at all earlier times. |
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I have failed to understand that why should acceleration be held constant throughout to get max. time when we can decrease or set acceleration to zero as well
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| « Last Edit: Oct 15th, 2013, 9:16pm by titan » |
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