

Title: Particle Time Post by tim on Aug 15^{th}, 2002, 5:56am 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? 

Title: Re: Particle Time Post by AlexH on Aug 15^{th}, 2002, 7:45am 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 (BA)) 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. 

Title: Re: Particle Time Post by S. Owen on Aug 15^{th}, 2002, 8:31am Yeah, I am assuming that this is a 1dimensional 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. 

Title: Re: Particle Time Post by AlexH on Aug 15^{th}, 2002, 8:55am If you assume nonnegative 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 

Title: Re: Particle Time Post by S. Owen on Aug 15^{th}, 2002, 11:00am 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. 

Title: Re: Particle Time Post by Chronos on Aug 15^{th}, 2002, 11:27am Should we assume, based on the lack of a red M under the problem, that we're not considering special relativity? 

Title: Re: Particle Time Post by tim on Aug 15^{th}, 2002, 5:49pm Yes, a onedimensional 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. :( 

Title: Re: Particle Time Post by tim on Aug 15^{th}, 2002, 7:08pm 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 :) 

Title: Re: Particle Time Post by tim on Aug 15^{th}, 2002, 10:46pm 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 = (gamma1) c^2, where gamma is the usual c/sqrt(c^2v^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 worldline of the particle. The elapsed time from B's point of view is t' = (c/a) sqrt((a X/c^2 + 1)^2  1) 

Title: Re: Particle Time Post by James Fingas on Aug 28^{th}, 2002, 1:40pm William has revised the question to include a zerovelocity 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 vectorvalued 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? 

Title: Re: Particle Time Post by AlexH on Aug 28^{th}, 2002, 2:47pm 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). 

Title: Re: Particle Time Post by LuisFigo on Sep 21^{st}, 2002, 11:38am 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....? 

Title: Re: Particle Time Post by Icarus on Dec 3^{rd}, 2002, 9:49pm On the offchance Luis is still around to read this after 2.5 months: Quote:
No, D=(integral t=0 to t=T)v(t)dt = (V/T)(T^2)/2 = VT/2. [e]Eleven months later I reread this thread and see that the concern I brought up had in fact been addressed early on.[/e] 

Title: Re: Particle Time Post by titan on Oct 15^{th}, 2013, 9:16pm on 10/15/13 at 21:15:05, wrote:
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 

Powered by YaBB 1 Gold  SP 1.4! Forum software copyright © 20002004 Yet another Bulletin Board 