wu :: forums (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi) riddles >> hard >> Particle Time (Message started by: tim on Aug 15th, 2002, 5:56am)

Title: Particle Time
Post by tim on Aug 15th, 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 15th, 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 (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.

Title: Re: Particle Time
Post by S. Owen on Aug 15th, 2002, 8:31am
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.

Title: Re: Particle Time
Post by AlexH on Aug 15th, 2002, 8:55am
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

Title: Re: Particle Time
Post by S. Owen on Aug 15th, 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 15th, 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 15th, 2002, 5:49pm
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. :(

Title: Re: Particle Time
Post by tim on Aug 15th, 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 15th, 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 = (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)

Title: Re: Particle Time
Post by James Fingas on Aug 28th, 2002, 1:40pm
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?

Title: Re: Particle Time
Post by AlexH on Aug 28th, 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 21st, 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 3rd, 2002, 9:49pm
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

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.

Title: Re: Particle Time
Post by titan on Oct 15th, 2013, 9:16pm

on 10/15/13 at 21:15:05, 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.

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