- March 31st, 2010, 9:08 am
- Forum: Brainteaser Forum
- Topic: meeting point of two trains
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<t>I think I have the answer :in the continuous case in fact it's simple, you have just to calculate the distance E(d) as in the deterministic case and then make the average over the speed. So with K the value of the prequoted integral.This formula works for the limit cases I have presented before ....

- March 31st, 2010, 5:14 am
- Forum: Brainteaser Forum
- Topic: meeting point of two trains
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<t>Intuitively I think that the formula should satisfy to some limit cases (I only deal with the continous case here). For Vm1=Vm2 by reason of symmetry, E(d) should be equal to L/2.When Vm1>Vm2, E(d) should be superior to L/2.When Vm1<Vm2, E(d) should be inferior to L/2.When Vm1 tends to 0, E(d) sh...

- March 30th, 2010, 8:35 pm
- Forum: Brainteaser Forum
- Topic: meeting point of two trains
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*correction: I forgot a term 1/2 in E(d) so so at the continuous case it's E(d)=L(1/2+K). I hope I didn't make any other mistake. So the formula works if K<1/2 and it seems (I've made a very quickly graph on Mapple) that it doesn't correspond to all values of Vm1,Vm2.

- March 30th, 2010, 8:16 pm
- Forum: Brainteaser Forum
- Topic: meeting point of two trains
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<t>I agree with you about the integral. It's just I'm not really sure of the validity of the method I used.Anyway by taking your formula for E(d) and by assuming that the value of the integral is K,I find:so in the continuous case L(1+K). I think there is somewhere a mistake because the integral is ...

- March 30th, 2010, 7:10 pm
- Forum: Brainteaser Forum
- Topic: meeting point of two trains
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<t>Thanks dunrewpp to clarify the problem and pose the basis of the calculus. I agree with your calculus reaverprog, the hard point is maybe the calculus of the double integral. So to avoid the singularity at the origin, I replace 0 by a small eps, calculate the double integral and make eps tends to...

- March 30th, 2010, 3:51 pm
- Forum: Brainteaser Forum
- Topic: meeting point of two trains
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<t>De rien reaverprog so first I think it would be interested to calculate the time that the train 1 reachs the interval #K. with L/N the step size, is an event of speed on the interval i.For the train 2, the formula would bequite the same except that the train comes from the other train station soS...

- March 30th, 2010, 11:52 am
- Forum: Brainteaser Forum
- Topic: meeting point of two trains
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<t>Hi reaverprog, sorry for being not really clear. It fact what I mean is that for each train when you pass from an interval N to N+1, you choose another speed but in the same uniform distribution. Example, imagine that V1 is U([0,1]). At the first interval, the speed value is for example 0.7, so t...

- March 30th, 2010, 7:25 am
- Forum: Brainteaser Forum
- Topic: meeting point of two trains
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<t>Let assume two train stations A and B distant of L. This length is divided into N intervals of the same length. So here the problem, the train 1 is going from A to B with a random speed V1 (uniformly distributed between 0 and Vmax1). Each time, the train change of interval (passing from N to N+1)...

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