[Guest]

A train travels from station A to station B which is 300km away. It has to stop at 8 other stations along the way and wait for at least 5 minutes to let passengers get on/off. The stations are well spaced out.

An expert train driver is able to drive the train as efficiently as possible so that it takes exactly 2 hours from A to B. Given that the train has a maximum acceleration of 0.5m/s^2, what is the maximum speed of the train to the nearest m/s?

(You can assume that acceleration is always linier up to its top speed)

Can you also find a gereral formula to find maxspeed v in terms of all the other variables?

Edited April 26, 2009 by psychic_mind

[Joe's Student]

...The stations are well spaced out.

Do you mean that the distance between each station is the same? Or that it is different between each station?

[Guest]

Do you mean that the distance between each station is the same? Or that it is different between each station?

No they are not necessarily evenly spaced. I just put that so that you know the train will reach its top speed between each stop.

[Joe's Student]

No they are not necessarily evenly spaced. I just put that so that you know the train will reach its top speed between each stop.

Ok thanks

[Guest]

One thing I also don't think I made clear was that the train decelerates (breaks) as fast as it accelerates.

[Joe's Student]

One thing I also don't think I made clear was that the train decelerates (breaks) as fast as it accelerates.

Was going to ask about that... so you're saying it's maximum deceleration is O.5m/s^2?

[Guest]

Was going to ask about that... so you're saying it's maximum deceleration is O.5m/s^2?

Indeed .

[Joe's Student]

Well I've been taken up with Bb's latest cryptogram (that I still can't get), but after giving up for a while I looked at this properly:

This is fairly straightforward I think. We know the train reaches its max speed between each train station, so it stands to reason the maximun speed it can reach is related to the distance between each station. As the max acceleration and deceeration is the same, the distance it takes to reach this max speed is the same as the distance it takes to come to a stop.

The optimum distance between each station for getting to the highest speed possible is 30km. If we break down AB into 10 identical distances that gives the biggest distance between each station in which the highest speed can be reached each time. If there were irregular distances between the stations, a higher max speed may be reached, but not between every stop.

So, because the distance needed to reach max speed is 15km (30/2 because its the same distance needed to reach it and come to a stop), the starting velocity is 0m/s, the acceleration is 0.5m/s^2, the simple equation:

v^2 = 0^2 + 2(0.5)(15000)

will give us v. Which leaves:

v= 122.47m/s or as asked 122m/s.

I think that's right . I'll do some work on the second part after giving Translator 25 some more thought.

[Guest]

This is fairly straightforward I think. We know the train reaches its max speed between each train station, so it stands to reason the maximun speed it can reach is related to the distance between each station. As the max acceleration and deceeration is the same, the distance it takes to reach this max speed is the same as the distance it takes to come to a stop.

The optimum distance between each station for getting to the highest speed possible is 30km. If we break down AB into 10 identical distances that gives the biggest distance between each station in which the highest speed can be reached each time. If there were irregular distances between the stations, a higher max speed may be reached, but not between every stop.

So, because the distance needed to reach max speed is 15km (30/2 because its the same distance needed to reach it and come to a stop), the starting velocity is 0m/s, the acceleration is 0.5m/s^2, the simple equation:

v^2 = 0^2 + 2(0.5)(15000)

will give us v. Which leaves:

v= 122.47m/s or as asked 122m/s.

Unfortunatly no.

Your calculations I’m sure are correct but I think you've made some assumptions which are wrong.

Firstly, when I said max speed, I meant the max speed it can travel at. (i.e. the engine can not make it go any faster). Once it reaches this speed it doesn't immediately slow down again, it maintains a constant speed until it gets near the next station.

Secondly, it shouldn’t really matter how far the stations are form each other. There isn't an optimum distance, only long enough so that the train has time to reach its top speed. And the train does not necessarily reach its top speed at halfway between them. Let's assume for arguments sake that no 2 stations are within 20km of each other.

Also I don't think you have considered the time in your calculations. Remeber the journey takes 2 hours.

I hope the problem makes sense. I'll post a hint later.

Edited April 26, 2009 by psychic_mind

[Joe's Student]

Unfortunatly no.

Your calculations I'm sure are correct but I think you've made some assumptions which are wrong.

Firstly, when I said max speed, I meant the max speed it can travel at. (i.e. the engine can not make it go any faster). Once it reaches this speed it doesn't immediately slow down again, it maintains a constant speed until it gets near the next station.

Secondly, it shouldn't really matter how far the stations are form each other. There isn't an optimum distance, only long enough so that the train has time to reach its top speed. And the train does not necessarily reach its top speed at halfway between them. Let's assume for arguments sake that no 2 stations are within 20km of each other.

I hope the problem makes sense. I'll post a hint later.

I see... I assumed that you meant the train had an infinite top speed, i.e it would keep gaining in speed until it was forced to decelerate because of an upcoming station. I didn't realise that you meant it would hit its top speed at some stage and it couldn't go any further than that. That's where the basis for there being an optimium distance between the stations came from.

Don't worry it makes sense , I think I just misunderstood it. I'll have a look at it again later. But I would be surprised if I managed to get the right answer .

[Guest]

Haven't done any algebra for a while and this looked like a challenge, so here goes...

100 m/s

The train accelerates at 0.5 m/s/s for 200s, then travels at 100m/s until taking another 200s to slow down. It waits 300s and then does the same again. It does this 9 times (9 intervals between A, the 8 stations and B). It doesn't really matter what the distance is in these intervals, as long as the train gets to max speed, it's the fact it is doing this speed that is important.

So, in summary it accelerates 9 times for 200s (1800s), decelerates 9 times for 200s (1800s), waits at stations 8 times for 300s (2400s) and spends the rest of the 2 hours (1200s) travelling at 100m/s. The distance travelled, respectively, is 9 sets of 10,000m, another 9 sets of 10,000m, 8 sets of 0m and a total at full speed of 120,000m which all adds up to 300,000m.

OK, now this is probably a complicated formula and I'm sure there are easier ones, but here goes...

If T is the total time of the journey, w is the wait time at a station, n is the number of stations and D is the total distance travelled, then the maximum velocity of the train is:

(T-wn +/- sqrt((wn-T)² - 8(n+1)D)) / 4(n+1)

That gives two solutions, which I then fed back in to the original problem to see which works. Not sure if there is an easier way to tell this, but out of time now.

[Guest]

There's a big error in the beginning that says the train waits at least five minutes at each station. If it stopped for 10 minutes at one of the stations, that would increase the max speed, correct?

I haven't had a chance to plug in equations, but I just feel the "at least" leaves us with a bit of an unknown.

[Guest]

Yes neida you are absolutely correct . I have to admit that I make up all my own problems and at the last minute I changed some of the values and forgot to recalculate the speed. I was amazed that it actually came out as a nice round number - that was not intended at all (hence why I said to the nearest m/s). I was also hoping the second value to be higher so that we could discard it because it is unrealistic, but never mind.

Anyway, your formula is exactly the same as I have and I don't think it will simplify anymore.

There's a big error in the beginning that says the train waits at least five minutes at each station. If it stopped for 10 minutes at one of the stations, that would increase the max speed, correct?

I haven't had a chance to plug in equations, but I just feel the "at least" leaves us with a bit of an unknown.

Well if the driver makes it as quick as possible then he would only wait for 5 minutes.

[Guest]

Well if the driver makes it as quick as possible then he would only wait for 5 minutes.

So a rewrite should read exactly 5 minutes and not at least. The way it reads now is that the waiting is variable, and not necessarily at the whim of the driver.

Good puzzle, though.