What is the Minimum Time for a Subway Train to Travel Between Two Stations?

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The discussion focuses on calculating the minimum time for a subway train to travel between two stations using parameters such as distance, maximum speed, acceleration, and jerk. The train must start and stop completely at each station, and the calculations involve determining the optimal acceleration and deceleration profiles. A key point is that the train can accelerate up to its maximum speed or may not reach it depending on the distance. There is a debate about the maximum acceleration value, with one participant suggesting that the specified 20 m/s² may be unrealistically high for trains. The original poster seeks assistance in identifying the missing mathematical elements in their calculations.
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I am working on a computer program that determines the minimum time it
will take for a subway train to go from one station to the next based
on a few parameters:

The parameters to the problem are all positive integers not greater than 1000.

* d - the distance between stations, in metres
* m - the maximum allowable speed of the train, in metres/sec
* a - the maximum absolute acceleration of the train, in metres/sec2
* j - the maximum absolute jerk, in metres/sec3

The train must be completely stopped at each station and must move in
one direction at speeds not exceeding m. Acceleration can be positive
(forward) or negative (backwards) but its absolute value must not
exceed a. The last parameter, jerk, is the rate of change of
acceleration in either direction. That is, acceleration cannot
increase or decrease at greater than this rate. This parameter
prevents toppling the standing passengers.

For example, if the inputs are:

d = 1000
m = 70
a = 20
j = 1

The correct answer to 3 sig figs is 31.7


I have gotten close to this, but I am missing something. Basically, I
am looking for the math to figure out the problem.

Thanks in advance.
 
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The minimum time will occur in the following situation:

The train accelerates from rest at the maximum jerk rate, up to its maximum allowed acceleration, which it then maintains until it gets to maximum allowed speed. This process is reversed at the other end of the trip, as the train comes to a stop. In between the accelerations, the train should travel at its maximum speed.

It is possible, of course, that there won't be time to reach maximum speed, in which case there will be no constant speed section of the trip.
 
James R said:
The minimum time will occur in the following situation:

The train accelerates from rest at the maximum jerk rate, up to its maximum allowed acceleration, which it then maintains until it gets to maximum allowed speed.

And as the train approaches its maximum allowed speed, it has to reduce its acceleration to zero at the maximum jerk rate. It has to start doing this at just the right time so that when the acceleration reaches zero, its velocity is at maximum.
 
x911gt2 said:
Acceleration can be positive
(forward) or negative (backwards) but its absolute value must not
exceed a.

I don't mean to be picky, but keep in mind that a positive or negative acceleration does NOT have to mean forward or backwards, respectively. A positive acceleration means it is accelerating in the direction of motion, while a negative acceleration can mean it is decelerating along the direction of motion. Since your problem stated that:

The train must be completely stopped at each station and must move in
one direction

It means it cannot go backwards. Thus, for this case, a negative acceleration can only mean a deceleration.

Zz.
 
Hm, are you sure the maximum acceleration is 20 m/s^2? That's about two "g" which is very very large for a train. I seem to remember reading somwhere that a typical maximum acceleration for a train is 0.1 "g", that is, 1 m/s^2. Perhaps your problem really specifies a maximum acceleration of 2.0 m/s^2?

I wouldn't want to stand up on a train accelerating at 2 "g"! A mass on a string would be hanging at an angle of about 63 degrees from the vertical. :eek:
 
Last edited:
x911gt2 said:
The correct answer to 3 sig figs is 31.7[/I]

I have gotten close to this, but I am missing something.

If you show us what you've done, someone can probably zero in on what you're missing. Otherwise all we can do is guess.
 
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