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Minimum time between stations
By Yiingleong Chin (P3505) on Saturday, December 30, 2000 - 02:08 am:

Suppose a rail car has a maximum acceleration of 24 ft/s2 and maximum deceleration from braking of a=-32ft/s2. Please find the minimum time it would take to start from rest, cover 1/4 mi, and come to stop at the 1/4-mi mark. This can be achived by accelerating as much as possible for part of the 1/4 mi, followed by a period of maximum deceleration to the final stop. What's this minimum time?

Thanks.

By Brad Rodgers (P1930) on Sunday, December 31, 2000 - 05:05 am:

We know that in a period of minimum time, all of the time will be spent either accelerating, or decelerating. Any other method would leave a time period of less velocity. So

call "a" the maximum acceleration
and "b" the maximum deceleration

call ta the time spent accelerating
and tb the time spent decelerating


call va the velocity achieved accelerating
and vb the velocity achieved by decelerating

First

dvy/dty=±y

Where y is either a or b

So,

vy=±yty+C

substituting intitial conditions and thinking about acceleration and decceleration:

va=ata

and

vb=ata-btb

We know that at the end of the journey

vb=0

So,

ata-btb=0

Also,

If xy is the distance travelled by having y happen then,

dxa/dxt=at

then xa=ata2/2

A similar process yields,

xb=atatb-btb2/2

As

xa+xb=
ata2/2+atatb-btb 2/2=D

if D is the total distance travelled.

We have the set of equations:

ata-btb=0
ata2/2+atatb-btb 2/2=D

Keeping in mind what the total time is, we can solve for ta and tb to get the total time taken. Then substitute what a, b and D are into the equation we get to get the answer in units. Keep in mind that you'll need to convert feet to miles or miles to feet.

Sorry for the overcomplexity/lack of good explanation in this; if you have any questions, just post again.

Hope this helps,

Brad

By Yiingleong Chin (P3505) on Tuesday, January 2, 2001 - 12:24 pm:

Wow! That's cool! I never thought I needed differentiation to solve that one. Thanks!