The Saga of the Two Trains

This page is probably not what you were expecting. 
In fact, it has nothing whatsoever to do with anything,
yet it took on such a life of its own, it could hardly be discarded.

It began innocently enough on October 5, 2004, when for perverse reasons of my own,
I decided to have a Newsletter with a train theme:

http://www.nnhs65.com/10-05-04-NNHS-Training-Counts.html

Just for fun and Old Times' Sake, I added a mathematical Train Word Problem.
Joe Madagan ('57) of FL immediately responded with another Train Word Problem.  It
lay dormant for several months, while we focused on The Great Reunion and the holidays
and such.  Then on January 24, 2005, Joe directed our attention back to the word problem,
and that's when the fun began.

LEST WE FORGET:

From Me ('65) of NC - 10/05/04:

For Old Times' Sake:

A Word Problem: A freight train leaves Chicago at 4:30 pm traveling at a speed of 60 mph. Two hours later
a passenger train leaves the same station traveling at 90 mph. How far will the first train get before the
passenger train catches up to it?

Translate: D = R x T

Answer: The freight train will get 360 miles away from Chicago when the passenger train catches up.

From Joe "Adonis" Madagan ('57) of FL - 10/06/04:

  HA-HA!!  HA-HA-HA!!!  Sweetie, I hope you don't think that I created that Word Problem myself!  There's a very good reason
that I majored in English rather than Math.  It's only been in the last 15-20 years that my Word Problem nightmares have ceased. 
So, don't worry, Joe.  I can assure you, there's not the slightest chance that I will spill the beans on this fun game before my return
from Illinois.  WILD GIGGLES!!!  Thanks!

From Joe Madagan ('57) of FL - 01/24/05:

   Ah - the question of the two trains!  Let's see - that was back in early October.... Ah yes, here it is:

From http://www.nnhs65.com/10-05-04-NNHS-Training-Counts.html  

A Word Problem: A freight train leaves Chicago at 4:30 pm traveling at a speed of 60 mph. Two hours later
a passenger train leaves the same station traveling at 90 mph. How far will the first train get before the
passenger train catches up to it?

Translate: D = R x T

Answer: The freight train will get 360 miles away from Chicago when the passenger train catches up.

From http://www.nnhs65.com/10-07-04-NNHS-Kindly-Thoughts.html

Another Word Problem: Now that the Passenger Train caught up with the Freight Train, how long will it take the
same Passenger Train to double the distance from the station they both left?

   And I've still no clue..... WILD GIGGLES!!!  Anyone?  Anyone??

From Frank Blechman ('65) of Northern VA - 01/26/05:

   WHAT?!?  Are you kidding me?!?  Where is that problem again, lemme see............

Another Word Problem: Now that the Passenger Train caught up with the Freight Train, how long will it take the
same Passenger Train to double the distance from the station they both left?

   Oh, for Pete's sake!  YOU GUYS!!  WILD GIGGLES!!!!  Thanks, Frank - and Joe!

From Joe Madagan ('57) of FL - 01/27/05:

From Joe Madagan ('57) of FL - 01/27/05:

   Thanks, Joe!  I know, I know!  I think we're just better at solving World Problems than we are at solving Word Problems! 

   Okay, back to the Word Problem:

From http://www.nnhs65.com/10-05-04-NNHS-Training-Counts.html

A Word Problem: A freight train leaves Chicago at 4:30 pm traveling at a speed of 60 mph. Two hours later
a passenger train leaves the same station traveling at 90 mph. How far will the first train get before the
passenger train catches up to it?

Translate: D = R x T

Answer: The freight train will get 360 miles away from Chicago when the passenger train catches up.

From http://www.nnhs65.com/10-07-04-NNHS-Kindly-Thoughts.html

Another Word Problem: Now that the Passenger Train caught up with the Freight Train, how long will it take the
same Passenger Train to double the distance from the station they both left?

   Well, I still don't know.  You know, no one ever mentioned which direction these trains were headed.  Where are they going
and what will they see?  How often will they brake for yard sales?  Will some bozo decide to park his car on the train tracks?
Will they be slowed by blizzards?  I'm just not at all certain that we have enough information to solve this problem.  I'm fairly
certain that I at least shall never solve this problem!  WILD GIGGLES!!!

HERE COME THE TRAINS:

From Fred Eubank ('64) of TX - 01/28/05:

Carol,

I showed my 6^th grader (Chris) the train ‘word’ problem and he offers the following:

The freight train (FT) will have traveled 120 miles (60 miles/hour x 2 hours) when the passenger train (PT) leaves the station
at 6:30.  The distance formula for FT is D = 120 + 60t.  The distance formula for PT is D = 90t.  When PT catches FT, the two
distances must be equal.  Then, 120 + 60t = 90t.  Solving for t = 4 hours, and substituting that into either distance formula gives
D = 360 miles.  Double this distance from the station is 720 miles.  PT will travel 720 miles in t = 720/90 = 8 hours.  Slowpoke
FT will have traveled only 120 + (60 x 8) = 600 miles by then.  Hope this helps. 

Keep up your great web work.

Fred Eubank, NNHS Class of 1964
Chris Eubank, Barbara Bush MS Class of 2007
San Antonio, TX

    Thanks, Fred - and Chris!

From Gail Kiger Bonsey (Ferguson HS - '73) of OR - 01/29/05:

Freight Train Math Word Saga continues!  

This from Jim Bonsey (Kailua High - Oahu - '74) - husband of Gail Kiger Bonsey (FHS '73): 

Carol/Joe: The passenger train would need to be traveling at least twice as fast as the freight train in order
to ever go twice the distance.  At 1-1/2 times the speed, it will NEVER get twice as far" (i.e.,  in 1 billion hours,
the freight train will travel 60 billion miles and the passenger train only 90 billion...always 1-1/2 times the distance). 
 gb

   Thanks, Gail - and Jim!

   Then just for fun I called in my own expert:

From my Friend, Mark Ratledge (Reid Ross HS, Fayetteville, NC - '73; NC State - '77) of NC - 01/30/05:

Answer: 4 hours. It took 4 hours to get 360 miles at 90 mph, it will take 4 more hours to get 720 miles,
double the original distance.

    Thanks, Mark!

   Okay, Joe - how are we doing???

From Joe Madagan ('57) of FL - 01/31/05:

   YEA!  I'm so glad we have resolved the problem!  And yes, it was fun!  But I must confess: I'll stick to crossword puzzles
for brain stimulation.  That math section of my brain atrophied long ago!  GIGGLES!

   Thanks, Adonis!

From Jean Poole Burton ('64) of RI - 02/02/05:

Carol,

I keep dreaming that we are in algebra class and Mr. Taylor won't let us out until we solve the word problem
about the trains! Call the station. They will tell you which arrives first.

   WILD GIGGLES!  Thanks, Jean!  The trains have ceased running - and I never did find out where they were going
when they left Chicago!

From Joe Madagan ('57) of FL - 02/03/05:

   Oh, Jean - pull up a pillow and get comfy there in Mr. Taylor's algebra class.  The trains are still running..........

From Dave Spriggs ('64) of VA - 02/04/05:

Carol,

As my name has been mentioned several times in connection with the train word problem, I can
no longer remain silent. Remember: You asked for this.

In the wonderful world of Algebra, the Laws of Physics are flagrantly disregarded in the interest
of mathematics, e.g. velocities are achieved instantaneously. This is not the case in the real world
of Newtonian mechanics. Both trains must accelerate from zero velocity to their final velocity, and
this takes time. This time varies with the mass of each train and the force applied to accelerate it ….
remember F=MA? Further complicating the issue is this: The continued of application of force will
result in continued acceleration which will result in ever increasing velocity. If the train engineer
wishes to reach and maintain a constant velocity, then he must reduce the force being applied as he
nears the desired velocity. (Just think about your foot pressure on the accelerator of your car as you
enter the interstate from the on-ramp. Push hard to get up to speed, then begin to ease off as you reach
the speed limit.) Even worse, the force required to sustain a constant velocity is affected by the wind
resistance to the train, and that resistance is not a constant nor is it even linear with velocity, i.e. the
resistance at 60 MPH is more than twice the resistance at 30 MPH. So now we have to consider the
acceleration curves from a standing start and the curves as each train approaches its desired velocity.
All of these factors involve time, and therefore, the distance traveled by each train as the passenger
train overtakes the freight train.

Once you lay out all the equations, you are well into differential calculus and, perhaps partial
differential equations, as well as a host of unknown values not provided in the original word problem.
Accordingly, my solution to the word problem is: INSUFFICIENT DATA.

   David, you're not only adorable, you're positively delightful!  Thank you so much for these elucidating insights! 

   I think the important thing to remember here is that I WAS RIGHT - INSUFFICIENT DATA!  WILD HYSTERICAL
GIGGLES!!!

   (By the way, Dave's degree in 1969 from the United States Naval Academy is in Marine Engineering; his Master's
from Old Dominion University is in Engineering Management.)

From Joe Madagan ('57) of FL - 02/04/05:

   Oh, Adonis - I disagree!  Had this problem been crystal clear from the beginning, we could never have had so much fun
with it!  I found it almost exhilarating to realize that an exact science could be so open to interpretation.  And I never knew
word problems were supposed to be fun!  They always frustrated the bee-bees out of me!  I thought they had a three-fold
mission: to give ulcers, raise blood pressure, and cause one to break out in hives.

   So thanks for the brain teaser!  It has been very educational on a number of levels!

From Linda Lane Lane ('64) of VA - 02/06/05:

From Jean Poole Burton ('64) of RI - 02/06/05:

NOT THOSE TRAINS AGAIN...

From Carol Buckley Harty ('65) of NC - 02/07/05:

   AWWW, Jean!  Trains are FUN!  I've only taken three train trips in my whole life, and they were all delightful and incredibly
memorable.  The first was the summer of 1949, when my parents, my sister, my daddy's mama and his two sisters, and my
cousin Cheryl White (John Marshall HS - '64), all took the excursion train from Richmond to Buckroe for a week.  Aside
from the unforgettable memories, this trip added two very useful phrases to our family's vocabulary, both courtesy of Cheryl:
"STOP THE TRAIN!!!", and "No, wait - my shoes are in there!"

   The second trip was in July of 1966, when my sister, my mama, her two sisters, and my uncle took the C & O from Richmond
to Chicago, and the Union Pacific from Chicago to Bismarck, ND for my cousin Clarke Booth's (VMI - '61) wedding.  What a
wonderful time!  I had never been anywhere at all before, so it was my first look at other parts of the country.  This, too,
introduced some rich family phrases, both from my mother, delivered in her own amazing stage whisper: "Mike! What time is it?"
and "Wake up!  Wake up, everybody!!  We're at Fargo!  FARGO!!  You know - where the STAGE COACH runs!"

   My third train trip was in October of 2003, aboard an Amtrak train from Fayetteville to Richmond, and was even more
enchanting than the first two.

   I just LOVE trains!

From Fred Eubank ('64) of TX - 02/07/05:

Carol, 

Being a former engineer like Dave Spriggs ('64 - of VA), I can’t let this go either.

Concerning the train problem, Dave is absolutely right in that there was INSUFFICIENT DATA given to arrive at a conclusive
answer, even a simple algebraic one.  Here is the original problem.

A Word Problem: A freight train leaves Chicago at 4:30 pm traveling at a speed of 60 mph. Two hours later a passenger train leaves the same
station traveling at 90 mph. How far will the first train get before the passenger train catches up to it?

Translate: D = R x T

Another Word Problem: Now that the Passenger Train caught up with the Freight Train, how long will it take the same Passenger Train
to double the distance from the station they both left?

When insufficient data is given, the only way to arrive at an answer is to make ASSUMPTIONS.   However, what one reader
posted about “assume” is generally true.

Assumption No. 1: Are the trains traveling on different tracks, or are they traveling on the same track?  If they are on different
tracks, then the passenger train will pass the freight train after it overtakes it 360 miles from the station.  At least one answer
was based on the assumption that both trains were traveling on the same track.  Therefore, the 90-mph passenger train would
have to slow down behind the 60-mph freight train, and not be able to pass.  If you assume they were traveling on the same
track, then that answer makes perfect sense.  The problem does not say if there were 1 or 2 tracks, so you have to make an
assumption.

Assumption No. 2: The second part of the problem involves what happens after the passenger train overtakes the freight train
360 miles from the station.  The question is asked “…how long will it take the same Passenger Train to double the distance
from the station they both left?”  One has to assume “the distance” referred to is either the distance of the passenger train
from the station, or the distance of the freight train from the station.  If it is the former, the question should have been worded
“double its distance”.  If it is the latter, the question should have been worded “double the freight train’s distance”.  Now it’s
fairly obvious that if we’re talking about the passenger train doubling its distance, the answer is 8 hours and 720 miles. 
However, if we’re talking about the passenger train doubling the freight train’s distance, then the answer is “never” since the
passenger train is traveling at 90 mph and the freight train at 60 mph.  The only way the passenger train can travel twice as far
as the freight train is if its speed is twice that of the freight train, or 120 mph.  Alternatively, the freight train could slow down
to 45 mph.  Again, the problem did not say which distance to use, so you have to make an assumption.

As far as Dave invoking the Laws of Physics and Newtonian Mechanics, not to mention all the other fields of Engineering and Science, I never realized train engineers had to be so SMART.
I just “assumed” they wore those neat little, blue-striped engineer caps, red handkerchiefs (not to be confused with “kerchief”, or “hankerchief”) and pulled down on their “Toot-Toot” whistles all day, kind of like Casey Jones.		The only known authentic photograph of Casey in the cab of an engine.	Of course we all remember the Casey Jones TV show back in the 1950’s with Alan Hale, Jr.

When is Casey Jones' birthday?

All Aboard!

  WILD HYSTERICAL GIGGLES!!!  Wake up, Jean - We're at Fargo!  FARGO!!

   Thanks, Fred!  I think I laughed for five whole minutes!

http://www.infoplease.com/ce6/people/A0826558.html

http://www.watervalley.net/users/caseyjones/casey.htm

http://taco.com/roots/caseyjones.html

http://taco.com/roots/caseyvillage.html

http://www.caseyjones.com/

   Oh, John Luther ("Casey") Jones was born 14 Mar 1864 near in southeast MO, but moved to Jordan, Fulton Co,, KY
(near Cayce - hence his nickname) when he was in his teens, and died 30 Apr 1900 near Vaughan, MS as a result of the
train wreck.

   But these trains - our trains - were not involved.  No, they just keep going and going and going .....

All right I give up...a train story from me!

If you can't beat 'em, join 'em...:  In l979 my youngest brother was getting married in Virginia two days after we
moved into a new house.  My husband could not go to Virginia with me due to work obligations and so I decided
to take the train rather than drive with my two children, who were five years old and fourteen months old.  My
husband, who had parked in a 15 minute parking zone in Providence, got on the train to put my suitcases on while I
had the two children in tow.  He turned and his head disappeared through the door of the car just as the train
started to move...I feared he had not gotten off the train.  Sure enough, back he comes, saying "I guess I will have
to ride to Kingston"  (the next station south)  In a few minutes he left to find the conductor...back he came, saying,
"I will have to ride to New London, this train does not stop at Kingston".  So he rode to New London, caught another
train back to Providence and collected his car and parking ticket.  When we got to Newport News my parents and
my mother-in-law came to meet us at the train station. 
My mother-in-law said, "I prayed all day that John would come with you". 
I replied, "Well, you either prayed not enough or too much because he actually rode with me for two hours!

   GIGGLES!  Oh, good for you, Jean!  Thanks!

From Linda May Bond Crayton ('66) of VA - 02/26/09:

This is my father at the age of 10, around 1912. To his right are his parents and to his left are the engineer with his wife. Note the name on the train, that was my grandfather's sister's husband! Wise Co., VA    COOL BEANS!!! Your own personal train!  Well, sorta - close enough for our purposes, anyway!  Thanks, Linda May!
		ca. 1912 - Wise Co., VA

From Jean Poole Burton ('64) of RI - 03/02/09 - "Oh, no, not the trains again!":

FYI I rode the train from RI to Newport News on the 18th of February and back on the 25th.  I do not know how fast it was going, how many times it stopped (a lot), or whether there were any other trains going the other way...

I do know this:  I had two whole seats to myself, a lot of floor space, room to stretch my legs out, could walk two cars back and get food and drink, did not have to take off my shoes or jacket, did not have to put anything in a plastic bag, had my luggage in sight at all times, no standing in line for security...it was lovely!!!

   Sounds delightful to me! Thanks, Lady!

(This page was created on 02/08/05 for no particularly good reason...)

"The Orange Blossom Special" midi courtesy of http://www.banjo.com,
at the suggestion of Dave Spriggs ('64) of VA - 07/04/03.
Thanks, Dave!

Amtrak Train Divider Line clip art courtesy of http://www.bravenet.com - 08/12/04

Two Trains Math Problem courtesy of http://www.msjc.edu/math/mathcenter/handouts/Five-Step%20Strategy%20to%20Solving%20Word%20Problems.htm - 10/04/04

Laughing Smiley courtesy of Janice McCain Rose ('65) of VA - 02/07/05
Thanks, Janice!  Just what I needed!

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