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### A Train Breaks Down

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Date: 06/30/2001 at 22:28:05
From: Steve
Subject: A train question

Hi there,

I can't figure out how to do this. I know there's probably some sort
of algebraic way around it, but I can't figure it out. Here it is:

A train breaks down an hour after starting its journey. The engineer
takes half an hour to make repairs, but the train can only continue
at half its original speed and arrives at its destination two hours
late. If the delay had occurred 100km farther on in the journey, the
train would only have been one hour late. What was the distance of the
journey?
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Date: 07/01/2001 at 14:32:36
From: Doctor Jaffee
Subject: Re: A train question

Hi Steve,

Here is how I solved this problem.

Draw a segment representing the path of the train from the beginning
to the end of the trip.

Let d = the distance of the trip
let r = the average rate of speed
let t = the time it takes to finish the trip

Now, draw another segment representing the trip when the train breaks
down after 1 hour. Put an X on the track where you think the breakdown
occurs. Since d = rt and the train has been travelling 1 hour, the
distance that the train has travelled so far is d = r(1) = r.

After the repairs, the train travels at a rate of r/2 and the time it
travels is 2 hours more than the trip would have been without the
repairs, less the time it took before the train restarted, which is
1 1/2 hours.  So the length of time for the second part of the trip is
t + 2 - 1 1/2 = t + 1/2.

The distance the train travels in the second half of the trip is
(r/2)(t + 1/2).

So, adding the distances of the two parts, we get
r + (r/2)(t + 1/2) = d

You can simplify this equation, replace rt with d and end up with an
equation in terms of d and r.

Now make a third line and plot the trip with the repairs being made
100 km down the track. I suggest you consider three parts of the trip:
First, the train travels r km in 1 hour, then the train travels 100 km
at which time it breaks down. Finally it travels the remainder of the
trip at r/2 km/hr.

Give it a try and if you want to check your solution with me, write
back. If you are having difficulties, let me know and show me what you
have done so far, and I'll try to help you some more.

Good luck.

- Doctor Jaffee, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Basic Algebra
Middle School Algebra
Middle School Word Problems

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