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A Train and Clock Problem


Date: 04/29/2000 at 07:33:32
From: Stan
Subject: VERY Confusing Word Problem

Here's my problem:

A train leaves the station at the top of the minute. If it averages 66 
miles per hour for the first 10 miles it will pass milepost 10 exactly 
at the moment that the minute hand of the clock is directly over the 
hour hand. At what time does the train leave the station?

How in the world am I supposed to understand and solve this? What 
steps do I take to get to the answer? PLEASE HELP me do this and get 
to the right answer; I'm really desperate to solve this.

Thanks a lot for previous help, Dr. Math!
Stan


Date: 05/02/2000 at 10:16:57
From: Doctor Rick
Subject: Re: VERY Confusing Word Problem

Hi, Stan.

This is certainly a tough problem. It calls for attacks from several 
directions.

First, you can find the time that elapses between the two events: when 
the train leaves (on an exact minute), and when it reaches the 
milepost (when the minute hand coincides with the hour hand).

Then you need to find two events of the specified types that are 
separated by exactly this time. There are lots of exact minutes in a 
day, but there are only 11 times in 12 hours when the minute hand and 
the hour hand coincide. Try to find these times. This is a good 
problem in itself!

Once you have found the 11 possible times for the train to reach the 
milepost, subtract the elapsed time to find when the train would have 
left the station in each case. Which of these times is an exact 
minute?

There is an outline of your plan of attack. There is still plenty of 
work to do.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   


Date: 05/02/2000 at 17:22:35
From: Stan
Subject: Re: VERY Confusing Word Problem

Okay, I partially get this, but how can you get an exact minute when 
you subtract, for example, 9.09 minutes. Are you saying that there is 
a time like 1:06:09?

Please help.


Date: 05/03/2000 at 08:43:04
From: Doctor Rick
Subject: Re: VERY Confusing Word Problem

Hi again, Stan.

Yes, time is continuous. We're assuming that the minute hand turns 
smoothly. There are dial clocks that are run digitally with stepper 
motors, so the hand will jump from minute to minute. This problem 
would have no solution with that sort of clock.

The time you have written, though, is 1 hour, 6 minutes, and 9 seconds 
(or 9/60 minute). It's easier to use decimal minutes, like 1:06.09. 
Better yet, though, is to stick with exact times like 9 1/11 minutes, 
rather than 9.09, which is only an approximation to 2 decimal places.

You did step 1 successfully. Now on to step 2: finding all the times 
when the minute hand and the hour hand coincide. This will take some 
algebra - as I said, it's a good tough problem in itself. I'll get 
you started.

One obvious time when the two hands are together is 12:00, when both 
hands are straight up. The other times are not going to be even 
minutes. The next one is sometime after 1:05, as I suspect you already 
see. At 1:00 the hour hand is at the 5-minute mark, but by the time 
the minute hand gets there, the hour hand will have moved on a bit. 
You could follow this back-and-forth, advancing the minute hand to 
where the hour hand is now, then advancing the hour hand again, but 
this will go on forever. Algebra is better.

Let's say that the hands are together at n minutes past 1:00. The 
minute hand is at the n-minute mark (remember, it won't be an exact 
minute, so there won't really be a mark there). Where is the hour 
hand, in terms of minutes? When n = 0, the hour hand is at the 
5-minute mark, as I said above. When n = 60, it's 2:00, so the hour 
hand is at the 10-minute mark. Can you write an expression that says 
where the hour hand is for values of n between 0 and 60?

When you have this expression, you can quickly write an equation that 
says that the two hands point to the same minute. Solve this equation. 
Then go on to the time between 2:10 and 2:15 when the hands are 
together again, and so on.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   


Date: 05/03/2000 at 16:52:35
From: Stan
Subject: Re: VERY Confusing Word Problem

Hi:

Boy am I confused! I understand what you said, but...

I tried to write an expression in all different kinds of ways, but 
couldn't get one that worked. (I thought of proportions, but that 
didn't make sense.) I know when one minute passes, the hour hand moves 
1/60 of its way between the 5-minute mark and 10-minute mark (I 
think.) Please help me write or fix or the equation; I'm trying!

Thank you, Dr. Rick, for coping with me  :)
Stan


Date: 05/04/2000 at 08:22:11
From: Doctor Rick
Subject: Re: VERY Confusing Word Problem

Hi again, Stan.

You're right, the hour hand moves 1/60 of the distance between 1:00 
and 2:00 each minute. In terms of minutes on the dial, that's 1/60 of 
5 minutes, since 1:00 is at the 5-minute mark and 2:00 is at the 
10-minute mark. So in one minute, the hour hand advances 5/60 minute 
on the dial. That fraction reduces to 1/12 minute.

Now we're ready to write an equation. The hour hand starts at 5 
minutes when n = 0 (at 1:00). It advances 1/12 minute when n increases 
by 1. Can you write a linear equation expressing the hour-hand 
position h in terms of the minute n?

When you have this equation, you can easily change it into an equation 
that states that the hour-hand position is the same as the minute-hand 
position. Then you're on your way!

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   


Date: 05/04/2000 at 17:30:03
From: Stan
Subject: Re: VERY Confusing Word Problem

I tried. Is

     h + n = 5 + (1/12n)

correct? It may not be.

Thanks, Dr. Rick!
Stan


Date: 05/05/2000 at 15:29:35
From: Doctor Peterson
Subject: Re: VERY Confusing Word Problem

Hi, Stan. I'm going to take over from my brother Dr. Rick for now, 
because he's not available to answer you today. It sounds like it's 
time for a change of direction anyway. I'd like to go back to the 
original question:

   A train leaves the station at the top of the minute. If it 
   averages 66 miles per hour for the first 10 miles it will pass 
   milepost 10 exactly at the moment that the minute hand of the 
   clock is directly over the hour hand. At what time does the 
   train leave the station?

Let's try to work through the whole problem; I won't give you all the 
details you have to fill in, but will try to guide you through the 
process. I'm not taking the time to go through all your interactions 
with Dr. Rick, but will try to focus on what has to be done, even if 
it repeats some of what he has said, and I will ignore any specific 
problems you have dealt with. You need a fresh start.

What do we know? The train started at some time that is an exact 
number of minutes, then went 66 miles per hour for 10 miles, at which 
time the hour and minute hand were together.

We can first look at the rate aspect of the problem. If he went 66 
miles per hour for 10 miles, you need to determine how long this took; 
I'd recommend finding it in minutes, and writing it as an exact 
fraction. This gives you the elapsed time of the trip.

We'll be looking for two times that are exactly that far apart, the 
first being an exact minute, the second being one where the hour and 
minute hand are in the same position. That's the second part of the 
problem, for which we have to think about how a clock works.

The minute hand moves 360 degrees in 60 minutes, or 6 degrees per 
minute. The hour hand moves 360/12 = 30 degrees in 60 minutes, or 1/2 
degree per minute, 1/12 as fast as the minute hand. The position of 
the minute hand at M minutes after any hour is 6M degrees; the 
position of the hour hand at M minutes after hour H is (60H+M)/2 
degrees. (Notice that I'm assuming that H is a whole number, but M 
doesn't have to be.) The hands will be together when these two angles 
are equal, or

    6M = (60H+M)/2

You should solve this to find M in terms of H; H could be any integer 
from 0 to 11, so this will give you 12 (actually, it will turn out, 
11) different values of M. From this you can find all the times at 
which the hour and minute hand are in the same position. Work them 
out, again, exactly, giving a fractional number of minutes where 
necessary.

Now, does the fractional part of any of these values coincide with 
that of the number of minutes you calculated for the elapsed time? 
That's what you are looking for, since you know that the starting time 
was an exact number of minutes.

Once you find the time that fits, you will know the ending time, and 
can subtract the elapsed time to find when the train started.

Of course, this problem is terribly unrealistic; we're depending on 
the knowledge that the minute and hour hands are together exactly, 
down to a fraction of a minute (and no mention has been made of the 
second hand!); any slight error could make a huge difference in the 
answer. But it does bring in some interesting math.

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

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