## Wednesday, January 30, 2008

### Too Much 'Time on Your Hands?' Classic Clock Problems as an Application of RxT = D

Have you submitted your vote yet in the MathNotations poll in the sidebar?

Target audience for this investigation: Our readers and algebra students (advanced prealgebra students can sometimes find a clever way to solve these).

Let's resurrect for the moment those ever popular rate/time/distance classics. Hang in there -- there's a more interesting purpose here!

We'll start by using fictitious presidential candidates running in a 'race.' Any resemblance to actual candidates is purely coincidental.

R and J are running on a huge circular track. J can run a lap in one month whereas it takes R twelve months to run the same lap. To be nice, J gives R a 3-month head start. After how many months will J 'catch up' to (overtake) R?

Are those of us who were trained to solve these feeling a bit nostalgic? Do you believe that our current generation of students has had the same exposure to these kinds of 'motion' problems or have most of these been relegated to the scrap heap of non-real world problems that serve no useful purpose. Well, they still appear on the SATs, a weak excuse for teaching them, perhaps, but I can certainly see other benefits from solving these. Can you?

Ok, there are many approaches to the problem above. Scroll down a ways to see a couple of methods (don't look at these yet if you want to try it on your own):

Method I: Standard Approach (using chart)

..............RATE ...x.......TIME .....=.....DISTANCE
............(laps/mo).....(months)............(laps)

R.........1/12....................t........................t/12

J................1....................t...........................t

Equation Model (verbal): At the instant when J 'catches up' to R:
Distance (laps) covered by J = Head Start + Distance covered by R

Equation: t = 1/4 + t/12 [Note: The 1/4 comes from the fact that R covers 1/4 of a lap in 3 months]

Solving: 12t = 3 +t --> 11t = 3 --> t = 3/11 months.

Check:
In 3/11 months, J covers 3/11 of a lap.
In the same time, R covers (1/12) (3/11) = 1/44 lap. Adding the extra 1/4 lap, we have 1/4 + 1/44 = 12/44 = 3/11. Check!

[Of course, we all know these fractions would present as much difficulty for students as the setup of the problem, but we won't go there, will we!]

Method II: Relativity Approach
Ever notice when you're zipping along at 65 mph and the car in the next lane is going the same speed, it appears from your vehicle that the other car is not moving, that is, its speed relative to yours is zero! However, if you're traveling at 65 mph and the vehicle in front is going 75 mph, the distance between the 2 cars is ever increasing. In fact, the speedier vehicle will gain 10 miles each hour! This 75-65 calculation is really a vector calculation of course, but, in relativity terms, one can think of it this way:
From the point of view of a passenger in the the slower vehicle, that person is not moving (speed is zero) and the faster vehicle is going 10 miles per hour. We can say the relative speeds are 0 and 10 mph.

Ok, let's apply that to the 'race':

If R's relative speed is regarded as zero, then J's relative speed will be 1 - 1/12 = 11/12 laps/month.
Since R is not 'moving', J only needs to cover the head-start distance to catch up:
(11/12)t = 1/4 --> t = (1/4)(12/11) = 3/11 months. Check!
[Note: Like any higher-order abstract approach, some students will latch on to this immediately and others will have that glazed look in their eyes. It may take some time for the ideas to set in. This method is just an option...]

There are other methods one could devise, particularly if we change the units (e.g., working in degrees rather than laps). Have you figured out how all of this will be related to those famous clock problems? Helping students make connections is not an easy task. One has to plan for this as opposed to hoping it will happen fortuitously.

Here is the analogous problem for clocks:

At exactly what time between 3:00 and 4:00, will the hour and minute hands of a clock be together?

Notes:
(1) I will not post an answer or solution at this time. I'm sure the correct answers and alternate methods will soon appear in the comments.
(2) A single problem like this does not an investigation make. How might one extend or generalize this question? Again, these are well-known problems and I'm sure many of you have seen numerous variations on clock problems. Share your favorites!
(3) Isn't it nice that analog watches have come back into fashion so we can recycle these wonderful word problems!
(4) For many problem-solvers, part of the difficulty with clock problems is deciding what units to use for distance (rotations, minute-spaces, some measure of arc length, degrees, etc.). This is a critical issue and some time is needed to explore different choices here.

MathFr3d said...

I did the first and the 2nd problem and I came to the coclusion
that it's best to choose the desired result unit as a unit for distance.

In the first example it would be 1 Unit = 1 lap on the track.
In the clock example the unit could be minute or second.

A solution with t in minutes as unit:

The hour hand starts at the 3 o'clock (= 15 minutes) mark and reaches the 4 o'clock (=20 minutes) mark after 60 Minutes.

h(t) = 15 + 5*(t/60)

(Note: 15 + 5*1 = 20)

The minute hand starts at the 12'o clock mark (= 0 minutes) and reaches
it again after 60 minutes:

m(t) = t

The solutions can now be determined by finding t such that h(t) = m(t):

h(t) = m(t) <=>
15 + 5*(t/60) = t <=>
15/t + 5/60 = 1 <=>
15/t = 1 - 5/60 <=>
15/t = 55/60 <=>
t/15 = 60/55 <=>
t = (60*15)/55 <=>
t = (60*3)/11 <=>
t = 180/11 <=>
t = 16.3636363636....

This means that the hour and the minute hand meet 16.36 minutes
past 3 or to be more exact 16 minutes and 21.6 seconds past 3.

---

I think the general pattern of this type of problem is obvious.
Each Player has a starting condition (either a head start t0 and/or
possibly a head position x0) and each player has a speed v(t).

From this one can construct a function that calculates the player
position over t:

f(t) = v(t + t0) + x0

Similary for the 2nd player

g(t) = v'(t + t0') + x0'

Set g(t) = f(t) and solve for t to find the solution. Pretty simple!

One only has to watch out for sentences like "player b starts 3 months
later" ... this would mean t0' = -3 for player b. If you dont convert
this into "player a starts 3 months earlier than b" such that t0 = 3
(and t0' = 0) you will get a result relative to the starting point
of player b (which is 3 months later) and not of player a.

MathFr3d said...

I should think before I type ... I where i said "velocity" above I meant "traveled distance" ;)

Dave Marain said...

Nice analysis mathfr3d!
Using minutes made it fairly simple. Some students prefer degree-units or rotations and parts thereof. What can be confusing to students is distinguishing between a unit of time like 'minute' and a unit of arc-length like 'minute-space'.

Now, how many variations or generalizations can one imagine? The 180/11 answer can be turned into a general formula, yes?

Here are a few more:

(1) How many times in the 12-hour period from 12 Noon to 2 Midnight (inclusive) will the hands of a clock be together? Can this be done intuitively without algebra?

(2) Determine all times in that same 12-hour period will the hands of a clock form a right angle? Form an angle of a given measure such as 10 degrees?

(3) Mirror image times? I'll let you imagine what that means...

MathFr3d said...

(1)

That one is simple, in each hour of
the 12 hours the minute hand has
to meet the hour hand because
the minute hand goes from 12 to
12 during one hour.

This means there are 12 positions
where the hands are equal plus
the 12 o'clock position after the
12th hour. So the result is 13.

(2)

The hands form a right angle twice
each hour. There are 12 hours so
the result seems to be 24.

(3)

Maybe this means calculate the mirror time
from a given time h:m ?!

Mirror m to (m+30) mod 60 and h to (h+6 +(m+30)/60) mod 12.

(4)

The one when a clock falls into a singularity in space.

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