Mundane Pointless Algebra Problem

Miscellaneous and Personal Stuff I Must Share
1

This is not homework. I got a free college algebra book at a library book sale some time ago. The other day I was looking at some of the problems and I came across this problem:

“Jim and Pam leave a campsite, Jim biking due north and Pam biking due east. Jim bikes 7 km/hr slower than Pam. After 4 hr, they are 68 km apart. Find the speed of each bicyclist.”

I am approaching this problem as if the physical quantities are exact. If the physical quantities were measured values, then, depending on the uncertainties in each of the measurements, a simplified mathematical model may be adequate.

First, do Jim and Pam leave from the same exact spot in the campsite? Can Jim and Pam occupy the same space at the same time? I suppose we could contrive some unusual combination of a long, tall bicycle and a short, low bicycle that would almost allow that except for differences in distance from the surface of the campsite. Otherwise, there would be some displacement in an unknown direction between the two at the very start of their bike rides and this must be accounted for.

Second, I am assuming that Jim and Pam will be traveling over the surface of the Earth and not in a straight line through the Earth. However, I am assuming that when the problem states “68 km apart” it means in a straight line. As we all know, the shape of the Earth is approximately that of an oblate spheroid. The National Geospatial-Intelligence Agency has developed the World Geodetic System 1984 (WSG-84) which describes a spheroid with a semi-major axis of 6378.1370 km and a semi-minor axis of 6356.7523142451 km. I am further assuming that Jim and Pam are traveling on the surface of or at some constant height above a geoid exactly described by WSG-84. If Jim and Pam are traveling on the surface of the Earth above this geoid, i.e. at an elevation above sea level, then this must also be accounted for. Furthermore, since we are talking about an oblate spheroid, the latitude of the campsite must also be accounted for.

I am guessing that by “4 hr” they mean 4 hours as measured from the frame of reference of the campsite. Because Jim and Pam are moving at different speeds from the stationary campsite, special relativistic effects cause the 4 hours to be determined by each of them to be different from each other and different from the 4 hours determined at the campsite.

There may be general relativistic effects to consider because of the changing distribution of mass in the Jim, Pam, bicycles, and Earth system. However, such matters are beyond my pay grade.

I have five questions:
(1) Does anyone see any errors in my approach so far?
(2) Does anyone have suggestions on how to further refine my (as yet to be derived) really stupid mathematical model for this problem?
(3) Are there in fact three different solutions to this problem depending on the frame of reference used to determine the time interval of 4 hours?
(4) Are special relativistic effects further complicated by the rotation of the Earth?
(5) Do you think I can waste an entire weekend on this one algebra problem?

2

I don’t see where you’ve taken into account that Jim and Pam are not zero-dimensional points. Since they occupy space, “68 km apart” is ambiguous. 68 km between what points on their bodies, exactly?
(You’re the kind of algebra student that a teacher would either love or hate.)

3

He’s quite mad, you know.

4

Fortunately we don’t know exactly where they are or Heisenberg would keep us from ever knowing their (exact) speed.
I saw nothing in the problem about their speeds remaining constant or taking into account that we are to assume, I guess, instantaneous acceleration to that speed from the campground.

5

How could you use a barometer to measure the height of a building?
That ought to keep him busy for a while.

6

One way is to drop it from the roof and time the fall.

7
  1. Yes, you are overthinking it. It’s clear you’re supposed to make a right angled triangle and use Pythagoras.
  2. No
  3. I don’t know.
  4. I don’t know this either.
  5. I think you could spend longer than that - have you considered asking Sheldon Cooper to help?
8

We know how fast they bike, but presumably only on straight, level roads. Shouldn’t you use google maps to determine if there are any places on the Earth where this is possible?

9

[ol][li]On a sunny day, set the barometer on the ground. Measure the length of the shadow cast by the barometer and the building, and use simple trig to compute the height. [/li][li]Go up to the roof. Tie a long string around the barometer, lower it to the ground, and measure the string length used.[/li][*]Find the building’s janitor and offer him a nice barometer if he will tell you how tall it is.[/ol]

10

Why limit it to just you? I hypothesize that a large subset of message board members can waste an entire weekend on this one algebra problem.

11

“… Pam biking due east …”

A geodesic which starts out eastward will remain eastward only at the equator. Therefore we must ask whether it is the “due” or the “east” which has priority in the sentence. Since the “due” is otherwise redundant we can assume Pam is describing a geodesic. In the absence of any colored bear to help us hone in on the latitude, we should be able to jump immediately to a conclusion of “Insufficient Information”.

This is just as well, as Pam does NOT enjoy being approximated as a spherical cow.

12

We might need to know the exact time the travel occurred, in case the definition of the meter changed mid-trip. Per Wikipedia: " the International Prototype Metre remained the standard until 1960, when the eleventh CGPM defined the metre in the new International System of Units (SI) as equal to 1 650 763.73 wavelengths of the orange-red emission line in the electromagnetic spectrum of the krypton-86 atom in a vacuum".

13

But the foot was silly because it was based on the length of the/some king’s foot. : old rolleyes :

Up with Imperial! Down with metric!

14

Are you taking into account the accuracy of your calculator? Or are you calculating by hand? Irrationals will do you wrong.

15

The problem is intrisically flawed.

“Jim bikes 7 km/hr slower than Pam.”

This creates a problem. What speed are they starting out at? If zero then once they begin riding Pam (at least) has to instantaneously accelerate up to at least 7km/hr and then somehow both Jim and Pam accelerate in such a way to keep their velocities in relative sync.

Speaking of acceleration. The problem doesn’t mention constant speeds, just constant speed difference. Using cell phones and GPS they could be changing their velocities, possibly by a quite large degree, in union. And all this acceleration makes things even more uncertain.

And what if this camp site is near the North Pole? Jim is heading towards the pole but Pam may be be partially circling it. The great circle distance would be hard to calculate without knowing the initial latitude.

Are we even sure that the “4 hours” is absolute for each rider? What if Pam crosses a time zone? (Jim could cross one as well but that occurs in fewer places.)

16

It is clear that the original problem does not give enough information to calculate a solution that is just two physical quantities. At best, the solution for Pam’s speed would be a function of several variables, including, but not limited to, the latitude of the campsite, the elevation of the campsite above (or maybe below in some place like Death Valley) the theoretical geoid described by WSG-84, and the initial displacement between Jim and Pam at the campsite. Once the solution for Pam’s speed is determined, the solution for Jim’s speed is trivial. It is Pam’s speed minus 7 km/hr.

17

Please correct me if I’m wrong, but, ignoring possible relativistic effects, I don’t think the instantaneous velocity or acceleration of Jim and Pam really matters. The problem only concerns the average speed of Jim and Pam over each of their entire journeys. I do not interpret “Jim bikes 7 km/hr slower than Pam” as requiring that the magnitude of Jim’s instantaneous velocity be 7 km/hr less than the magnitude of Pam’s instantaneous velocity at any given moment.

18

At the risk of excess caution, I think inquiries are needed to determine how the participants calculate “average.” Suppose Pam travels at 12 km/h for two hours and then 20 km/h for the remaining two hours. I would call that 15 km/h average, but it’s not hard to imagine zealots who would insist that the average here is 16 km/h.

19

I’m compelled to say that I’m tired of Ynnad’s efforts to absolve Jim and Pam of the murder. After all, Jim was carrying the victim’s severed head in his bike’s saddle bag. And Pam was wearing a blood-stained T-shirt emblazoned with “I Killed [insert victim’s name here]”.

20

I find it very odd that Jim, a male, pedals 7 km/hr slower than Pam, unless Pam is exceptionally athletic, a fact which is not noted in the problem statement. This may sound sexist but the problem itself presents us with a case of reverse sexism. One must therefore conclude that Pam’s advantage stems from the fact that Jim is either quite old or otherwise enfeebled. That being the case, I find it inconceivable that either of them were capable of pedaling to the extent of achieving a 68 km separation in any period of time, let alone in just 4 hours, and that such an effort would probably cause Jim, in his enfeebled state, to drop dead from exhaustion within the first ten minutes.