Speed Race: Newsletter Challenge (April 2013)

If points A, B, and C are not equidistant, then the puzzle would not work as worded. That is, the third time they cross is at point A.

Imagine that the three points are situated such that point B is nearly opposite point A and slightly closer to A for car # 2 than it would be for car # 1.

Point C is somewhere between them. This is the non-equidstant situation you describe.

Now car # 1 must be slightly faster than car # 2 for the first cross to take place at B, but for the next pass to take place at point C mandates one or both cars radically change speeds because the distance from B to C is much, much proportionally less than the distance just covered from A to B for car # 2.

For Car # 1 it now must cover proportionally larger distance than it did at the first pass at B. So, this violates the constant speed requirement of the puzzle.

The only way the puzzle works is if all three points are equidistant along the track.

I understand what you're saying, and it's true, as worded, that the 3 points are equidistant, but that information is almost irrelevant. All you need to know is that car 1 goes around twice, while car 2 goes around once, when they meet at the starting point again. There's no point in complicating the math to arrive at the answer that car 1 is going twice as fast.

Can you prove that these points are equidistant?

Yes. First consider the requirements or constraints as given by the puzzle:

1. Two cars start moving in opposite directions from point A.
2. They are traveling at constant speeds on a closed race track.
3. The cars cross for the first time at point B,
4. then they cross at point C,
5. and a third time they cross is at point A.

In order for the above to happen one car makes one full lap and the other must make two. There are only three times the cars pass and because the speeds are constant the only way that works is if the points are the same driving distance from each other.

If you want to see more, look at post # 43 for a more in-depth analysis.

YES

Car 1 covers twice the distance (A-C-B anticlockwise) as car 2 in the same time (A-B clockwise)to meet at point B
Then again at point C (again twice the distance...B-A-C anticlockwise)
ERGO: car 1 travels twice as fast as car 2
50mph * 2 = 100mph

Three laps is the minimum car 1 will do.

one lap to B

one lap to C

one lap to A

Car 1 will leave A then pass A twice finaly to meet and return to A.

so 125 mph

Math is perfect, but does it say that Car 1 must be faster than Car 2? I think the answer is 25 mph.

edit. I didn't have the picture in my email. If the positions are fixed int he order shown then 100 mph is the answer.

Posted my first comment only having seen the text of the puzzle, without the diagram. If the diagram is official, and not added by a contributer later, Car 1, 25 mph doesn't work, and we are back to Car 1, 100 mph.

The diagram could be "official" - the real question is whether it is officially definitive or officially illustrative only ie showing what one solution might look like.

For reasons given before (see posts #86 and point 3 of #92) and additionally because this sketch does not have a unambiguously defined system of reference co-ordinates, I believe that without further clarification from the person who submitted the problem, we need to interpret the illustration merely as illustrative.

I think JD got it.

The logic here is that there are three points where the cars cross and the key clue is that they again cross at point A after their start.

Since the speeds are constant, the distance covered must also be constant.

This means that the points A, B & C are equidistant from each other.

Since the number of cross points is three, then the track is divided into three equal distance sections.

If the velocity of car 2 is 50 mph and gets 1/3 of the way around the track, Car 1 must cover 2/3 of the track in the same time period to make the cross.

That means car # 1 must also be going twice the speed of car # 2, or 100 mph. So, car # 2 gets to point A in one lap while car # 1 makes two laps in the same time.

This totally ignores the apparent fact that both cars can not start at point A with instantaneous acceleration (i.e., 0 mph to its steady state speed) in the real world, but this is not the real world or the track supervisor would black flag the car going around the track in the wrong direction.

I think JD got it.
I'm afraid he missed the word "minimum".

What kind of tires are on the cars?

90 mph... A to B is longer than B to C and C to A which are equal....It is longer by 1/3rd

er,,, a,,, 75 mph.....

What no hills to even the distance between A,B and C? (Going same constant speed up and down?) 75 mph uuuuuuuuuuuum, will have to think about that one?

Regards JD.

Maybe it's different constant speeds between different points of measurement....

what is the minimum speed of Car 1?

I think you may have a point? The answer relates to both cars arriving back at point A at the same time, with car 2 having a constant speed and car 1 a minimum overall speed? suggesting that the speed of car 1 may vary, but the answer is tied to starting and finishing at point A. And as the speed of car 2 is constant and car 1 only passes car 2 twice, car 2 is restricted to one lap at 50 mph, and as car 1 passed car 2 twice, its overall minimum speed is twice that of car 2. Then the answer for car 1 is 100 mph.

Regards JD.

Well if they both start and end at the same time, then the average would be 100 mph for car 2....but the speeds between points of measurement need not all be the same ....just constant....A to B could be 75, B to C could be 116.5, and C to A 116.5...constant speed meaning no allowance for time of acceleration...

An interpretation of constant? If the speed of car 2 varied passing points B and C and arriving back at A, having done one lap, it could be described in terms of its average speed. That a car had done one lap at X mph, is a reference to a constant speed, and as car 2 had only done one lap it was done it at an average or constant speed of 50 mph. But as car 1 has passed point A once, having done two laps, then its constant speed is averaged over two laps giving a minimum constant speed, therefore as you say the times and distances do not necessary have to be the same. But in relative terms one could think of them as being the same? as the answer is the relative relationship between car 1 and car 2.

Regards JD

Happy Easter High Flyer

Regards JD.

So that's how they get those ......

The second sentence of the puzzle states both cars are moving at a constant speed. Therefore, neither car can change velocity during the circuit.

I think the drawing is simply an abstraction. There is a mathematical solution to the problem, not geometric.

Since the rate of the vehicles is constant, albeit different from each other, the key is that they both arrive back at point A at some point, which implies that one car is going faster by some number of intervals over the other.

However, the velocity of car # 1 being 2X the velocity of car #2 is the only solution where they could cross for the first time at point B because they subsequently meet again at Point A after a second pass at point C.

The distance between point A and point B must be the same as point B to C for car # 2, just as the distance between point A and point B must be the same as point B to C for car # 1 going in the opposite direction.

The minimum speed for car #1 can only be met if all three points are equidistant around the track (that, divide the track length into thirds).

There are other solutions where points A, B, and C are close to each other on one side of the track, but car # 1 would need to be going even faster to make that scenario work.

Remember, the first pass must be where both cars have not yet completed a full lap. It's the only physical way that can happen. Subsequent passes are not restricted to a single lap, they could be any number of laps.

However, if the puzzle constrains us to finding a solution for the minimum speed, then the only solution to do that requires car # 2 to make one lap and car # 1 to make two laps to meet at point A for the final pass.

That requires point C to be the same distance from point B for car # 2 and therefore the distance from point C back to A for car # 2 is also the same distance, dividing the track into three equal segments. It's the only solution possible for a minimum car # 1 speed.

Car 1: 100mph

I think you need to show your work. :)

A-B-C-A are equal distances
Car 1 covers twice the distance (A-C-B anticlockwise) as car 2 in the same time (A-B clockwise)to meet at point B
Then again at point C (again twice the distance...B-A-C anticlockwise)
ERGO: car 1 travels twice as fast as car 2
50mph * 2 = 100mph

Prove it!!

The operative part here may be the first part of the question. "Two cars start moving".

Therefore at some stage, the minimum speed of both vehicles will be zero or very close to it after having left the starting line.

It is obvious that they used a rolling start as is used in 90% of races today so the timing would not start till they crossed the start/finish line at the constant speed.

For simplicity that us believe the track is 50 miles long. Car 2 takes 1 hour to complete 1 lap. Car 1 covers that distance 2 times in that same hour. There you have it, car 1 to car 2 = 2:1. Now if car 2 was driving the whole time with his left turn signal on (or right turn signal, for those in places where driver sits on the right) Car 1 would have slowed down 2 times (first at point B, then at point C) thinking that car 2 was going to turn across in front of him, that would change things up. That would have been something to think about.

As they "start moving" from point A, the minimum speed of car 1 is zero mph, just at the start of the "Speed Race".

Assuming no trick question (1st April) when the minimum speed is zero when they start, then by inspection of the diagram Car 1 must be going faster than Car 2 if it is to pass at B and again at C.

Car 1 will meet Car 2 again at A at the end of it's second lap just as Car 2 completes it's first lap.

That looks like twice as fast to me = 100 mph (or zero).

I am taking the assumption that both cars "start" at their steady state speeds. In reality you could do this with the track set with two extensions from point A where the cars could reach terminal speed via an acceleration lane and still technically "start" at point A.

The puzzle does not make this clear (any surprises?), but I would think that the larger picture here is the distribution of points A, B, and C and the steady state velocity of Car # 1.

In the diagram, car 1 going counterclockwise outside circle ABC would travel a greater distance than car 2 going clockwise inside circle ABC. Therefore, for both cars to meet at points a, b and c, car 1 would have to be traveling faster than car 2. If both cars drive directly on the circle ABC they would crash - probably before the race starts as they meet at point A the first time.

It says the cars are traveling at constant speeds. The minimum and the maximum are the same. One would have to precisely measure A to B clockwise and A to B counterclockwise. I get a ratio of approximately 11 to 18 with my compass.

X/50 = 18/11

11X = 900

X = 81.8

The answer is approximately 81.8 mph.

car2: A B C A B ...

car1: ACBACBA...

Car1 completes 2 loops while car2 completes 1 loop. Car1 must be going twice as fast as car2 or 100 mph.

...therefore it doesn't matter about the location of points B and C, as they only happen to be the places where the cars pass each other. Point A and the number of times they pass it defines the cars' speeds relative to each other, and points B and C just happen to be 1/3 and 2/3 the way round on account of the differing speeds.

Back in #15, my minimum speed of the other car was 100mph. But I could be wrong. Very wrong. Look below. So who missed the flying factor of the faster car? I think when it broke the speed of sound barrier, (about 13 loops depending on the height of the crossing loop tracks), and I ran out of fingers and had to slip off the shoes and socks, then I knew I should have gone to Walmart and bought those replacement batteries for the handheld (non solar assisted) calculator. Oh the life of a misunderstood engineer can be tough.

Yeah, then there is quantum tunneling...

Missed deadline? "posted later this month" - It's April now.

The fast one is the Russian Soyuz.

The Slow one is made by NASA.

Did you notice that the track is a big Easter egg painted yellow? I think it's an April fools joke, worded so that any answer can be called wrong.

Lets first review the requirements as laid down in the puzzle:

1. Two cars start moving in opposite directions from point A.
2. They are traveling at constant speeds on a closed race track.
3. The cars cross for the first time at point B,
4. then they cross at point C,
5. and a third time they cross is at point A.
6. Car # 2 goes clockwise (Red).
7. Car # 1 goes counter clockwise (Blue).
8. Car # 2 speed = 50 mph
9. The requirement is to find the minimum speed for car # 1.

There are three possible scenarios for this track:

1. Point B lies closer to point A for Car # 2 and point C lies somewhere after point B, but before point A.
2. All three points are equidistant from each other along the track length.
3. Point B lies further from point A for Car # 2 than it does for Car # 1 and point C lies somewhere after point B, and after point A.

Here is a schematic for scenario # 1. The problem with scenario 1 is that it fails the 5th requirement that they meet at point A for the third pass. In fact, they meet at a point called D, which lies somewhere in-between points B and A.

Scenario #2 meets all requirements and Car #1 has a speed of 100 mph or 2X car #2 because car #1 covers twice the distance as car #2. The third pass is exactly at point A.

The third scenario is where Car #2 must travel more than half way around the track to reach point B. This is interesting because car #1 must have a speed of less than car #2. This means that requirement #9 is met by beating scenario #2.

However, scenario #3 has the same problem as scenario #1 in that it once again violates the 5th requirement that they meet at point A for the third pass. the third pass takes place at point D on the picture, which is not at point A.

The conclusion is that only one scenario satisfies all requirements (in particular that the third pass must occur at point A) and that is where all three points are equidistant along the track - scenario #2.

In your third scenario you've conceded that A, B, & C might be in the opposite order around the track, but, you haven't gone to the logical placings at 120° and 240° (or 1/3rd and 2/3rd if any one's quibbling about the track not being circular) for C and B respectively.

There really isn't a difference for scenario 3 with equidistant points than scenario 2, except that Car #1 would not have a speed of 25 mph.

The fly in the ointment with that scenario is that it conflicts with the drawing in the puzzle by transposing points B and C. If the puzzle is really written as it should be/meant to be, then transposing the points is not an option.

As far as track shape goes, it doesn't make any difference is the track is circular, square, or looks like Sebring. The speed is constant, so it is strictly distance that counts, not shape.

What if the track is the entire surface of a 3D egg shape free floating object, and the requirement is just to cross at points on the surface but no path is designated only that they must start in opposite directions....?

See my similar argument in #29.

Of course, what if one car ran on the inside of the shell and the other on the outside. The fact that the cars actually crossed 3 times could be purely by chance, and what about the fluid drag inside the egg. Still need those batteries for the calculator.

Inside the egg!?! Why, why ,that's just, crazy talk! Next you're going to tell me the entire track is in a 'black hole' and the answer is relative.....

http://web2.0calc.com/

Yokes on you.

There are two answers to the question posed by the text. If the diagram is only there to help, and, might not be topologically accurate: then because the question asks for the minimum speed (implying more than one possible answer); I'm voting for 25 mph.

Just swap points B and C in the given diagram for the rationàle.

You may be right. If the original poster took a word problem and them embellished the problem with a cute Easter egg track picture, then they did so and spoiled the puzzle.

The fact that a diagram is included implies that the diagram contains relevant and non-conflicting information.

Since the order of the points are given in the picture, it then follows that this is the orientation of the points for the problem (unless it specifies otherwise in the word problem).

I think the diagram was included only to illustrate the problem, not to add additional constraints.

In general I take diagrams solely as advice unless the wording of the problem specifically refers to the details of the diagram. In this particular case we have already established that it is not to scale. If we can't trust the relative distances between A B and C as shown on the diagram, why should we trust the relative orientation of A, B and C, or that the directions shown for the cars as they move around the track are authoritative regarding the correct solution?

As well, if we accept the orientation information found in the diagram, there is no longer a need in the word problem to specify that the "minimum" speed is the desired solution - the diagram and problem description would then allow only one unique solution. Specifying the "minimum" solution technically over-constrains the problem, though fortunately a solution is still possible within that constraint.

I'm inclined to agree; the word "minimum" seems to clearly suggest two (or more) possible solutions, only one of which is the correct answer.

As an engineer I see it differently.

If someone describes hardware in writing and provides a conflicting schematic, then there is a defect in the document. How many times have we been slammed for that in a design review?

I suspect the original poster used or modified an existing word problem that never had a picture and then added their own diagram which had unintended consequences.

I agree that the requirement for minimum speed is a valid clue that the diagram is not really part of the problem, but we can only arrive at that conclusion after working the problem and discovering that there is only one solution that satisfies all the problem's requirements. At first one might naturally infer that non-equididtant points might generate other solutions, but they don't.

What was intended as something cute has turned into a wild goose egg chase. :)

I agree completely. If only they omitted the direction arrows.

I think we have more fun on these challenges. deconstructing imprecise or ambiguous challenges and then building new versions that are precise and unambiguous.

Would you expect less from engineers than to nit-pick things to death? :)

Respectfully, these challenges are not engineering specs. They're more like a barroom bet.

Or more precisely, this is a riddle that requires some facility with math rather than merely words. (the words are still important, at least, as we're translating from English to Math, and back)

As a riddle, I view the diagram as depicting a conceptual red herring - the "obvious solution", not the correct one. The name of the challenge as a "Race" is also a red herring - they're not racing, they're maintaining a carefully orchestrated spatial relationship.

True, they are not engineering specifications...

However, this is an engineering forum and we have been trained, and in some cases beaten, to be comprehensive, consistent, and correct with our data.

I still think this problem is ill conceived, particularly for an engineering forum.

if we accept the orientation information found in the diagram, there is no longer a need in the word problem to specify that the "minimum" speed is the desired solution - the diagram and problem description would then allow only one unique solution.

No, 100 mph is the minimum, but 250 mph is also a solution. see my #70.

Car 1 (C1) moving at V1=250mph is not a valid solution to all the verbal constraints.

The crossing/passing at B must be the first one. Similarly, the meeting back at A must be the third crossing.

If C1 is moving at V1=250 mph and C2 at V2= 50 mph the first instance of the cars crossing would not be located at B, but a point halfway between A and B. B would be the location of the 2nd crossing of paths and they would not meet back at A until their 6th crossing, not their third as specified in the problem statement.

It depends on how you define crossing. I think the OP meant "crossing at a letter".

"The cars cross for the first time at point B, then at point C, and a third time at point A."

Interesting take.

This posting contains three different threads of thought; one addressing additional solutions gained by allowing crossings not at points A, B and C to go uncounted, a second being a discussion of the definition of "crossing" with added thoughts and the third being another way of looking at these challenges.

1) Non-counted crossings occurring away from points A, B and C also allows for slower speeds than 50 mph using a similar methodology to that which gave 250 mph. If C1 goes 2/3 of the way around and C2 goes 7/3 of the way around they will meet at B (after going past each other twice at points between A and C then between C and B respectively). The speed ratio V1 to V2 is 2:7 for a V1 of ~14.3 mph. This method will also give diminishing speed ratios of 2:13, 2:19, 2:25 with the minimum V1 being a limit approaching zero. I don't think this is what the problem intended.

(note in the previous paragraph that I excluded ratios of multiples of 2 such as 2:10 and 2:16; such ratios would cause the cars meet first at C rather than B)

2) I do find it interesting that you were considering points B, C and A to have an existence apart from the fact that they are starting and crossing points; for me the only reason why they are of enough interest to label is because they are the first, second and third crossing points respectively.

"Crossing" has a perfectly useable default definition - any time the two cars go past each other. It is unnecessary - and presumptive - to assume a special definition in order to permit additional solutions to the equations.

3) One might accuse me of the same thing - how I have assumed that the diagram is not prescriptive in order to obtain the 25 mph solution. The difference is that my assumption is based on some experience; I have seen similar brain teasers which have re-drawn their diagrams to illustrate the actual solution. It's part of the genre, sort of the same way SciFi books and movies frequently involve faster-than-light travel.

Consider it this way: This question does not come from a client who needs our help and so has worked to make things as clear as possible, but a challenger who is actively trying to best us. If the problem was illustrated as per the 25 mph solution it would give the puzzle away - not much of a challenge.

The word problem is completely solvable (ie has a reasonable unique solution) without the diagram.

Good answer but for the 25mph solution the first time they pass each other would be at point C.

Agree, you can also swap the direction the cars take and leave the points as they are, and you get the 25mph. Whoever planted this Q' is enjoying the comments....just for show. Very vague!

The minimum speed of car 1 is the speed it started at, which is zero.

And what if it rolled backwards .... at the start ... is that 'less than zero'?

Or the track was also moving?

I like your thinking outside the box .. or should I say egg?!

It would be good if some of our fellow travellers were also a bit more open in their thinking, particularly when dealing with today's (engineering design) challenges.

I used to get a lot of these word problems in school and I really hated them. That must be why the schools would keep pilling them on.

However, one thing that got drilled into me after some time was that these word problems did have some undocumented rules.

One, all the relevant information was there and no leaps of faith were required, you just needed to read them carefully.

Two, there was only one correct solution for the problem.

Three, if the solution seemed to be excessively complex you were probably on the wrong track.

The good news is these word problems did help us develop out-of-the-box thinking habits.

A friend sent me this. I actually liked word problems, but I can sympathize. I had a lot of friends who saw word problems this way:

depends on the effectivness of the fire you light witih the pencils!

looking from (moving) track or above the track?

If both cars started at zero, then they would not have been traveling at constant speed.

At least one car was going 50 mph.

The intuitive answer is correct in this case but the arguments or proofs are interesting.I looked at it as three distinct stages; we can say that the start occurs at the instant when the cars pass point A simultaneously until the first pass at point B, the second stage is a duplicate of the first stage except for the node point designators and then the third stage is another copy. This is proof that the course is divided equally into three lengths. Later insight says that this was not necessary, given the sequence, it is obvious that the #1 car travels twice as far as the #2 car in the same time.

150 m.ph.

"Air is for free, Thunder, you pay for! live life!

I performed the same calculations as JD but no where does it state which car is moving faster. From our analysis the faster car is moving twice the speed of the slower car. The minimum speed of car 1 would be if it is the slower car, thus the minimum speed would be 25 mph.

The original problem was stated without a diagram. It says nothing about acceleration. It merely says that Car 2 is moving at 50 mph and that Car 1, moving in the opposite direction on a closed track (not necessarily round) passes Car 2 twice before both pass each other at the origin. Assuming constant speed for both cars, given that there is no mention of varying speed, and in order for both cars to eventually meet at the origin and twice before that, the three points, A, B & C must be equidistant. There are two possible results to this problem. If Car 2 covers 1/3 lap when Car 1 passes, then Car 1 must have been going twice as fast as Car 2. When Car 2 covers another 1/3 lap (2/3 lap total) then Car 1 will pass again where it will have covered 1-1/3 (4/3) laps. By the time both cars meet again, Car 2 will have traversed 1 lap and Car 1 will have traversed 2 laps (6/3 laps). To do this, Car 1 will have to have been moving at 100 mph. If Car 2 travels 2/3 lap when it meets Car 1, then Car 2 will cover 4/3 laps when it meets Car 1 again and finally 6/3 or 2 laps when the two cars meet for the 3rd time (point A). In this scenario, Car 2 will have covered 2 laps in the time it took Car 1 to cover 1 lap. Therefore Car 1 will have been moving at 1/2 the speed of Car 2. In the second scenario, Car 1's speed is 25 mph. 25 is less than 100, therefore, the minimum speed of Car 1 is 25 mph.

Ah, Ha!

So, the diagram is the author's embellishment!

Without the diagram the puzzle makes much more sense and I agree that 25 mph is the right answer.

Good catch!

Good catch!
I'm not sure that it is... it is more of an opinion. The word "minimum" just seems to suggest two solutions.

Even if the diagram is the posters embellishment it is what this particular puzzle is based on. In this particular puzzle the direction of each car is shown, the order of the passing points is dictated and the speed of 1 car is stated. Given these conditions your original speed of 100mph is the only answer I can see.

Jim

I agree but only IF it is word puzzle alone. The addition of a diagram ( a picture saves a thousand words ) changes things. Without the picture to illustrate salient points the word puzzle would have to read:-

Two cars start moving in opposite directions from point A. They are traveling at constant speeds on a closed race track. Car 2 travels clockwise and encounters point B first. The cars cross for the first time at point B, then at point C, and a third time at point A. If the speed of Car 2 is 50 mph, what is the minimum speed of Car 1?

Admittedly not a thousand words but i still believe a picture helps clarify the question.

Jim

You wrote, "I still believe a picture helps clarify the question."

Or, as I have claimed, confounds it!

It is probably intended that the "start" at point A is with the cars already in motion at their constant speeds. Then, acceleration is taken out of the question. The cars subsequently meet again at B, C, and A. So one is going twice as fast as the other (travelling one third of the course while the other travels two thirds). Therefore, the minimum speed of car 1 would be 25 mph, half the speed of car 2.

The word "minimum" is a key.

I would like to change my answer to, car 1 needs to travel at a minimum of 99.999996

MPH.

Well done K_Fry: The question uses the word "minimum". The ratio is 2:1 so although there are 2 possibilities, the "minimum" is 25kph

My 2c worth, if it is worth 2c???