# Is zero speed a constant speed?April 21, 2005 7:33 AM   Subscribe

Mathematically, can a speed of zero be considered a "constant speed"?

I teach sixth grade math. Recently, on one of the district tests, there was a multiple choice question asking students to choose which one of four graphs matched the description "Maria biked at a constant speed." All four graphs had time on the x-axis and distance on the y-axis.

Graph A was a zig-zag of different slopes - clearly not constant.

Graph B was a straight diagonal line - roughly y=x - clearly constant.

Graph C was an exponential increase - roughly y=x^2 - clearly not constant.

But Graph D was a flat line - roughly y=4. (Maria's speed is zero).

The answer key for the test said that Graph B was the (only) correct response. In some ways, that makes sense to me, for in the real world we wouldn't say "Maria biked at a constant speed" if Maria is standing still like she is in Graph D. But I was concerned, because in all the definitions of "constant speed" I'd seen (for example, "the object will cover the same distance every regular interval of time") I'd never seen anything that demanded the distance for every interval be non-zero.

I was inclined to treat either Graph B or Graph D as a correct response. The school district math coordinator disagreed, refering me to Newton's First Law. I'm assuming by this she meant that Newton's First Law implies that an object in motion and an object at rest are two different things. Is she right and I'm wrong? Or is this a question on which reasonable mathematicians can disagree? (It's only an academic question now, since in fact none of my kids chose Graph D on their tests).
posted by Chanther to Education (64 answers total)

You're correct, zero speed is constant speed.
You're supervisor's talking out of her 455.

But please don't call y=x2 "exponential increase".
posted by Wolfdog at 7:40 AM on April 21, 2005

Yep. 0 is just as good a constant as 1 (better, sometimes). You're right, mathematically. "Real-world--wise", we probably wouldn't ever say that Maria was "biking" if she wasn't going anywhere (except that I have this mental image of her doing that thing that showoffy cyclists do at stoplights, where they stand up on the pedals and balance in place, so as to save the extra half-second it would take them to get going when the light turns green).

<pedant>y=x^2 is polynomial, not exponential</pedant>

Ah, out-pedanted by Wolfdog.
posted by gleuschk at 7:44 AM on April 21, 2005

Graph B described a situation where Maria's speed was some unchanging number, C.

Graph D also described a situation where Maria's speed was some unchanging number, C; that case we happen to know C = 0.

And it's not merely an academic question: "your" kids may not have chosen D, but they may have had to spend more time on the question, trying to figure out why the promised test constraint, that only one answer per question was correct, was violated.

I remember as a kid being really annoyed by questions crappy questions like this; they usually appeared in similar cases, where the test-preparers decided that one answer was so obviously correct that they didn't have to think about any of their alternate, made up wrong answers. And inevitably, I'd waste time on them, trying to figure out what the trick was, when it was just a stupid, lazy ass test-preparer.

So thank you for going the distance on this one for your kids.

(And yes, I was one of those kids who always did real well on any standardized test.)
posted by orthogonality at 7:54 AM on April 21, 2005

well, if you want to pull rank, in the opinion of someone with a first class degree (natural sciences - the old fashioned name for their science degree, which in my case meant physics) and a phd (astronomy) from cambridge, she's wrong. a value of 0 is completely valid as a speed (i'd worry about -ve speeds though - if such things occur you really need to start introducing velocity and vectors).

in a sense, she's right, in that in newton's time it might well have been thought that "moving" was fundamentally different from being "at rest" and not simply an accident depending on the observer. but things have changed since then - it is a fundamental tenet of relativity that inertial frames are equivalent (basically, that you can add or subtract any value from a constant speed - including the value you started with, leaving you with something "at rest" - and you don't change the physics).

now she might come back and argue that you're not teaching relativity. that's not the point. we keep newton's laws not because we're interested in the ancient history of science, but because the bits we keep are consistent with relativity in everyday situations (more accurately, because they're consistent with our best understanding of how the world works, whether that's special relativity or soemthing else). in other words, we keep the bits that we believe are right. we don't teach kids the associated errors (for example, it's quite possible that newton believed in alchemy, but we don't teach that at school). so newton may well have thought that a speed of zero was somehow special. but current scientific opinion does not, and you are teaching science, not history.

however, having said all that, my experience of school was that teachers and tests just aren't that smart. and anyone who is smart enought to recognise that zero is a valid answer should also learn that it probably isn't the answer that the dumb teachers/test wanted. there's a difference between being right and getting things done, and this difference is soemthing it's worth learning as soon as possible.
posted by andrew cooke at 7:55 AM on April 21, 2005

we keep the bits that we believe are right

no, not exactly. we keep the bits that we believe are a useful approximation to what is right.
posted by andrew cooke at 7:57 AM on April 21, 2005

pedants: is y=2^x exponential? If not, what is?

And Chanther-- I too salute you.
posted by Kwantsar at 7:58 AM on April 21, 2005

Would she be "biking" is she was sitting still?
posted by cillit bang at 7:58 AM on April 21, 2005

Speed is the magnitude component of velocity (a vector). An object can have a constant velocity of zero relative to a point. Thus, its speed is zero. Therefore, its speed is constant.
posted by mischief at 8:00 AM on April 21, 2005

oh yeah, that exponential reference is a blooper. the phrase you want is "polynomial" or "quadratic".

on preview: yes, 2^x is exponential (or "power law"). the thing you twiddle has to be on top.
posted by andrew cooke at 8:01 AM on April 21, 2005

(fwiw, 2^x = e^(log 2 + x), or something like that.)
posted by andrew cooke at 8:02 AM on April 21, 2005

arrgh no. ignore that.
posted by andrew cooke at 8:03 AM on April 21, 2005

ok, i *think* it's e^(log 2 * x). sorry!
posted by andrew cooke at 8:05 AM on April 21, 2005

ok, i *think* it's e^(log 2 * x)

You're right this time. The real question is why e^{stuff} is supposed to make any more sense than 2^{stuff}. The answer has to do with an amazing series of coincidences1: that defining

exp(x) = Σ0\infty \frac{x^n}{n!}

gives a function2 that has all the properties we usually associate with exponentiation; that in fact, it has all the properties that we associate with exponentiation from a certain base, somewhere in the neighborhood of 2.718; which specific number also happens to be the limit of (1+ 1/n)^n as n gets very very large; and on and on.

1Yes, of course they're not coincidences at all, but inevitable consequences of the way we've set up our axioms. But still, they're fantastical. Whoodathunk, ya know? I don't think we as math professors do any kind of adequate job conveying how unbelievably cool this circle of ideas is.

2Even this part sometimes freaks me out.

posted by gleuschk at 8:18 AM on April 21, 2005 [1 favorite]

you can add or subtract any value from a constant speed
should be "velocity", not "speed" there, really. sorry to keep posting, just don't want you to have holes in your argument.

posted by andrew cooke at 8:20 AM on April 21, 2005

What cillit bang said. A speed of zero isn't "biking at a constant speed," it's standing (or possibly balancing) at a constant speed. No greek letters required.
posted by revgeorge at 8:23 AM on April 21, 2005

oh, there's two arguments here. one, that if the question says "biking" then she must be moving. the other that "zero speed" is a valid concept in physics.

i'm only arguing the point of physics - particularly the response of the coordinator. but if it comes down to arguing about whether "biking" can include track stands then this is a joke, not a physics paper.
posted by andrew cooke at 8:37 AM on April 21, 2005

"then this is a joke" : Most standardized tests are jokes. ;-P
posted by mischief at 8:42 AM on April 21, 2005

yeah - see the last para of my first comment. on reflection, i think the "biking" argument is a pretty strong argument for the question given to the students having a single answer. pity that's not what the coordinator said, though.
posted by andrew cooke at 8:50 AM on April 21, 2005

I remember doing something like this on a test. I couldn't balance an equation properly, so I multiplied both sides by 0 and moved on. I think I got credit. I know I was told not to ever do that again.
posted by trondant at 8:51 AM on April 21, 2005

As others have said, speed should be described as velocity.

Velocity is a vector, which needs to have a direction and magnitude.

According to this page, "A vector which is not null is a proper vector."

If a null vector is not a vector, and speed needs to be represented by a vector, then I would propose that the zero speed is not a speed.

I'm willing to debate this.
posted by jonah at 8:52 AM on April 21, 2005

"proper vector" doesn't mean "a decent, upstanding, correct vector". "proper" is a technical term that means "non-zero". so what you read as "verctors are non-zero" was really a definition of what "proper" means when used to describe vectors. so your argument is "non-zero vectors have none-zero speeds", which doesn't help.
posted by andrew cooke at 8:57 AM on April 21, 2005

No, I am now completely convinced that I am correct, that zero speed is not a speed.

A vector must have magnitude and a direction. Without a magnitude, there is no vector. How would you describe the direction of a zero vector? You can't give it an x value and a y value, because they are zero as well.
posted by jonah at 9:03 AM on April 21, 2005

Zero is most certainly a speed. It seems like many people in this thread (and your district math coordinator) need to take (or retake) high-school physics.

And trondant, please don't multiply both sides of an equation by zero. That hurts my brain.

On preview: jonah, a vector with magnitude 0 and an arbitrary direction has a magnitude and a direction, and is a vector.
posted by oaf at 9:06 AM on April 21, 2005

Do you consider zero force a force oaf?
posted by jonah at 9:14 AM on April 21, 2005

Newton's first law doesn't come into analyzing this problem: You're looking at a bunch of functions graphed in a coordinate plane and determining which ones have d/dx (function) = C over the whole graph. This isn't physics, it's basic calculus [albeit at a level sixth graders can understand, and without all that fancy derivative stuff.] All the bits about Maria are just flavor text, unless you feel like quibbling over whether one can "bike" at 0mph. The key to the problem is this: d/dx (C*x + b) = C, so you know you're looking for straight lines. Written in full linear form, the relevant equations are y=1*x+0 and y=0*x+4. Thus, in one case C=1; in the other, C=0. The coordinator's wrong.

And jonah, I must disagree. When I took multivariable calculus with vectors a few years ago, we learned that the absolute value of a vector's magnitude was >= 0. A vector can have a direction and a magnitude of zero. [In fact, among the basic algebraic vector laws we learned: the cross product of a vector with itself is a new perpendicular vector with a magnitude of zero, and the cross product of any vector with a vector that has a magnitude of zero is a new vector [perpendicular to both] with a magnitude of zero.]
posted by ubersturm at 9:15 AM on April 21, 2005

A force of magnitude 0 is still a vector in the mathematical sense. It's not really up for debate. It's still a vector. And moving nowhere is still moving with constant speed.
posted by oaf at 9:24 AM on April 21, 2005

And yes, zero force is still a force. We may, in everyday life, refer to a force [or speed, or whatever] of zero as being no force [or no speed, etc.]. However, physics and mathematics require a much more rigorous and formal way of approaching things. In those cases, a magnitude of zero is still a magnitude [although the fact that it's zero may obviously have slightly special implications, as in the cases of the vector cross-product rules that I quoted above.] Getting back to the AskMe question: the problem posed to the kids is a math problem, but the testmakers seem to want the kids to approach it from a commonsense, rather than a mathematical angle. And there's the conflict.

[Er, to be clarify my previous post: A vector can consist of a magnitude of zero and a direction. I wasn't implying "a direction of zero", whatever that means.]
posted by ubersturm at 9:28 AM on April 21, 2005

Okay, I understand that a zero magnitude vector is still a vector. Can you explain to me in a formula how you would describe a zero velocity vector?
posted by jonah at 9:37 AM on April 21, 2005

However, physics and mathematics require a much more rigorous and formal way of approaching things. In those cases, a magnitude of zero is still a magnitude

Expanding on that, in mathematics and Newtonian Physics, you must be able to describe, absolutely, the direction and magnitude of a velocity vector, right? How do you do that if the magnitude of the vector is zero?
posted by jonah at 9:40 AM on April 21, 2005

jonah, one typical notation is to use (x,y,z) for vectors. so (0,0,0) is a vector with zero magnitude. it doesn't have a direction. maybe you think that's against the rules - as far as i'm concerned "magnitude and direction" is another of those guides you learn at school that over-simplifies things (when you're teaching this, the direction is important because, compared to speed, that's what's "different" about velocity, hence the emphasis).
posted by andrew cooke at 9:56 AM on April 21, 2005

ubersturm is right on with the calculus definition of "constant speed" ie. the first time derivative of the position = a constant.

jonah - I think your argument might be valid in set theory, where there is a difference between the null set (has no value) and a set where the only value is zero. But in this case, there is no distinction between a vector with no magnitude and a vector with zero magnitude. In practical terms, a physics/engineering/math student would simply write the vector as 0i + 0j + 0k (if they were in 3-space).
posted by mbd1mbd1 at 9:58 AM on April 21, 2005

jonah: I'm sorry, but you're simply wrong.

A vector space is a set over a field (such as the real numbers) that has two operations (call them + and *) and a number of properties, but the important one here is that the space is a commutative group under +.

This implies (a) that there is an additive identity: the zero vector (b) that for any vector v, there is an additive inverse -v such that v + (-v) = 0. c) the vector space is closed: you add two vectors v + w and the result is always a vector, which, combined with b should show, again, that 0 is a vector.

So the 0 vector is a perfectly valid vector, whether you like it or not. (I encounter the same sort of objections -- albeit less sophisticated -- when people insist that zero's not a number.)

In any case, it's just confusing the issue here; we don't even need to deal with directions, we're just dealing with magnitudes. D is a valid answer, unless you get into semantics over what "biking" means. (And if you insist that "biking" means a non-zero velocity, I'll ask whether Lance Armstrong is "biking" when you're keeping pace with him in a car at a constant velocity of 40 mph.)
posted by cgs06 at 10:07 AM on April 21, 2005

argh! - the dreaded watching lance from a car argument! :o)

you have to remember context. this is for children. it has to be expressed in their language using concepts they are familiar with. on the other hand, tests should be written so that they don't confuse in the future. ideally, someone with a lot more knowedge than even the smartest child should write the test, because of traps like this - something that is non-intuitive, that a small fraction of the students will learn later in life.

is zero speed "ok" as physics? yes.
was there a moderately obvious best answer to the question? yes.

ps thanks for a decent description of vectors - i tried writing something and realised i was out of my depth so deleted it.
posted by andrew cooke at 10:19 AM on April 21, 2005

All this stuff about the definitions of vectors is pretty irrelevant. If you're not moving, you're not biking.

Next.
posted by recursive at 10:31 AM on April 21, 2005

What I find most disturbing about this is the math coordinator's apparently shockingly poor understanding of Newton's First Law. A body will remain at rest or in uniform linear motion unless a net external force acts upon it (really just a special case of the Second Law, where F=0).

The question, as asked, is kinematics, requiring no knowledge or application of mechanics. The mechanism for maintaining constant speed is irrelevant to the correct answer.
posted by normy at 10:45 AM on April 21, 2005

If you're not moving, you're not biking.

So, if you're performing a track-stand, what are you doing, if not "biking"?
posted by normy at 10:51 AM on April 21, 2005

Well, maybe she's biking as hard as she can, but it's windy and the ground is slippery so she's being pushed back as quickly as she would be going forward, giving her a speed of zero. Or maybe she's on a stationary bike.

It's not like the test tells you these things.
posted by mai at 11:06 AM on April 21, 2005

Maybe I have a shockingly poor understanding as well, but if Newton's First law says that a body in motion stays in motion and a body at rest stays at rest unless an outside force is applied to the body, doesn't that mean that the two states can not be the same?

I tend to interpret Newton's First as saying that bodies have motion or they don't have motion. They don't change that state unless acted upon.

What confuses me is that when I use vectors for things like forces or shear stress, I would never consider a zero value as a constant.

The basic F=ma formula requires that the force have some value, or it doesn't exist. You wouldn't say that there is a force in the amount of zero being applied to a body, you would say that there is "no" force being applied to a body. There exists two cases, a case where a force is being applied and a case where no force is being applied. Just as with the moving body. There is a case where it has a velocity and a case where it does not have a velocity.
posted by jonah at 11:09 AM on April 21, 2005

p.s. I am enjoying this thought process, I'm not (intentionally) being a pain in the ass.
posted by jonah at 11:10 AM on April 21, 2005

If you're not moving, you're not biking.

How often do you measure the distance travelled? Start from point A, bike 50 yards in any direction and return to point A. Measure your distance from your starting point. Have you moved? Have you biked?
posted by forrest at 11:12 AM on April 21, 2005

So, if you're performing a track-stand, what are you doing, if not "biking"?

Waiting for the light to turn, I'd guess.
posted by recursive at 11:17 AM on April 21, 2005

This comment is mostly to jonah, despite his already being enlightened (I hope) by andrew. In mathematics, the "right" definition is the one that makes the theorems easiest to state and remember. So for example, we prefer to say that 1 is not a prime in order that we get the theorem "every positive integer can be written as the product of a unique list of primes, up to reordering" rather than the uglier "..up to reordering and adding or removing a bunch of 1s". The mathematical content is the same -- people could work with either definition, and make the same discoveries -- but one is prettier.

In the case at hand, "a vector has a magnitude and a direction" would work perfectly well for many things; it would be a little more subtle than the definition most people like (which is better encoded by things like (x,y,z)) because there'd be many different flavors of "zero vector". The two annoyances would be that (1) you'd have to decide, whenever computing something like A+B+C a sum of vectors, which zero vector you get if the magnitude of the sum is zero. But fine, say you come up with some rule to make that choice. (2) when you had an equation A+B=C+D, you'd probably want to say "either A+B=C+D, or both sides have zero magnitude".

So, it would work, but it's yucky. Hence people don't do it.

There are many examples where people have worked with one definition for decades, and eventually realized it's yucky and changed it, but they're probably not worth explaining for a general audience.

On preview, again to jonah: yeah, Newton splits those up into two laws. Which, nowadays, we regard as ugly. A body without forces applied to it "tends to remain" at its original velocity -- zero or nonzero.
posted by Aknaton at 11:23 AM on April 21, 2005

You wouldn't say that there is a force in the amount of zero being applied to a body,

You might not, engineers might often talk like that, however.

you would say that there is "no" force being applied to a body.

You're making a semantic distinction, not a physical one. Physically, they're the same.

I would never consider a zero value as a constant.

Maybe that's your problem. Zero is just as much a magnitude as any non-zero value. It just happens to be zero.
posted by normy at 11:25 AM on April 21, 2005

Mathematically, a speed of 0 is a constant speed, and there seems no possible argument against this, so the attack is on the word "Mathematically" to make it mean "conceptually" or "in Physics" or "in well-formed meaningful physics, kind of". I guess these are okay for conversation -- is a book resting on a table falling at a constant speed? Is it moving at a constant speed? Since its a is 0, does that mean that the sum of all Fs on the book is 0, or does that mean that there are no Fs acting on the book?

But this isn't a test of physics or semantics. Chanther's right, and we should all go to the district math coordinator's house and jeer "bad question bad question".

Which of these functions is constant?
a) f(x) = | sin x |
b) f(x) = 1
c) f(x) = 2x
d) Chicken
posted by fleacircus at 11:36 AM on April 21, 2005

You might not, engineers might often talk like that, however.

I am an engineer, and I haven't ever heard someone say "a zero force is being applied". I have heard that the sum of the forces is zero, a force goes to zero (and is removed from the equation), but never that a zero force is being applied to something. A zero force in an equation results in force being taken out of the equation.

On preview - fleacircus, that's an easy answer about the book, the sum of the forces is zero, there are forces being applied to the book.
posted by jonah at 11:41 AM on April 21, 2005

I am an engineer, and I haven't ever heard someone say "a zero force is being applied".

Sure you have. People have said "A force F is being applied, and blah blah blah". Then they get a result that is valid for all F -- even zero! Don't single out cases as being special unless you have to.
posted by Aknaton at 11:49 AM on April 21, 2005

Jonah - For a vector A, A is the sum of three vectors, A1, A2, and A3. Each of those is a magnitude [i.e., a1, a2, and a3] multiplied by one of the unit vectors [i, j, and k.] A can thus be represented as a1*i +a2*j+a3*k. In multivar., the special case where A is the zero vector is by definition just described as the zero vector, but you can also write it as 0i+0j+0k. If you do the calculations, you start to see that andrew cooke is right - it doesn't actually matter all that much whether you can give the zero vector an absolute direction. After all, the dot product of a vector and the zero vector is zero, the cross product of a vector and the zero vector is a zero vector, a scalar quantity multiplied by the zero vector is the zero vector... And so on. Due to the magnitude of zero, pretty much any interaction between the zero vector and another vector gives either zero, a zero vector, or the original other vector. So "direction" is pretty much a moot point in this case.

Regarding Newton's law: when you think about it, a better way of putting it would be that the state of the thing doesn't change. Without an outside force, you will have constant velocity, whatever that velocity is [including zero.] The traditional way of stating the law creates a false dichotomy between stationary and moving things, given that the point of the law is that in both cases, the object's acceleration is zero without the help of an outside force. As for your other comments: you're again confusing "magnitude of zero" with "doesn't exist." When I say that the force on an object is zero, I'm making a statement that the object's acceleration is zero, just as when I give a non-zero value for the force, I'm telling you [quite accurately] that the object is accelerating at a non-zero rate. If I refuse to believe that there is some sort of value called "force" attached to the object, I can't come to any useful conclusions. Dragging a huge mental wall between "zero velocity/force/etc" and "non-zero" is like saying that zero can't be a number because if you have zero things, you have nothing, whereas with [positive] numbers, you have a countable number of things. It's a semantic difference that's tripping you up, and I'm not really sure what I can say to convince you that in physics and mathematics, "zero force/velocity" and "no force/velocity" are equivalent statements. Similarly, if you can't regard a velocity of zero as constant [even when the function describing the velocity, and the derivative of that function which proves that the velocity is constant, follow all of the rules of calculus, arithmetic, etc], you're choosing to disagree with the way math works. Your prerogative, I guess, but unfortunately, you're wrong. In practical applications, you're free to ignore forces that are zero, but then there are always differences between the way things are done in theory and in real life.
posted by ubersturm at 11:55 AM on April 21, 2005

Forrest: How often do you measure the distance travelled? Start from point A, bike 50 yards in any direction and return to point A. Measure your distance from your starting point. Have you moved? Have you biked?

I don't know about you, but I don't even need to measure. I can just tell. It is just intuitively obvious to me when I'm biking. I guess I must have good genetics.
posted by recursive at 11:57 AM on April 21, 2005

jonah:I would never consider a zero value as a constant.

Ah, this is indeed the problem. How about this: do you agree with the statement "A function is constant if and only if its derivative is zero"? Apparently not. You'd insist "No, it's either jonah-constant (meaning constant and nonzero), or it's zero."

I don't mean to claim that this way of thinking is "wrong" -- I'm sure you can do all the same math anyone else could, using this version of the definitions. But I will claim that it's nonstandard, and I'll also claim that it's yucky. To get back to the original thread, please don't teach it this way to sixth graders. I know jonah is not Chanther
posted by Aknaton at 12:05 PM on April 21, 2005

and I haven't ever heard someone say "a zero force is being applied"

Maybe that exact wording isn't often used in the definition of a problem, but when I was in college our mechanics professor used to deliberately say this sort of thing all the time, in various ways and contexts, just to make us stop to think about it. He was right to. Regardless of language, the ability to conceptualize forces of zero magnitude is essential to the understanding of mechanics.
posted by normy at 12:07 PM on April 21, 2005

If you're not moving, you're not biking.

Bullshit. Three times a week I go to the gym and bike for a half hour, and the bike never moves an inch. Unless the question specified that Maria is not riding a stationary bike, she could easily be biking without moving.
posted by mr_roboto at 12:14 PM on April 21, 2005

All the talk about vectors is meaningless since velocity is a vector, and speed is the magnitude of that vector. You can't have a negative speed, but you can have a negative velocity.

You sure as hell can have a zero speed. And it's constantâ€”saying zero is not constant is crazy talk.

It's obvious Graph B is the right answer despite its problems. And the question is badâ€”graph D should be left out.
posted by grouse at 12:34 PM on April 21, 2005

grouse, I don't think graph D should be left out. It's probably there to test if students understand the difference between constant speed and constant position.

I'd say the question simply needs to be reworded to indicate that Maria was actually moving, that she had a non-zero speed.

But it's definitely a poorly constructed question.
posted by ldenneau at 12:53 PM on April 21, 2005

it struck me, on the metro just now, that in some odd corner of physics (evanescent waves in quantum mechanics?) you could argue that speed is imaginary (or, if you want to take "magnitude" in a certain way, negative :o)
posted by andrew cooke at 1:15 PM on April 21, 2005

jonah - i think newton may have thought that way too. unfortunately he was wrong. :o) see my earlier comment.
posted by andrew cooke at 1:17 PM on April 21, 2005

I don't know about you, but I don't even need to measure. I can just tell. It is just intuitively obvious to me when I'm biking. I guess I must have good genetics.

That "I can just tell" is just measurements being taken by your sense of balance, sight, hearing, or touch. Such measurements may not be accurate enough to tell you exactly how far you've moved, but they're measurements all the same.
posted by juv3nal at 1:22 PM on April 21, 2005

Chanther, you are absolutely correct. Zero is just as valid a constant as any other value either in mathematics, physics or engineering. The graph represents d as a function of t. Mathematically, speed is the first derivative of the function and is represented by the slope at any point on the graph. Any straight line on the graph (except vertical) means that the slope and derivative are constant and therefore the speed is constant. A slope (speed) of zero is perfectly valid.

All this talk about vectors is irrelevant. Speed is a scalar representing a magnitude. It tells you nothing about direction. In fact, she could be travelling on a circular track.

Likewise, Newton's First Law is irrelevant. After all, there is nothing in the problem about force or mass. How could F=ma be involved?

And if you want to talk about acceleration, that would be the second derivative of the function, and that is zero, exactly the same, for both graph B and graph D.

Since we know that a horizontal line represents a constant speed, the argument boils down to the semantics of the word "biking". If that is the case then this is a very poorly worded mathematics problem.

This could become a lesson for students that life is not always fair, not every wrong will be redressed, people make mistakes, what you understand and believe is more important than what someone else says, the teacher is not always right, etc.
posted by JackFlash at 2:14 PM on April 21, 2005

i don't understand all the math here. it seems a pretty simple matter of definition.

Zero is the lack of speed. Thus, she is not biking at any speed if she is not moving. Zero ? speed. So I don't understand how zero could possibly = any definition or adjectival variation of speed.
posted by luriete at 4:06 PM on April 21, 2005

You math junkies are sick. SICK!

Get help.
posted by SlyBevel at 4:43 PM on April 21, 2005

I don't know about you, but I don't even need to measure. I can just tell. It is just intuitively obvious to me when I'm biking. I guess I must have good genetics.

I don't know which is more simpleminded here: The Force applied to scientific reasoning or The Theory of Genetic Superiority.
posted by forrest at 4:55 PM on April 21, 2005

i don't understand all the math here. it seems a pretty simple matter of definition.

I understand all the math here, and completely agree with you.

Zero is the lack of speed. Thus, she is not biking at any speed if she is not moving. Zero ? speed. So I don't understand how zero could possibly = any definition or adjectival variation of speed.

I'd much rather say "zero is the amount of speed you have when you're not moving". It's just another speed: 55 mph, 20 mph, 0 mph.

Say I have a car on top of a train in a Galilean, not Einsteinian universe. I'd like to say "the speed of the car relative to the ground is the sum of the speed shown on the car's spedometer, plus that shown on the train's spedometer". Nice, simple, formula, no? I don't want to have to make four cases, depending on whether the car or train have a spedometer showing zero. Why should I?
posted by Aknaton at 5:03 PM on April 21, 2005

I don't need to invoke my Harvard degree, my vast cycling experience, my long and brilliant career of humanitarian public service, or Isaac Newton to know that you're right and your supervisor's nuts.

0 is a constant speed, answer D suggests the speed is zero, and even though B might be the "best" answer, answer D's confusing and should be removed from the question. I've aced enough standardized tests in my life to know how freaking annoying it is when you come across something like this and have to spend a moment congratulating yourself for being smarter than the morons who wrote the test.
posted by ikkyu2 at 6:53 PM on April 21, 2005

Response by poster: Many thanks for all the replies - I appreciate it. I'm glad to know I'm not off my rocker. Hopefully I'll be able to advocate to get the question changed, clarified, or deleted to eliminate the disconnect between the mathematical interpretation (where zero is a perfectly valid constant speed) and the semantic one (where the word 'biking' seems to imply non-zero speed).

And thanks, genuinely, for the correction about "exponential increase." I hate it when I find out I've been teaching something incorrectly, and I'm afraid that's what I've called y=x^2 in the classroom. I now know differently.

Thanks for the great discussion.
posted by Chanther at 7:18 PM on April 21, 2005

You're welcome. I wish I had had more math and science teachers like you. So many do not care about the details, which are important.
posted by grouse at 2:24 AM on April 23, 2005

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