Is Velocity Increasing Below the X-Axis?

[TBEV5]

In my 5th year teaching Physics but 10th of sciences, and just curious about other physics educators' take on the following: in a graph interpretation exercise in AP Physics 1, my most advanced student had an opinion my admin (also Physics certified, though his background is very high level math, and this year is the first year he's actually taught physics) agrees with. While I agree from a "pure math" standpoint, I disagree when it comes to the "applied math" context of physical phenomena.

The student stated that the line below the x-axis shows that speed is decreasing (fine so far), but velocity is increasing. On this last point I disagree - while in pure math terms it's correct, and agreed it's an unbroken positive slope, since +/- is directional and thus arbitrary, I believe claiming velocity is "increasing" below the x-axis in that first 2/3 of a second is inadvisable. (If this were a pure math context I'd be fine with it.) I would say instead that the object has decreasing velocity in the negative direction; I'd also say acceleration is constant in the positive direction while initial velocity was in the negative. Arguably most importantly, everyone's on board as to what's actually happening in the real world (object has an initial velocity, slows, instantaneously stops, speeds up in the opposite direction, all the while under constant acceleration until the 2 second mark); I'm just uncomfortable with the "velocity increasing" bit before the x axis is crossed. I'm also unclear how an exam reader would grade an answer that includes this approach. Hoping just semantics, though semantics can be important. Thoughts?

 

[kuruman]

If you compare $-5$ and $-2$ which is the larger number?
Answer: Clearly, $-2 > -5$ because to get to $-2$ you have to add $+3$ to $-5.$

I think that it is important to make a clear distinction between speed (a magnitude) and velocity (a vector) in the minds of beginners right from the start to avoid confusion later. They are separate ideas and the mathematical description helps keep them separate.

In one dimension the mathematical description is that velocity can be positive or negative but speed is only positive. Acceleration is also a vector that can be positive or negative. This description reflects the "real world" and I disagree that it is a "pure math concept". Physics uses math concepts to describe the physical world. Here, the signs of the one dimensional velocity and acceleration vectors can be used to summarize, rather compactly, what is "actually" happening according to the graph.

The rule is simple: If the product of the velocity and the acceleration is a positive number, the speed is increasing; if it is a negative number the speed is decreasing. In other words, regardless of which direction one chooses as positive (left, north, up, etc.) , if the acceleration is parallel to the velocity the speed is increasing and if it is antiparallel to the velocity the speed is decreasing.

In the first leg of the graph, the acceleration is positive throughout. Following the rule, when the velocity is negative, the speed is decreasing and when it is positive the speed is increasing. In the middle leg the speed and velocity are not changing hence the acceleration is zero. In the third leg the velocity is positive throughout, the acceleration is negative throughout hence the speed is decreasing throughout. Simple, no?

 

[Baluncore]

Welcome to PF.

What a lovely question.

In a similar vein.
If I drive in a circle at a constant, but non-zero speed, my velocity is continuously changing. Over which parts of that circle is my velocity increasing?

Clearly, only the direction of my velocity vector is changing. Can a direction vector increase or decrease while maintaining the same magnitude or speed? Or do we first have to resolve the 2D vector into orthogonal x and y directions before claiming that a direction component is increasing or decreasing?

How does the terminology change if we use a polar coordinate system, where the radius remains constant, but the speed changes.

 

[Dale]

There is no $>$ operation for vectors

velocity is increasing

This means $$\frac{d\vec v}{dt}>0$$ which doesn’t make sense. It is “not even wrong”

 

[PeroK]

In my 5th year teaching Physics but 10th of sciences, and just curious about other physics educators' take on the following: in a graph interpretation exercise in AP Physics 1, my most advanced student had an opinion my admin (also Physics certified, though his background is very high level math, and this year is the first year he's actually taught physics) agrees with. While I agree from a "pure math" standpoint, I disagree when it comes to the "applied math" context of physical phenomena.

View attachment 360568
The student stated that the line below the x-axis shows that speed is decreasing (fine so far), but velocity is increasing. On this last point I disagree - while in pure math terms it's correct, and agreed it's an unbroken positive slope, since +/- is directional and thus arbitrary, I believe claiming velocity is "increasing" below the x-axis in that first 2/3 of a second is inadvisable. (If this were a pure math context I'd be fine with it.) I would say instead that the object has decreasing velocity in the negative direction; I'd also say acceleration is constant in the positive direction while initial velocity was in the negative. Arguably most importantly, everyone's on board as to what's actually happening in the real world (object has an initial velocity, slows, instantaneously stops, speeds up in the opposite direction, all the while under constant acceleration until the 2 second mark); I'm just uncomfortable with the "velocity increasing" bit before the x axis is crossed. I'm also unclear how an exam reader would grade an answer that includes this approach. Hoping just semantics, though semantics can be important. Thoughts?

There are only two possible answers, IMO. One is that velocity is a vector and only the magnitude of a vector can increase or decrease.

The other is that motion in one dimension can be reduced to a single component of velocity and that component is increasing below the x-axis.

 

[PeroK]

There is no $>$ operation for vectors


This means $$\frac{d\vec v}{dt}>0$$ which doesn’t make sense. It is “not even wrong”

The answer is not so clear for a 1D vector. $\mathbb R$ is both a vector space and an ordered field of scalars.

One could argue that in the introductory mechanics of SUVAT we do have an order on the velocity vector. That said, the exact mathematical basis of SUVAT is generally not specified.

 

[Borek]

There is no $>$ operation for vectors


This means $$\frac{d\vec v}{dt}>0$$ which doesn’t make sense. It is “not even wrong”

More or less my thought. Everyone uses some intuition to compare velocities, and the problem stems from the fact these intuitions are different. To avoid problems they should start by defining and agreeing on a metric used for comparing.

 

[TBEV5]

Thanks for all the input. And yes, it's obviously an increasing slope, and in terms of derivatives, it was readily apparent velocity is increasing no matter what, even below the x axis. (I routinely reinforce the distinction in graphs between speed and velocity, in that a speed v time graph has nothing below the x axis, identical to |v| over time).

But there's a lot about non-calculus algebraic physics (including AP Physics 1) that's not strictly speaking "correct," e.g. "gravity is a force" (which at least is Newtonian!) but it's helpful to look at it that way, just as it can be helpful to model an atom as a little solar system even though that's not literally correct; I would have lumped "in the negative direction" in with the "helpful but not strictly speaking correct" category. (Though honestly, back in high school and night school chemistry, poorly-explained contradictory models of atoms, still in use - except plum pudding - actually stymied me at the time, which may inform my teaching of what we're discussing here!)

So long as a student who claims "increasing velocity" understands the object is slowing in the first 0.67 seconds, I guess I'm fine with it, and I'm probably changing my approach going forward.

 

[A.T.]

since +/- is directional and thus arbitrary, I believe claiming velocity is "increasing" below the x-axis in that first 2/3 of a second is inadvisable.

Not just the orientation of the axes, but the reference frame as such is arbitrary. So based on your logic, any claim about velocity, or speed, or any other frame-dependent quantity, would be "inadvisable". This would make doing any physics rather hard.

The only "issue" with the student's statement is, as noted by others, that it's not the "velocity" that is increasing, but "velocity component along the spatial axis being considered". However, when only one spatial axis is being considered, this is often shortened and gets potentially confusing. Here, that ambiguity is already introduced by the provided graph, with the axis-label reading 'v', not 'v_x' or 'v₁'.

 

[Dale]

Not just the orientation of the axes, but the reference frame as such is arbitrary. So based on your logic, any claim about velocity, or speed, or any other frame-dependent quantity, would be "inadvisable". This would make doing any physics rather hard.

I don’t know about that. Velocity is a vector, and although it can have different components in different reference frames the geometric object is the same. So as long as you speak of vector operations it can be meaningful. And that notion carries through to manifolds and tensors and such as well.

I do agree that the “policy” you objected to does limit some things, but maybe not as bad as you might think.

 

[Dale]

The answer is not so clear for a 1D vector. R is both a vector space and an ordered field of scalars.

While it is true that R is both a vector space and an ordered field of scalars I would say that as a vector space it does not have an order. And velocity is a vector. So the specific property that gives an ordering to R isn’t one that velocity has.

 

[PeroK]

While it is true that R is both a vector space and an ordered field of scalars I would say that as a vector space it does not have an order. And velocity is a vector. So the specific property that gives an ordering to R isn’t one that velocity has.

Whether you can order a set doesn't depend on whether it satisfies the vector space axioms! It doesn't stop being ordered when you consider additional properties.

PS the concept of a vector as a frame-independent geometric object is beyond high-school level.

 

[Dale]

Whether you can order a set doesn't depend on whether it satisfies the vector space axioms!

Yes. I agree.

 

[Herman Trivilino]

This means $$\frac{d\vec v}{dt}>0$$ which doesn’t make sense. It is “not even wrong”

True. But what is really being talked about is the x-component of the velocity. An endless source of confusion for students.

Note that if by velocity they were really referring to $\vec{v}$ they wouldn't be able to draw a graph of the velocity. Yet one sees these graphs throughout introductory physics textbooks and tests.

 

[Herman Trivilino]

The student stated that the line below the x-axis shows that speed is decreasing (fine so far), but velocity is increasing. On this last point I disagree - while in pure math terms it's correct, and agreed it's an unbroken positive slope, since +/- is directional and thus arbitrary, I believe claiming velocity is "increasing" below the x-axis in that first 2/3 of a second is inadvisable.

But what is really being graphed is the x-component of the velocity. Of course, at this point in the course vectors haven't been introduced, so it would make no sense to speak of vector components.

Edit: So, just to be clear, the x-component of the velocity is increasing, therefore the x-component of the acceleration is positive.

Once it's understood (at least by the instructor) that in equations like $v=v_o+at$ what the symbols $v$ and $a$ refer to are the x-components of $\vec{v}$ and $\vec{a}$, respectively.

Once vectors are introduced it is up to the instructor to make this point clear. And it cannot be done with a single series of statements, however well stated. It must be repeated throughout the remainder of the course whenever the opportunity arises. Otherwise the students will forever be confused by this issue, even if they are not able to articulate the confusion.

 

[Herman Trivilino]

And yes, it's obviously an increasing slope,

No, it's a positive slope. Positive and constant, not increasing. If the slope were increasing the graph would be a curve, not a straight line.

 

[Dale]

But what is really being talked about is the x-component of the velocity. An endless source of confusion for students.

Yes. This was an “interpret the graph” question. What was graphed was the x-component of the velocity and the statement “the x component of the velocity is increasing” is a correct interpretation of the graph. In math $$\frac{dv_x}{dt}>0$$ is both meaningful and true.

 

[sophiecentaur]

Yes. This was an “interpret the graph” question. What was graphed was the x-component of the velocity and the statement “the x component of the velocity is increasing” is a correct interpretation of the graph. In math $$\frac{dv_x}{dt}>0$$ is both meaningful and true.

There are so many instances when launching into the vernacular can only make things worse. Determined 'non-mathematicians' make life hard for themselves with their "but what really happens, in Physical terms?" questions.