Is velocity ever a scalar quantity?

[paulb203]

TL;DR Summary

Is the v in the suvat equations a scalar or a vector?

I thought velocity was always a vector quantity, one with both magnitude and direction.

When it came to the suvat equations, where v = final velocity, and u = initial velocity, I thought both of those were vector quantities, e.g;

v (final velocity) 112km/hr North

u (initial velocity) 0km/hr (I'm now asking; what do we put for direction when the object is initially stationary?)

But in a Khan Academy question they ask what does the letter v (lower case with no arrow above it, or anything else) stand for, and whether it’s a vector or a scalar.

I answered ‘velocity’ (it was multiple choice with no option for ‘final velocity’) and that it was a vector.

Their answer was;

“The symbol v represents speed, a scalar.”

I know speed is a scalar, but thought v stood for final velocity.

 

[Ibix]

I would say it's a vector in that case, since it can be positive or negative and those have sensible meanings (which direction you are moving). $|v|$ is a scalar, the speed.

 

[Orodruin]

Definitely a vector. Khan academy is wrong if it says it is not.

TL;DR Summary: Is the v in the suvat equations a scalar or a vector?

u (initial velocity) 0km/hr (I'm now asking; what do we put for direction when the object is initially stationary?)

The zero vector does not have a well defined direction.

 

[Orodruin]

It should also be noted that the SUVAT equations are typically written down for one-dimensional motion. However, they are easily generalized to several dimensions, where the vector structure is clearer. All of the equations except the one relating $u^2$, $v^2$, $\vec s$, and $\vec a$ are then vector equations.

 

[PeroK]

Definitely a vector. Khan academy is wrong if it says it is not.

That's a big if.

 

[Steve4Physics]

TL;DR Summary: Is the v in the suvat equations a scalar or a vector?

Common conventions are:
Use an arrowed or bolded letter to represent a vector, e.g. $\vec v$ or $\bf v$.
To represent the magnitude of a vector, use a plain letter or bars, e.g. $v$ or $|\bf v|$.

However, at an introductory level (typically 1D, constant acceleration contexts), you will often see a plain letter used to represent a vector. E.g. you will see $v= u+at$ rather than $\vec v=\vec u +\vec a t$, presumably for simplicity and neatness. And also maybe because, for 1D, a vector can be represented by a signed scalar. This is ‘bending the rules’ a bit and opinions about its validity will vary.

So (like it or not!) you sometimes have to decide from the context whether a plain letter such as ‘$v$’ represents a scalar or a vector. IMHO.

 

[weirdoguy]

That's why I always say to my students, that 1D motion may be simple in general, but it's also difficult because it's hard to grasp what is the difference between some of the concepts, like speed and velocity, or distance (length of curve) and displacement. 2D and 3D are better for that.

 

[paulb203]

I would say it's a vector in that case, since it can be positive or negative and those have sensible meanings (which direction you are moving). $|v|$ is a scalar, the speed.

Thanks. Regards the vertical bars either side of your 'v'; does that mean, The absolute value of v (final velocity) is (effectively?) the speed of the object, which is a scalar quantity?
I've put 'effectively'? there because I'm often told that velocity is similar to speed but not the same as speed.

 

[Orodruin]

Speed is - by definition - the magnitude of velocity.

 

[paulb203]

Definitely a vector. Khan academy is wrong if it says it is not.



The zero vector does not have a well defined direction.

Have they just neglected to put absolute value vertical lines either side of their v?

 

[Orodruin]

Have they just neglected to put absolute value vertical lines either side of their v?

Where? You have not provided an actual equation to check.

In one-dimensional motion the distinction between vector and scalar quantities is a bit floating notation wise as vectors only have one component. In more dimensions it would be quite customary to denote a vector $\vec v$ or $\mathbf v$ and its magnitude $v$.

 

[Ibix]

Have they just neglected to put absolute value vertical lines either side of their v?

Actually, I think @weirdoguy's comment is the most important here. If we are talking 2d or 3d, everyone writes $\vec v$ or $\mathbf{v}$ or $\underline{v}$ and then we understand that $v=|\vec v|$. Unfortunately in 1d people tend to leave off the vector signs. Then you get a lot of confusion between the signed quantity (which is a 1d vector) and the absolute value (which is the scalar).

Is the question you are looking at using only 1d motion? Or is it considering 2d or 3d motion? If the latter, and they are writing the SUVAT equations like $\vec v=\vec u+\vec a t$, then I have sympathy for their answer that $v$ is a speed. If they're doing 1d motion and writing $v=u+at$ then I disagree with them. (Edit: 1d was what I was assuming in my previous post, but perhaps that was a mistake on my part.)

Incidentally, this kind of thing is why Homework Help requires exact problem statements. Quite small stuff that gets lost in a paraphrase can matter a lot to the answer.

 

[Orodruin]

Incidentally, this kind of thing is why Homework Help requires exact problem statements. Quite small stuff that gets lost in a paraphrase can matter a lot to the answer.

Indeed. A not insignificant portion of questions that go on for multiple pages could have been resolved in one reply if only the exact statement would have been provided. Instead, those threads go on and on with different people chiming in with their interpretation of the OP’s paraphrasing of the problem only for it all to be realized at the end that OP misinterpreted and therefore misrepresented the problem statement.

 

[paulb203]

That's a big if.

Thanks, PeroK.
Here's what they said.

 

[Ibix]

In that case, my assumption of 1d motion was in error and I agree with them. $\vec v$ is the vector and $v=|\vec v|$ is a scalar.

 

[paulb203]

Common conventions are:
Use an arrowed or bolded letter to represent a vector, e.g. $\vec v$ or $\bf v$.
To represent the magnitude of a vector, use a plain letter or bars, e.g. $v$ or $|\bf v|$.

However, at an introductory level (typically 1D, constant acceleration contexts), you will often see a plain letter used to represent a vector. E.g. you will see $v= u+at$ rather than $\vec v=\vec u +\vec a t$, presumably for simplicity and neatness. And also maybe because, for 1D, a vector can be represented by a signed scalar. This is ‘bending the rules’ a bit and opinions about its validity will vary.

So (like it or not!) you sometimes have to decide from the context whether a plain letter such as ‘$v$’ represents a scalar or a vector. IMHO.

Thanks. Really helpful answer.
Regards the letter with the arrow above it ('arrowed letter'); does that arrow always point to the right? Is it just there to tell you it's a vector quantity, rather than having anything to do with right being positive, left being negative, etc?
Also, I'm wondering now if, given the context, they are correct in saying that their 'v' means speed, a scalar, the context being; an arrowed x, for position, an arrowed v, for velocity, and an arrowless v, for speed.

 

[PeroK]

Thanks, PeroK.
Here's what they said.View attachment 355012

As long at they are consistent in their course material then that is fine. Also, that may be in a section on 2D and 3D motion. In the section on 1D motion, they may say that in this case it's common to drop the vector symbol and use $v$ for velocity (may be +ve or -ve) and $|v|$ for speed.

Notation in all maths and physics is specific to the context. You can't assume that universally in all cases the same notation applies. Taking statements out of context can be a problem.

 

[PeroK]

For example, as you progress in your studies you will find that speed is not a scalar! It's the magnitude of a vector. A scalar is something like mass of a particle or electric charge of a particle. But, for now, within the context of the subject matter at hand, Khan are correct to say that speed is a scalar.

I'm not sure what I was thinking of there!

 

[paulb203]

Speed is - by definition - the magnitude of velocity.

What about round trips, where the magntitude of velocity is zero, but the speed is >zero?

 

[PeroK]

What about round trips, where the magntitude of velocity is zero, but the speed is >zero?

That's average velocity over a period of time. Average speed is average magnitude of velocity. Not magnitude of average velocity.

 

[paulb203]

Where? You have not provided an actual equation to check.

In one-dimensional motion the distinction between vector and scalar quantities is a bit floating notation wise as vectors only have one component. In more dimensions it would be quite customary to denote a vector $\vec v$ or $\mathbf v$ and its magnitude $v$.

 

[PeroK]

PS confusing average velocity and velocity can lead to serious confusion!

 

[paulb203]

Actually, I think @weirdoguy's comment is the most important here. If we are talking 2d or 3d, everyone writes $\vec v$ or $\mathbf{v}$ or $\underline{v}$ and then we understand that $v=|\vec v|$. Unfortunately in 1d people tend to leave off the vector signs. Then you get a lot of confusion between the signed quantity (which is a 1d vector) and the absolute value (which is the scalar).

Is the question you are looking at using only 1d motion? Or is it considering 2d or 3d motion? If the latter, and they are writing the SUVAT equations like $\vec v=\vec u+\vec a t$, then I have sympathy for their answer that $v$ is a speed. If they're doing 1d motion and writing $v=u+at$ then I disagree with them. (Edit: 1d was what I was assuming in my previous post, but perhaps that was a mistake on my part.)

Incidentally, this kind of thing is why Homework Help requires exact problem statements. Quite small stuff that gets lost in a paraphrase can matter a lot to the answer.

Thanks. I don't think it mentioned whether it was 1d, or 2d etc. I just tried to go back to the question but it's now giving a slightly different version of it.
Lots of good answers here. I think I've just about it got it now. Context. Arrowed letters. Etc.
Just double-checking though; the arrow above the letter is merely letting you know it's a vector, it's not alluding to rightwards = positive, yeah?

 

[paulb203]

PS confusing average velocity and velocity can lead to serious confusion!

Thanks. When you say velocity on it's own here do you mean instantaneous velocity? I did get into the habit of putting a bar over s to remind me I was dealing with average speed, and a bar over my v to remind me I was dealing with average velocity. Now I'm putting an arrow over my v to get into the habit of that way of doing notation. I'm guessing you never put a bar AND an arrow over a v ! :)

 

[Steve4Physics]

Regards the letter with the arrow above it ('arrowed letter'); does that arrow always point to the right? Is it just there to tell you it's a vector quantity, rather than having anything to do with right being positive, left being negative, etc?

Yes - the arrow over the letter only tells you it's a vector quantiity. The arrow always points right by convention. It does not tell you anything about any direction(s).

Also, I'm wondering now if, given the context, they are correct in saying that their 'v' means speed, a scalar, the context being; an arrowed x, for position, an arrowed v, for velocity, and an arrowless v, for speed.

Yes they are correct. Since there is an 'arrowed v', that is unambiguously a vector. It follows that a 'plain v' is its magnitude (the speed in this case). (As described in Post #6.)

 

[PeroK]

Thanks. When you say velocity on it's own here do you mean instantaneous velocity?

Yes, velocity by definition is a derivative. The only ambiguity comes when it is constant. And then velocity and average velocity are the same.

I did get into the habit of putting a bar over s to remind me I was dealing with average speed, and a bar over my v to remind me I was dealing with average velocity. Now I'm putting an arrow over my v to get into the habit of that way of doing notation. I'm guessing you never put a bar AND an arrow over a v ! :)

An alternative notation for average velocity is: $\vec v_{avg} \equiv \frac{\Delta \vec x}{\Delta t}$. Average speed is harder to define formally. It would have to be:
$$v_{avg} \equiv |\vec v|_{avg} = \frac 1 {\Delta t}\int_{t_0}^{t_0 + \Delta t} |\vec v(t)| dt$$And something very important is that:
$$v_{avg} \neq |\vec v_{avg}|$$

 

[Steve4Physics]

Temporary divergence [Edit - I mean digression] and reminisce…

When teaching/introducing this stuff (a very long time ago) I would would walk one complete circuit around the lab’ and ask the students to write down (anonymously, on a scrap of paper) their own estimates for the distance covered, time taken, displacement, average speed and average velocity.

We’d collect the scraps and skim through the (disconcertingly wide range of) answers. Then we’d go through the ‘correct’ answers.

It was a very useful teaching exercise as you could easily assess the extent of (mis)understanding and (though less important) mis-estimation.

The nice thing was that individual students were actually engaged – wanting to compare their own results to the rest of the class’s.

 

[Orodruin]

For example, as you progress in your studies you will find that speed is not a scalar! It's the magnitude of a vector.

The magnitude of a vector is a scalar.

For example:

A scalar is something like mass of a particle or electric charge of a particle.

The mass of a particle is the magnitude of its 4-momentum.

 

[Orodruin]

Temporary divergence [Edit - I mean digression] and reminisce…

When teaching/introducing this stuff (a very long time ago) I would would walk one complete circuit around the lab’ and ask the students to write down (anonymously, on a scrap of paper) their own estimates for the distance covered, time taken, displacement, average speed and average velocity.

We’d collect the scraps and skim through the (disconcertingly wide range of) answers. Then we’d go through the ‘correct’ answers.

It was a very useful teaching exercise as you could easily assess the extent of (mis)understanding and (though less important) mis-estimation.

The nice thing was that individual students were actually engaged – wanting to compare their own results to the rest of the class’s.

Nowadays you’d have students answering on their smartphones and then immediately get nice histograms for displaying with a projector…