Teaching Methods/The Current School Science Curriculum

Naughty teacher should have said speed not velocity for a start.
400m in 40s that's a damn fast runner. (Actually a World record!)
The answer of 0ms^-1 is probably right as runner has doubtless stopped by the time you read the question.
Can I go home yet?
Watt was the engine driver's name.
Del

There were examples about speed as well - which seems to be = Distance ÷ Time.

They started off OK, with a few examples of speed calculations, then (quite correctly) introduced velocity as speed and direction (example: 10ms^-1 North-East).

I think their answer to the runner question, with the word "velocity" unqualified (by, e.g., "average", "instantaneous", "tangential", "linear" or "angular") leaves a bit to be desired.

OMG! We are doomed!

Fire the teacher. What school system is this? (country?)

England.

I stand corrected. After doing a bit of poking around this is indeed how average velocity is defined. I'm so used to thinking in terms of instantaneous velocity (as I imagine we all are) that it didn't compute at first.

The worst you can say is that the teacher didn't qualify their question by asking for the average velocity.

You can re-hire the teacher with my apologies.

Think teacher still needs a slap on the wrist for not using the word "average".

Except that velocity is always defined as a vector, speed is just a magnitude.

You know, after thinking a bit on this, maybe it is being used by the teacher to demonstrate critical thinking.

Everything here hinges on the word displacement. If you use displacement rather than distance on a oval track, winding up at the place you started, you can say that your displacement is zero. Used this way, the answer indeed becomes zero.

But does that make sense, class?

No.

Why?

Let's start over.

Should we be using "displacement"?

Does displacement = distance?

Etc.

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Did your daughter have anything to say anything about discussion on this problem?

That is the way it appear to me. Displacement = Ø. You end up at the start line when you run around 400m track.

From physicsclassroom.com:

"Velocity is a vector quantity that refers to "the rate at which an object changes its position." Imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. While this might result in a frenzy of activity, it would result in a zero velocity."

Is this not the whiteboard problem? There is another example on the tutorial with circumstances similar to the track question as well.

I think these discussions all miss out the important word - "average" - which makes sense of it.

The net velocity was zero. Average implies you've summed a series of velocities and divided by the number of measurements. This could be done, by recording the velocity at different points along the track -- but unless one is very careful in choosing the locations to measure, one could wind up with a non-zero 'average'.

In this case, the teacher provided the exactly correct answer, but should have used the term 'net velocity'.

nah...shirley 'Net velocity' is a fishing term?
Del

No, it's a volleyball term! And don't call me Surely!

"Did your daughter have anything to say anything about discussion on this problem?" - not at the time, except to say that she was certain that it's what the teacher wrote. Teacher also said something to the effect that had the runner only gone half-way round, the velocity would have been the diameter of the track divided by the speed [edit] time. She was into her Latin homework and I didn't want to distract her - I'll have to pin her down at the week-end.

Teacher also said something to the effect that had the runner only gone half-way round, the velocity would have been the diameter of the track divided by the speed [edit] time. I was gonna say but decided not to.

Assuming that the track was circular, (implied by the mention of diameter) no, it wouldn't. It would be ( * Diameter)/(2 * 40s).

However, it was not revealed in the OP whether the 400 metre track was described as being it's diameter or it's length. Further, the geometry of the track was not revealed, whether circular or oblique straight oval (which is the shape of most tracks).

Formula is Velocity = displacement / time. I used "diameter" as a point diametrically opposite the start point (irrespective of shape - as long as it's symmetrical). Displacement = linear distance between start and finish points.

In haste...

Mikerho said:

"It would be ( * Diameter)/(2 * 40s).

However, it was not revealed in the OP whether the 400 metre track was described as being it's diameter or it's length..."

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I think that the object of this Physics lesson is precision in terminology. Given that, shouldn't Mikerho have said "circumference" and not "length?"

Now, had the OP's question been "What Work was done by John pushing against a 4-ton weight for 1-hour, trying unsuccessfully to move it?" I think all responders so far would be agreeing with the "0" answer (although with different units).

IF this is the case, I find the teacher's (apparent, I think) lesson in terminology refreshing and encouraging.

I love this forum.

Regards.

There is a very important lesson here.
It is that teachers are falible, sadly they don't often allow critical questioning/discussion.

It is the system that does not allow critical questioning/discussion. I'm sure many teachers would love to get back into it, but the current examination system and schools policy does not allow it.

The bright & successful kids these days are the ones who learn to adapt to, and play the system, not necessarily those replete with a knack for critical thinking and a questioning minds. It's a shame.

I'm with the two who said there was no displacement. The answer was correct. Average makes no difference here. Check whether it was deliberate or not.

So did the runner have any velocity 0.1 seconds into the run? 20 seconds in? 39 or 39.9 seconds in? Did the runner stop instantaneously? Did he decelerate to zero (speed and) velocity at the tape?

The same could be argued about a sprint runner on a straight. "What is the velocity of a runner who runs along a 100m straight track in 10s ?" Obviously zero, because he has now stopped.

OK, the velocity is now zero, because he's sitting in the pub discussing with his mates how fast he was round the circuit or along the straight.

Or does the velocity of the runner who ran the straight persist (gradually reducing as the clock ticks by)?

I believe that the question and answer were both phrased badly.

Well, JohnDG, I am not agreeing completely. I may be going a little easy on the instructor, or you may be a little rough on the instructor. I am keeping in mind these are 13 year old students.

From the educational preamble of the site I referenced earlier:

"As we focus on the language, principles, and laws that describe and explain the motion of objects, your efforts should center on internalizing the meaning of the information. Avoid memorizing the information; and avoid abstracting the information from the physical world that it describes and explains. Rather, contemplate the information, thinking about its meaning and its applications."

While I do agree the solution to the question is not quantifiable due to ambiguities, the essence of the concept of displacement versus distance is the lesson. No? I believe a number of students would answer 10m/s, thus beginning the discussion regarding the difference between vector and scalar quantities.

"... thus beginning the discussion ..." - as long as there is a discussion, and it continues, I'm right with you. Watch this space (I will be, and her teacher'd better watch too!).

"So did the runner have any velocity 0.1 seconds into the run? 20 seconds in? 39 or 39.9 seconds in? Did the runner stop instantaneously? Did he decelerate to zero (speed and) velocity at the tape?"

Of course he did. He had an instantaneous velocity at each point. It sounds like they're trying to teach the concept of average velocity. According to Wikipedia:

The average velocity v of an object moving through a displacement during a time interval (Δt) is described by the formula:

They've just simplified the problem as much as possible to make it something that grade-schoolers can address. I suspect many here are overthinking it.

Reminds me of a joke:

The mob wants to fix a horse race to get a certain horse to win. They bring in an doctor, an engineer, and a physicist.

The doctor comes in first and says "I can give the horse an injection of adrenaline that will make him run faster.

The engineer comes in next. He proposes: "I can put a shock device under the saddle that the jockey can use to give the horse a jolt to speed him up."

The physicist comes in and starts out: "Let's assume a spherical horse and an infinitesimal track . . ."

(Sorry if that came over as an attack on your answer - it was actually much more generalized).

I teach a class not too different from that. The trouble is the way they've said it. If you define algebraic velocity as

v = (final position - initial position)/(final time - initial time)

or v = Δx/Δt, Δt ≠ dt

then it stays pretty clear which one you're talking about. However, because I'm simple-minded, I usually write it as v_avg so that only a liberal arts major can screw it up.

Suppose I drive from A to B at 20 mph and return at 40 mph. What was my average speed for the round trip?

If we assume that the runner returns to the starting point, then the classic definition of velocity, which is the displacement from a specific point divided by time is indeed zero. No displacement from the initial point.

Of course, this is an example of the nitpicking *^#& that some educators love to dump on their students. They want to be such exact a**hole* about specific terminology that they pull these tricks. They claim velocity and speed aren't interchangable, because most of them don't live in the real world. If I run 400m back to a starting point, or point A to B, I don't give a *^#& what they say. I still got my exercise.

the teacher is from flatland!

The important thing here is the difference between SPEED and VELOCITY, and the necessity in physics to accurately define your terms.

Velocity is a vector, and the velocity of and object over a period of time will be zero if an object returns to its starting position, that's all part of the Physics definition of VELOCITY.

In this example the teacher is just trying to teach the necessary basics of Physics to the students.

plbplb.

PS. I must ask my daughter how she teaches the difference between speed and velocity.

PPS. Don't start on the Physics definition of WORK.

When teaching accuracy ... it is first necessary to be accurate. IE ... if the question fails ... were does that leave the discussion.

Okay Folks ... is this teacher trying to be tricky??? Displacement is zero??? Are they saying that since the runner returned to the starting point, the displacement was zero? Okay, the NET displacement is zero, but ...

Maybe try another experiment ... drop two teachers of different mass from the gym roof and see if they hit the ground at the same time (oooo, did I just say that??? )

Okay, now my head hurts ...

Back to work

Kind regards ...

Physic and Applied Physic…

Physic: Velocity = Distance/Time.

Applied Physic: Speed = Distance traveled/Time taken.

To me it seems that the difference between a scientist and a layman is that for scientists words have specific meanings that are not in complete agreement with the common usage. For the layman 'speed' and 'velocity' are interchangeable. A big part of learning science is to understand that while scientists use many of the same words as laymen, they are in fact speaking a different and more precise language. A scientist thinks in pictures (graphs, diagrams, etc.) and then translates those pictures into words. I think most people think only in words (as in):

1. The government collects taxes.
2. People are terrified that they will go broke paying their taxes.
3. Government is terrorism.

or

1. People invent new products or processes.
2. They start businesses, hire workers, sell things, and get rich.
3. Rich people create jobs.

Much of what passes for logical thought is nothing but wordplay, made possible by the ambiguity of language. It makes the person who makes the statement appear thoughtful. Appearances can be deceptive and scientists aren't (or at least shouldn't be) interested in appearances - they want results.

To me it seems that the difference between a scientist and a layman is that for scientists words have specific meanings that are not in complete agreement with the common usage.

problem also is, when one looks at it long enough, it becomes philosophical......soon after...it makes no sense.

I doubt any 13-14 year old will look at it long enough to become philosophical ... but many will look at it short enough to be confused & only to the brightest of them, will it make no sense.

How about the points I raised in #15?

"The teacher's answer is correct. That's how velocity is defined in physics." *** In prehisteric pharmacy, chemists knew that physics were a group of chemicals that caused bowel movements. I don't want to discuss velocity about that! *** "It's a bit worrying that so many people (in an engineering site) don't seem to understand this basic concept. Ffej." *** I understand your basic concept. I also understand my acidic concept, but I remain rather neutral about the results of a well balanced equation. *** Basic concepts are used in chemical engineering too, although the terms may precisely mean something else. *** Ester still uses alcohol and acid. *** Lewis makes his own acid. *** Poly is a kleptomaniac. *** Thomas Crapper invented the flush toilet. *** Does, "You are held in high esteem," mean they plan to cook you over boiling water in a tall building?

very funny... ga. welcome to cr4. keep the humour coming.

Response from a Physics Teacher.

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Other than the fact that it should be "average velocity" rather than just "velocity", the question and answer are fine. Surely an engineer should know this.

I'd presume it was part of a lesson learning about the differences between speed/velocity and distance/displacement.

cheers,

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Lol, he must be a high schoool physics teacher.

When I first read the question ... I intuitively said ... there is no way it can be zero ... even if the average velocity was implied. A vital assumption is missing. The more you know about the real world, the harder it is to accept the answer as zero.

To illustrate the point ...

Design of machine requires small key-way ...

Laborer builds machine, accidentally omitting key-way ... it's just a little piece of metal bar it's not really necessary is it?

Users turns machine on ...

Machine fails to perform ...

User calls engineer ...

Engineer suggest little metal bar to solve issue.

...

OR ...

Question fails to perform ...

User calls engineer ...

Engineer finds flaw in design ...

For answer 'ZERO' to be imagined as correct ... 'Design' of question requires runner to START & STOP ... IN THE EXACT POSITION & POSE ...

Engineer suggests alternative answer ...

If you ignore the velocity & position of runner crossing the line, the answer can be imagined as being zero.

...

SO WHY DO I SAY IMAGINED? The earth rotates, orbits, and otherwise goes it's merry way through the cosmos ... doesn't it? I don't have to explain further do I?

I guess you have to start somewhere, refining concepts as you advance your understanding of the real world ... but it brings back memories of my first college physics teacher ... who on the first day of class, told us that every answer to every question on every test he gives, will be proved wrong in the real world.

Hopefully this teacher has made the same disclaimer.

...

In an imaginary world, an imaginary (runner) ... mysteriously (crosses the line).

So to the person that said he was surprised that those on an engineering site would falter ... my response is ... we deal with the real world ... hypothetically speaking that is. :o)

That sounds fair enough to me. Teacher might have intended "what was runners velocity when he/she had breakfast", or "what was runners velocity as he/she stood on a podium. I'd expand on this, but somebody chucked a therapist over the wall and he had a heck of a deceleration. _{cat made me say it}

I don't get this, the techer is correct. Did you explain to your daugther the difference between Speed and Velocity and did she understand? If so then it what exactly is the issue?

If a boat travels in a straight line across a lake at a speed of 20 mph and returns to the starting point on the same path at a speed of 40 mph, what is its average speed on the round trip? Anyone?

Average speed is 30 mph.

Average velocity is zero.

Well, I'm going to argue again - the boat turns around (maybe)

Sorry Steve, your average speed is incorrect. I didn't ask about velocity, but you are correct there.

Damn! Pulled the trigger too fast on that one.

Average speed is 80/3 = 26.67 mph.

Absolutely right! Congratulations.

This all depends. I did the speed arithmetic right away (400m ÷ 40s = 10m/s speed, which may be faster than realistic), but then I also immediately got that the net velocity was zero.

If the teacher gave some examples beforehand that cleared up the distinction between speed (scalar) and velocity (vector), then this was a legitimate question. If not, then it may be considered as a trick question.

I think trick questions are sneaky and illegitimate, but I can't tell if that is the case in this instance.

'Trick' might not be quite the word. To leave students arguing over the meaning of speed/velocity might be fair enough - thinking for themselves is more effective than being told something as a fact - but maybe 13/14 y.o. is a bit young for such.

It's that old chestnut about the difference between 'teaching' (ie ram it down the throat) versus 'learning' (encourage thinking and discussion). My tuppence is the particular method is not apt for such an age, though it's entirely possible the teacher didn't have a clue. In these days, teachers of sport and arts subjects are often delegated to lead maths/science classes. Things could be worse.

Could one of you Brits explain your education system to the rest of us? For the most part we have no idea what A levels, fifth forms, etc are.

We don't either - that's the problem. Perhaps I should enroll in a local Academy to obtain a GCSE in this topic, I'd be a cert for grade A++ in it, just for the fact of turning up.

Are you sure you understand the question?? It is a no brainer - 400m in 40s - come on now. The answer is 10m per second.

That would be average speed, not average velocity. Since a track is circular, the runner actually has traveled nearly nowhere when complete. I am guessing you have missed the whole discussion above.

Thank you! At last, someone^[1] has recognized the importance of the word "average" in this context.

Have a GA on me.

[1] Even if you don't agree, the folks who posted the info. in your linked site do . Thanks again.

[Edit: To be fair, having reviewed (some of) the posts, I think Steve was the first to mention "average". Most folk just ignored the concept.]

JohnDG,

I agree that instantaneous speed and instantaneous velocity vary throughout the trip, the latter being numerically equal to the former, but having direction. Near the start and near the end, they must be variable in order to be zero at each end. If you stop for a cup of tea in the middle, your average speed is going to suffer, but if you return to the starting point, the average velocity is zero no matter how speedy you were along the way.

The attempted "trick" to this question is running "round" the track. After completing one lap, the runner's position at the end of the lap is the same as the position at the beginning of the lap. This is the grounds for saying "displacement" = 0 (i.e. x_f - x_i = 0). I am not sure why the term has been introduced to physics curricula, but I've seen it in one US 8th grade text as well. Calculating velocity in terms of "displacement" seems rather meaningless to me. It's like hypothetically measuring what one's velocity would have been if one had travelled as the crow flies. We really ought to reserve "displacement" for things like buoyancy.

I wonder if it is just meant to get students paying careful attention to detail by making sure they read carefully enough to discern between distance and displacement. If so, I wish they'd just work with paying attention to more meaningful terms like average vs. instantaneous velocity.

My answer will depend whether I Am the runner or the pedagogue who's watching?

Hmmmnnn, there is one BIG assumption missing, for the answer, 'zero', to be correct.

Without this assumption ... the question must be termed trickery.

Hint: On second thought, I'm not giving a hint ... Sometimes you just have to draw the line and put your foot down.

This thread seems to be one that goes round in circles.

You almost used the right formula and only one wrong term, which lead to using an invalid value and the wrong answer. Velocity = Distance ÷ Time = 400m ÷ 40s = 10m/s.