Teaching velocity and acceleration

David Hogg writes of teaching intro physics:

The time derivative of velocity is acceleration, both vectors of course. But I [Hogg] was reminded in office hours today of just how hard it is to get across the idea that the velocity vector and the acceleration vector can point in totally different directions. And some students have trouble seeing this when a ballistic stone is going upwards along some (parabolic) arc, some have trouble seeing it when it is going down, and some have trouble seeing it at the top. That is, different students have very different problems visualizing the differences of the vectors over time.

I wonder if it would help to use the business cycle as an example? When you’re at the top of the cycle, the position is high, the velocity is zero, and the acceleration is negative. And so on.

Would that example work or would it just confuse things further because you still have to make the transition to physical motion (in which the dimensions are positions in space, rather than the state of the economy)?

P.S. This sort of problem reminds me of why it was a good thing I left physics. I think I could figure out the answer if I worked at it, but I just don’t have the intuition, in the way that I have when thinking about statistical problems.

21 thoughts on “Teaching velocity and acceleration

  1. Having been in a car at some point in my life I’ve never had a problem imagining negative acceleration. Jerk, the third derivative of velocity, is trickier, but the name helps a lot. I can barely conceive of jounce.

    • Good point: and I think the idea is to imagine yourself in the situation: where are you located, how fast are you going, and in which direction are you being pushed by the seat/seatbelt?

  2. I have been thinking about a statement made in a math class I was attending by Peter Rosenthal – http://www.maa.org/careers/rosenthal.html

    “Most mathematicians find it difficult to be concrete”

    His point was that they were more comfortable being abstract.

    Now I will wonder what the difficult/comfortable dimension might be for statisticians…

    p.s. Also, when at my summer internship at MBA school, I calculated the change in change in sales for some reported data and outraged my fellow MBA student erased this muttering “there is no such thing as change in change.”

  3. I agree with keith’s first comment: how does decelerating in a still-forward-moving car not get the job done in a familiar way?

  4. I think it is a very small minority of students for whom business cycles are more intuitive than simple mechanics.

  5. @keith. Not surely — distances of things in front of you are positive, so we could reasonably answer “infinitely large” (the limit from the positive side), or, given measurement error, “very large compared to the distance between your eyes”.

    The key insight, for those not following, is that rainbows are not localized in space, but always appear at the same angle from the anti-solar point. Thus, they display no parallax. The formula gives distance = x / theta where theta is zero (since theta is the change in angle between your eyes, and that angle is constant). keith and I are discussing how to deal with the divide-by-zero problem when asked for a meaningful answer on a physics problem set.

    • @Ralph

      Hmm… I can’t say I’ve thought too hard about it, but I’d imagined there would be parallax (albeit very small), but in the opposite direction from that expected for a fixed object, giving a “negative” distance of ~150m km (assuming suitably accurate measurements).

  6. Regarding the rainbow parallax… I’m just going by intuition here, without working it out in detail, but following on Ralph’s insight above, I think the answer is probably “the distance to the sun”.

  7. @Andrew: definitely not, for the reason you mention (physical space is more familiar to the vast majority of students than abstract space) and also because in physical problems derivates are well behaved (velocity is finite and continuous; acceleration is constant for the case in question).

    @Ralph: isn’t the point that notion of parallax distance is defined only for objects that are localised in space? It doesn’t make sense to measure the distance to an “object” that has no position. Alternatively, one could say that the rainbow does have a position but that it moves (becoming a different rainbow) whenever the observer does; for a given observer position (and hence rainbow position) one can only ever get one measurement – this makes the parallax distance undefined for a reason other than divide-by-zero.

    • Konrad: I think you’re intended to compute the parallax distance as if you didn’t realize that there was no physical object corresponding to the rainbow. Otherwise, it’s just a trick question.

  8. Intuitively acceleration is caused by application of force, hence a simple manner in which we can resolve the differing directions issue is by simply realizing that the current direction of motion and force applied can be different and the same effect thus translates into acceleration and velocity having a different direction.

    Also it is because of the concept of force that we don’t measure the rate of change of acceleration , the rate of change of change otherwise is always well defined or for that matter higher moments.

  9. Another explanation is that velocity is multidimensional, hence change in one direction(dimension) can be be different form the change in other giving rise to differing directions for velocity and acceleration.

  10. Andrew, In answer to your question, one issue is that in kinematics you have three-dimensional vectors, whereas in the market, you are usually talking about scalars, so it might be useful but doesn’t get at the whole directional thing in general. @Wayne’s suggestion is great, and exactly what we do in class, but it is very very hard, it turns out, for the people to fully “get”.

    @konrad and @Radford: It is not a trick question, there is an answer with no divide-by-zeros or anything, but @konrad brings up a deep philosophical point: How can we even know that there are physical objects out there? Doesn’t Andrew teach us that all models are wrong? In astronomy you don’t detect objects, you detect photons coming from particular angular directions with particular energies, so there physical-objecty-ness is a model property, I think.

    Andrew, thanks (I think) for distracting your audience with physics!

  11. Here is an example I have used in class, to explain what is a Inflection point:

    Imagine you are driving a car which approaces a crossing with traffic lights. You see the light are red so you are
    reducing speed, preparing to stop. Then in the moment you would have stopped the light switches to greenso you are speeding up again. That “almost stopped point” is an inflection point of position as function of time.

  12. I think the economic notion of a ‘business cycle’ is an different, more difficult, and much much greater, abstraction than the vectors of velocity and acceleration of a particle in space.

  13. Pingback: Direction of velocity and acceleration - Ask Physics - Doubts in Physics Answered - Ask Physics

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