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Participation Requested: Anticipating Student Approaches

December 13, 2011

An undergraduate student I work with gave the following problem to a bunch of students enrolled in two different calculus-based physics course.

An object undergoes constant acceleration. Data from the object’s motion is shown below.

Time (s)

Velocity (m/s)

0

10

1

8

2

6

3

4

4

2

5

0

How far did the object travel during those 5 seconds?

 

First:

I’m curious, how you initially think about / approach this problem.

Second:

I’m curious what you would predict are all the different things students did (both correct and incorrect).

Third:

I’m also curious if there are any viable methods that you think were not likely be taken by students.

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12 Comments leave one →
  1. December 13, 2011 2:23 pm

    Here’s one incorrect approach I would expect to see students take—simply multiplying the velocity by the time for each instant and then summing them, (10m/s)*(1s)+(8m/s)*(1s)+….

    A viable method I would love to see students take, but I don’t think they would is to draw a graph. Then see that there’s a box (10m/s)*5s that represents how far the object would have gone if it was constant velocity, from which they could have removed the area of a triangle (1/2*(-2m/s^2)*(5s)^2), which represents the distance the object did not travel because it was accelerating.

    • December 17, 2011 12:19 pm

      So, yeah the 10(1) + 8(1)… approach is the most common “incorrect” approach… and yeah, no students took an explicitly find the area approach.

  2. Adam James permalink
    December 13, 2011 3:24 pm

    My first thought would be to take half of the velocity change and multiply by the time interval. This is because I’ve been calculating the areas of v-t graphs for 2 weeks with my college prep physics students, and this data set translate to a triangle area on a v-t graph in my head. Before I learned to teach through modeling, I’d have found the acceleration from two points and use 1/2at^2 – v0t to get displacement,

    I would second John’s incorrect sum of v*t for each instant.

    Depending on comfort with the idea of average velocity, some students might actually find the Vavg*time between successive pairs of points. [(v0+v1)/2]t+[(v1+v2)/2]t+… and add those numbers up. Others might find acceleration and use the kinematic equation like I would have coming out of my physics degree. I have one student who would draw a velocity time graph on graph paper, with a straight-edge, find the size of one square of graph paper and then count number of squares that make up the area underneath the line.

    None of my students would find the total average velocity for the whole 5 seconds and multiply that by the time interval. None of my students would find the linear equation describing those points and take the integral. None of my students would find the linear regression of those points and use their graphing calculator to find the area underneath the regression line.

    This is beyond the prompt, but if there was a 7th data point at (6s, -2m/s) some of my students would try to find the area of a whole triangle, using the height of 12 m/s and the width of 6s, for a response of 36m.

    • December 17, 2011 12:25 pm

      In one class, there were many students taking a “find acceleration plug into 1/2 at^2 + vot to varying degrees of success.” Some were not able to find a. Some were not able able to recall the formula. In this class, no students took an average velocity approach.

      In the other class, many students used the average velocity approach.

      The difference in the two classes’ approaches is really interesting.

      You do bring up to uncommon but present approaches:

      One student actually took the average velocity for each interval. He inadvertently, mis-added the sum.

      One student found the linear equation describing velocity and took an integral. This was interesting because this students’ approach was completely mathematical. There were no physically relevant symbols or talk of physics… writing instead all equations in terms of y and x.

  3. Becca permalink
    December 13, 2011 3:53 pm

    I pictured the velocity vs. time graph, and then recognized that the area under the curve is the distance traveled, and it’s a triangle, so A = (1/2)*10*5 = 25 m

    Then I also thought: well, you could find the average velocity in each time step, multiply by time (which is just 1) and add that up, so I checked my previous answer by saying 9+7+5+3+1 = 25 m. I doubt that students would do that though – I think they are more likely to do what John said and add 10+8+…

    One other approach I can imagine students taking is multiplying each velocity by the time written to its left, so they do: 10*0 + 8*1 + 6*2 + …

    • December 17, 2011 12:28 pm

      Becca, no student took the triangle approach explicitly.

      The last approach you predict is really interesting, and definitely some students (at least in one class) took this approach, which (in my mind) is the only physically meaningless approach. Multiplying the velocity by a clock reading is meaningless… It’s cool that you spotted this one.

  4. Becca permalink
    December 13, 2011 4:18 pm

    Oh, I wrote my post before I saw Adam’s, and now I want to add that even though I thought of finding the average velocity for each time interval, it did not occur to me that I could just find the average velocity for the whole 5 seconds. Kind of a dumb thing to miss on my part, but it probably says something interesting about the way I think about motion (not sure what though).

    • December 17, 2011 12:31 pm

      I don’t think it’s dumb you missed it. I think it’s cool to see and recognize how many different ways there are to think about this problem.

      One thing we are seeing in our data is how one class clustered around one approach and another class scattered about many many different approaches. But, still, very few students took some of the approaches we have discussed… such as thinking about area under the curve (as triangle or as rectangle minus triangle).

      One of the things I was talking about with the undergrad student (who is a future teacher) is how the kind of conversation that is happening is here about the problem is very rich– both thinking about the physics and in knowing about how students might think and approach physics.

  5. December 13, 2011 5:05 pm

    I’m going to not read the previous replies first – so apologies if I just repeat.

    How I solve: “Oh, this is uniform acceleration from 10mps to 0mps in 5 seconds… so avg. speed is 5 mps for 5 sec … 25 m” – that is, I ignore most of the data.

    Second:
    The chart implies we should use all of that data, so I’d expect a lot of students to do that.

    But! I find that it’s hard to think about intervals v. “moments.”- Like, there are 4 time zones in the continental US, but that means I add 3 hours to my time to find the time on the east coast. That never fails to trip me up, and I always count it out on my fingers (not joking). So I would expect that would show up here – 5 intervals, 6 measurements – that’s confusing. So I would expect lots of grappling with manifestations of that problem.

    Since these are calc-based physics and presumably have seen the equations for uniform acceleration before, they might be more inclined to find an appropriate equation – I’d be curious if the use an equation 5 times, 6 times, or 1 time!

    I’d be impressed if they graphed the points, used that to find the average speed or the distance. Less likely but possible – though it seems if you know to do this, you’d know to do this in your head and spit out “25m”. Maybe.

    I’d love it if they added up the velocities (times 1 sec) to get 30! – I can’t put my finger on why, but that seems sophisticated somehow – particularly, of course, if they then note that this is an over-estimate.

    Third:
    Hm. surely there are, but not coming to mind!

    • December 17, 2011 12:39 pm

      Interval and moments is certainly a big thing. Makes me think we need to run the velocity from 10 to 2…. (not to 0), so that we can better distinguish how many things they sum up. As is, they could leave the last sum out from their work because they think they should, or just because they know it won’t contribute (being zero and all). The confusion between moments and intervals also comes up in the approach discussed by Becca in which students just take the two column numbers and multiply.

      I like how you identify the crux of the matter has having too much information. I think that’s right. The undergraduate student I am working with is interested in looking at how students think about continuous motion given discrete data. This all started with an interest in how student learn from video-analysis of motion.

      I agree that the sum to 30 or to 20 are reasonable approaches, but especially if you recognize them as over and under estimates. They are just Riemann sums. Also, the “take the average of each interval” is just the midpoint Riemann sum, which happens to get you perfect here because it’s constant acceleration.

      We are going to do some interviews in spring with this question and some similar ones.

  6. Andy "SuperFly" Rundquist permalink
    December 13, 2011 6:00 pm

    My first approach to get close is to add them up – basically John’s incorrect approach. I know that that’ll be a little high as several have mentioned, but it’s not bad. What’s interesting, though, is how do we know it’s 10 m/s at time = 0? I wouldn’t be surprised if a student thought that meant the avg speed during the interval from t_i to t_i+1 was listed, in which case adding them all is correct.

    • December 17, 2011 12:42 pm

      It’s a good question Andy… we could be more explicit about it being readings of instantaneous speed that occur at specific clock readings.

      We also asked students to describe the motion in words, and I don’t think any student interpreted it in the way you suggest. But we should definitely check. Since we’ll be doing some interviews in the spring, this is certainly something we can look into.

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