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Connecting kinematic eqns & the best way to eat PB&J sandwich

October 23, 2010
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Fresh off of yesterday’s awesome discovery of THE equation, I decided to push us a bit deeper into studying constant acceleration kinematic equations.

I started by putting this graph up again on the smartboard.

Then I asked for the big ideas that kids found from the graph. First they mentioned that the slope in the velocity graph is the acceleration. Pushing them deeper, I ask them “suppose that factoid is pulled from your brain, how could you get it back?” and they also recognize that slope is \frac{\Delta v}{\Delta t}.” I talked about how in physics survivor, understanding the meaning of slope is like having a knife. And you can make fire with a knife and flint, which we’ll see later.


but someone else says v=a\Delta t +v_0

This lead to a great discussion of the difference between \Delta t, a time interval, and t an instant of time. And how if we assume our interval begins at time t=0, then the two equations are really the same.

Ok, onto bigger and better things. I remind them over the awesome video we watched a bit about the math salon, and the importance of seeing many ways to solve a problem. What were the other equations students came up with to find the displacement? And we start with the other big idea: in a velocity vs time graph, the area represents the displacement (this is the flint in physics survivor).

So students described 3 different equations:

  1. \Delta x= v_0\Delta t +\frac{1}{2}\Delta v\Delta t
  2. \Delta x= v_0\Delta t +\frac{1}{2}a\left(\Delta t\right)^2
  3. \Delta x= \frac{1}{2}\left(v_i+v_f\right)\Delta t

Then I asked them how these fit in with the big idea. Starting with equation 1, how do each of these terms represent areas?

They quickly realized that v_0t is the area of the red box, while \frac{1}{2}\Delta v \Delta t is the area of the blue triangle.

How do we get from equation 1 to equation 2? Giving the students time to think, someone eventually realized that you could write

\Delta v= a\Delta t, and consequently,

\frac{1}{2}\Delta v \Delta t= \frac{1}{2}a\left(\Delta t\right) ^2

Bang! Two seemingly different equations reduced to one.

Now let’s look a the third equation. It came from trying to find the area of a trapezoid. What is the meaning of v_i +v_f? It’s a height, which in this case is 9 tickmarks. Then \Delta t is a width, so we’ve just found a box that double the area (light red box) of the trapezoid we want (solid red trapezoid). ZOMG!

Now for a twist, I showed this graph, and said how can we use our our new understanding on this one. First, what’s different? Students see that the acceleration is negative, and because the inital velocity is positive, this object is slowing down. This also means that \Delta v will be negative. This will important later.

How can we write the area of the box and the triangle this time. Kids quickly say that the box should be v_f\Delta t (note the slight difference) and then the triangle is \frac{1}{2} \Delta v \Delta t.

So we look at the equation again, very carefully.

\Delta x= v_0\Delta t +\frac{1}{2}\Delta v\Delta t

Someone realizes that this won’t work. The area of the triangle is negative! So to fix it, we’d need absolute value bars, or to make it v_i-v_f. This seems kludgy, and I try to make a hackneyed analogy to learning zone defense, which I won’t even bother to repeat here.

But is there another way to see this? And someone says, sure—just take the bigger box ( v_0t) and then add the negative displacement of the triangle (\frac{1}{2}\Delta v\Delta t).

Hey, isn’t this isn’t the same thing we just did? Yup. Keep you eyes on the big idea (Area = displacement, cover your zone, not your man), and then the details (negative signs, foot placement) take care of themselves.

And so we come back to the big idea again, and give a bit more meaning to these terms

  • v_0t The displacement because of the objects initial velocity.
  • \frac{1}{2}\Delta v\Delta t=\frac{1}{2}a\Delta t The displacement of the object due to the changesin the object’s velocity.

Ok, time for the grand finale with PB&J. Here’s the graph we’ve seen a ton.

I ask the kids what they see. “The displacement is zero!” and they tell me how they can see the area of the positive triangle above the axis is equal to the negative area. Can we write this?

We quickly see that writing the area of that positive triangle is tough, without resorting to kludges. Hmm, is there another way?

Well, what about the rectangle v_0t?

Then we could just add the displacement \frac{1}{2}\Delta v\Delta t which is negative. And we should get zero.

But how can you see this? How is the area of that rectangle equal to the triangle? This is where the magic happened, and it was because a student pointed out that there’s an even bigger box.

And then it all clicked. Some people eat PB&J sandwiches like this:

The right people eat them like this:

But no matter what, the sandwich is cut in half. And for just a moment, it felt like everyone understood. Survivor—knife and flint makes fire, even when you lose your \Delta x=\frac{1}{2}\Delta v \Delta t+ \frac{1}{2}a\left(\Delta t\right) ^2 inst-o-flame.

I’ve attached the video me teaching this lesson for the second time here. As you’ll see, it’s pretty teacher led, and didactic. I don’t think that’s the best thing, and I’d appreciate any feedback/criticism you may have.


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