5.5. Multiplication and Geometry

Here is a neat way of thinking about algebra that will give us insight into many different circumstances. We’ll use it a lot. Let’s simplify our model to be just \(x = vt\). Do you notice anything familiar about the form of that equation? Maybe this will remind you.

Suppose my yard is chalked with foot-long markers. (You mean yours isn’t?)

Fig. 5.10 My yard: (a), now and (b) later.

What is the area of my yard? You’d quickly calculate for (a) that it’s \(20 \times 30 = 600\) square feet, right? That is, you remember that in order to calculate an area you multiply the width times the height:

\[A=E \times N.\]

Suppose now I want to sell an eastern parcel to a neighbor leaving me with only 10 feet of width. But I want the area of my yard to remain the same. That can happen if I can buy land from my northern neighbor. How much do I need to buy? What must the north - south dimension of my yard be in order to keep the area of 600 sqft? The yard drawing in (b) answers that question. We simply modify the area formula

\[N = A/E = 600/10 = 60,\]

making one variable on the right small and the other one large in order to keep the total area the same. Of course you could calculate the area in many ways: cut out the picture and reform it into a different rectangle. Or notice that there are dots in the squares of (a)… count them up and you’ll find there are six. So the new yard area also has to have six dots-worth of yard-squares. (Each rectangle is worth 10 sqft as you can see.) We’ll make use of this simple idea over and over.

Like now:

Notice that

\[x = vt\]

is actually the equation of a rectangle of “area” \(x\) with sides of in velocity (\(v\)) and time (\(t\)). We travel 300 miles at a steady 60 mph and it took five hours. We can make a geometrical model of this motion as in (a) below:

Fig. 5.11 Our trip’s model re-envisioned as a solution of areas. (a) is the safe trip in five hours and (b) is the unsafe trip in three hours.

  Pens out!

Now suppose we’re impatient and we want our trip up north to take three hours rather than five hours. The distance is still the same 300 miles, but now we have to go faster. How much faster? Count the dots (15), cut out each little rectangle and re-form them. See (b) in the diagram. Or, we could just do the simple math:

\[\begin{split}\begin{align} x &= vt \nonumber \\ v &= \frac{x}{t} = \frac{300 \text{ miles}}{3 \text{ hours}} = 100 \text{ mph} \nonumber \end{align}\end{split}\]

I don’t advise it. But you get my meaning.

We’ll be able to do this many times for many different circumstances. Yard acreage or even speeds seem obvious, but we’ll encounter more complicated ideas for which this geometrical thinking will actually lead to physics-insight. Trust me, I’m a doctor.

  Please answer Question 3 for points:

Area calculation of speed

**algebra and geometry ** I’m impressed by this sort of thing. Look what we did. We took our intuitive notion of a constant speed in an English paragraph, turned it into a simple algebraic equation, and then turned that into a graph that will give our location at any point along the way. And then we insight into the physics by thinking of our speed equation like an “area” equation. This happens often—some algebra leads to a geometrical relation. In physics we tend to treat the plot on par with the equation as an “explanation.”

The problem with this story is: how did we magically get to 60 mph? We accelerated and therein is another Galileo story and a fable.

But first, let’s visit Spacetime.