Super Advanced GraphingLevel 1: Finding Displacement on a Velocity GraphIn the last unit, we talked a lot about how to find the velocity from a displacement graph. We even used displacement graphs to create velocity graphs. What if we had a velocity graph and we wanted to know the displacement of an object? How could we do this?Believe it or not, some of you figured out this section by yourself. I know you did this because I snuck in a few of these questions in the last unit, just to see if you could do it. Here’s one:

Some of you very smart cookies realized that in Section B, you know the velocity (12 m/s, which you got from reading the graph) and you know how long the object was moving at this velocity (5 seconds, which also comes from reading the graph). You then remembered that:

velocity=displacementtime

You could plug these numbers in and solve:

12=d5

12×5=60m=d

If you did this on your own, great! If you didn’t, don’t worry, it isn’t immediately obvious at all. This problem solving works well as long as we have a nice, constant velocity, but I could also ask you to figure out how far something moves in much more complicated velocity vs. time graphs. Like these:

What do we do when we can’t just plug the numbers into an equation? Don’t panic. As it turns out, this is actually pretty easy if you know a trick. Here it is:Key IdeaThe area between a velocity vs. time line and the x axis is the displacement.

I know this sounds hard, but it is actually not that bad. Let’s look at an easy one together first.

Rectangular AreasThis graph shows an object that is moving at a steady 10 m/s for 0-30 seconds. We could plug this into an equation to find displacement (see above) but let’s try it the other way for right now.

First, shade in the area between the line and the x axis.

If you did that right, it should look like this:

0 5 10 15 20 25 30 350

2

4

6

8

10

12

Time (s)

Velo

city

(m/s)

Now, we need to find the area of this section. How can we do that? We can use one of the two area equations:

Arectangle=base×height=bh

Atriangle=bh2

Because we are looking at a rectangle, we’ll use the first equation. In the picture below, I show you the base and the height, in case you have trouble remembering.

0 5 10 15 20 25 30 3502468

1012

Time (s)

Velo

city

(m/s)

0 5 10 15 20 25 30 350

2

4

6

8

10

12

Time (s)

Velo

city

(m/s)

In this situation, the base was 30, the height was 10 so…Arectangle=bh

Arectangle=(30 ) (10 )

Arectangle=300

Which means…d=300m

Practice ProblemFind the displacement of the object in the picture:

Answer180 m

Triangular AreasWhat if we had triangular area? Then we use the other area equation:

Atriangle=bh2

Practice ProblemFind the displacement of the object in the graph below.

Answer313 m

Combination AreasFinding the areas of rectangles and triangles is nice, but what if we had a little more complicated motion? Maybe one like this:

This is the graph of an object moving at a constant velocity and then speeding up. Think of how you could find the area of a shape like this.Don’t worry- you don’t need to learn a new equation. Just take the area under the line and break it up into rectangles and triangles. It doesn’t matter how you do that, as long as you end up with rectangles and triangles under the line. It should look something like this:

Now, we follow the same steps we always take. We find the area of the rectangle and the triangle.

Add them together and we get the total displacement: 75 m.

Negative DisplacementOne last situation. What if you had an object that moved in the negative direction for a while? Would that effect the displacement? Yup! This is like someone walking forward for a while, slowing to a stop and speeding up in the opposite direction. Because displacement is how far you are from where the

object started, you need to think of the displacement in the negative direction as negative. How do you solve these?Look at this graph. The object is going in the positive direction at first, slowing down. Then it comes to a stop and speeds up in the opposite direction. In other words, it is moves forward, stops, then backs up. How do we calculate how far it is from where it started (its displacement)?

Shade in the two areas between the line and the x axis and find their areas.

For the area above the x axis, the displacement is positive (+9) for the area below the x axis, the area is negative (-8 m).

When we add these together, we find the object has a total displacement of +1 m. In other words, the object is +1 m away from where it started.

Level 2: Acceleration GraphsSo far, we’ve looked at displacement vs. time graphs that involved constant velocities. They looked like this:

In real life, however, most graphs look something like this:

Check out those curves. What does a curve on a position vs. time graph mean?Key IdeaA curve on a position vs. time graph tells us the object is accelerating.Remember that acceleration does not just mean the object is speeding up. It can also mean the object is slowing down or changing direction. Acceleration measures how much the velocity is changing. How can we describe the change in velocity by looking at a position vs. time graph? In higher level mathematics classes you can learn to do this quantitatively (with numbers). In this class, you only need to do this qualitatively (without numbers). I’m going to show you

how to do this. Nearly all my students are able to master this, but it takes a little practice. Be patient with yourself as you learn.

Finding the Velocity on a Curved Position vs. Time GraphLet’s start with a very typical curved position vs. time graph:

The second we look at this, we know that the object is accelerating, because the line is curved. We just don’t know whether the object is speeding up, slowing down, coming to a stop…How do we figure this out? We are going to follow a procedure that lets us make a velocity graph of this motion. First, keep in mind that the slope of a position vs. time graph is the velocity. But how do we find the

slope of a curve? As it turns out, this involves calculus, which is a branch of math that is stunningly beautiful but also fairly difficult. Have no fear- I will teach you baby calculus. Right now. In a page.Look at the graph. Let’s focus on that first point. If we wanted to know the velocity of the object at this point, we would have to know the slope. But the line is curved- that it makes this tough. What we need to do is imagine we were shrinking ourselves down on that point, smaller and smaller until the line at that point is huge.

Notice that if we super zoom in on that one point, the curve at that point no longer looks curved- it looks like a straight line. Look at that zoomed-in line. Is its slope positive, negative, or zero?It’s a little positive, right?Therefore, the velocity at this point is a little positive. We are going to use this idea of zooming in to figure out how the velocity is changing in this graph. I have a system that will work for you every time. Follow the steps below. Make sure you follow them exactly when you first start out. After you’ve done them a few times, they get to be second nature. You will need a ruler to do these.Our Goal: To create the velocity graph of the motion below.

1. Pick on point on the far left of the graph. It doesn’t matter where you pick, just as long as it is on the left. For this example, five points have already been picked for you. In the future, you will need to choose your own.

2. Lay your ruler down and move it until it just kisses that point. In other words, the ruler should only be touching the graph at that point. It shouldn’t be hitting multiple points.

3. Now, ignore the curve and just look at the ruler. Is the slope positive, negative or zero? Is it really positive? A little positive? A little negative? A lot negative? In this example, this point was a little positive. If we were to draw this on a velocity vs. time graph, we would put a dot at this point in time showing the velocity is slightly positive.

4. Repeat the process with the next point. Try it with your ruler now.

It’s still positive, right? Is it more positive than before, or less?

More.

This means the object is now moving a little faster in the positive direction.

5. Add another point to your velocity graph showing the new slope.

6. Repeat, moving left to right across the graph, until a clear diagonal line emerges.

If you did this right, you can clearly see the graph has an increasing velocity. The object is speeding up in the positive direction.

Take your time on the practice for this section. Don’t be afraid to ask for my help. Sometimes, it helps to have me act one out for your group.