Average Velocity - Problem 1
Let’s compute some average velocities. A pumpkin is launched into the air time t is in seconds, and height of the pumpkin, f(t) is in feet. Here’s my table of values for f(t) and for time. And part says compute the average of velocities on the interval t between 0 and 2 and t between 2 and 4. Let’s start with the interval between 0 and 2.
So I need to calculate the change in position which is f(2) minus f(0), over the change in time which is 2 minus 0. F(2) according to the table is 200. And f(0) is 4 and the change in time is 2. So 200 minus 4 is 196 and that’s in feet. And the denominator 2 is in seconds. So this is going to give me 98 feet per second.
Now what about between 2 and 4? The change in position here is f of 4 minus f(2). Remember it's final position minus initial over the change in time. F(4) is 268 and f(2) is 200 still, then we have 2 again in the denominator.
This is going to give me 68 feet over 2 seconds. So 68 divided by 2 is 34 feet per second. So you can see that the pumpkin is slowing down a little bit. Now let’s take a look at how we interpret this average velocities on a graph.
I’ve drawn a graph of the position the height if the pumpkin over time. Actually I should change this, it should be time t. It’s important to remember that a graph like this does not necessarily indicate the path of the pumpkin. The pumpkin is shot straight up into the air. And so all this tells you is what the height is at different times. And so we have to kind of lay our time axis and our height axis. This point represents the height of the pumpkin at time 0, it's 4 feet height. This point here the t equals 2, the pumpkin is 200 feet in the air.
T equals 4 to 268. Now think about what these quantities quotients look like. You’ve got 196 a change in position over 2 seconds a change in time. The change in time can be represent by that distance right there, 2 seconds. And the change in position could be represented by this vertical, 196 feet. And so the ratio of this divided by this could be thought as this is the slope of this line.
So the slope of this guy is going to be 98 feet per second. And that’s how we interpret average velocity. It’s the slope of a secant line drawn to the graph of position. Now what about the second interval?
Let me use another color here, red. From 2 to 4, we still have a 2 second interval but this time the increase is less. The increase was only 68 feet, but again the average velocity can be interpreted as the slope of that secant line. This time, the slope is 34 feet per second. Now this interpretation is important because it helps you look at a graph and tell approximately what the average velocity is, at least qualitatively. The fact that this secant line is steeper than this one, it means the average velocity is greater here than it is here.
And if I drew a secant line between two other points say here and here, the fact that it sloped downward indicates the average velocity is negative. So you could learn a lot about average velocity using its interpretation as the slope of a secant line on the position graph.