 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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Average Velocity - Problem 3

# Average Velocity - Problem 2

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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The average velocity of an object is the change in its position over a period of time. So, in order to calculate the average velocity of an object, simply find out what its position is at the desired times. This can be done by looking at a table of the values of the object's position and at what time it is at that position, looking at the graph of a function of the object's position and time, or by evaluating a function f(t), where f(t) makes the object's position a function of time (meaning that the position of the object varies, depending on the time). After finding the position of the object at the specified times, divide the difference of the position of the objects by the difference of the times. For instance, if at t=2, f(t)=7 and at t=5, f(t)=28, the average velocity will be (28-7)/(5-2)=21/3=7. Notice that finding the average velocity is just like finding the slope between two points.

We got another average velocity problem. In this situation, we have a motorcycle stopping at a stop sign, and the graph of its position over time is given here. Times in seconds and its position is given in feet. Find the average velocity over these intervals between 0 and 2, 2 and 3, 3 and 5 and 5 and 8.

So let’s start with the interval from 0 to 2. Remember that average velocity is, change in position over change in time. So this is my change in position. I need to know f(2) and f(0). So let’s take a look at the graph

F(2) is going to be 30. And f(0) is 0. So this is 30 minus 0 over 2, and that’s 30 feet over 2 seconds. So this simplifies to 15 feet per second. Now let’s see the interval from 2 to 3. It’s going to be f(3) minus f(2) that change in position over 3 minus 2. The change in time.

And remember, the change in position it's final position minus initial. So back to the graph. F(3) is 40 again f(2) is 30. So I have 40 minus 30 over 1. And this is 10 feet over 1 second. 10 feet per second. So it looks like the motorcycle is slowing down which it should be. And then from 3 to 5, f(5) minus f(3) over 5 minus 3. Let’s look at the graph. Now here’s t equals 3, here's t equals 5 and both of the values for f are 40. So we have 40 minus 40, over 5 minus 3, 2. And that’s going to be 0 feet over 2 seconds.

0 feet per second. And this represents the time the two seconds when motorcycles actually stopped right here. And now finally we want to go between 5 and 8. So we are going to need 5 of 5 which is 40 and f(8) which is 80. So this is going to be f(8) minus f(5) over 8 minus 5. 80 minus 40 over 3. 40 feet over 3 seconds. So this didn’t come out as nice but it happens to be 13 and 1/3 feet per second.

So remember average velocity is always change in position over change in time. And you can always read that from a graph. If your graph is drawn like this with a grid, you can approximate values for position in time and calculate average velocity that way. And also remember that average velocity can be represented as the slope of a secant line. So for example from 0 to 2, we got 15 feet per second. That would be the slope of this secant line, 15 feet per second.