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# Instantaneous Velocity - Problem 3

###### 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.

By finding the slope between two points that are very close together on a position vs. time graph, you can find an approximate value for the instantaneous velocity of an object. One method that can be used to find the instantaneous velocity is to use data points given in a table, and finding the average velocity of the object between two points where their times t are very close together. Instantaneous velocity can then be estimated using the same methods as finding the average velocity. However, this is closer to instantaneous velocity because the distance between the points is very small. One drawback to the method of using tables is that you can only use the values of time given, which may not be very close together. This would cause estimate for the instantaneous velocity to be less accurate. So, in order to estimate instantaneous velocity at a point, find the average velocity at that point one increment smaller than the point and one increment larger than the point. Then, by finding the average of those two average velocities, you find a better estimate for the instantaneous velocity at that point.

Let’s do another instantaneous velocity problem. This time we’re not going to be using a graph. We have a problem that involves a table. A motorcyclist position f(t) is given by, and here I have some values of time, 0 through 2.5 in increments of half a second. I have a distance in feet, 0, 4, 12, etcetera. Now I want to find the average velocity over the interval from 1.5 to 2. This is easy, this is review.

Remember average velocity is change in position over change in time. So f(2) minus f(1.5), over 2 minus 1.5. f(2) is 50 f(1.5) is 27 so that's 50 minus 27. Remember these are in feet. Over 2 minus 1.5, that’s 1.5 and this will be in seconds.

I have 50 minus 27, 23, over 0.5. I’ll leave out the units per second because I want to simplify this. I just multiply the top and bottom by 2, and I’ll get 46 over 1. This is 46 feet per second. That’s the average velocity for t equals 1.5 to 2.

Now let’s take a look at another part. Part b; Estimate the instantaneous velocity at t equals 2. Now this is going to be tricky because, the way we find instantaneous velocity, is we take intervals of time that are smaller and smaller. But if you look at the table, the smallest interval of time we have to work with is 0.5 seconds. So we’re not going to be able to get smaller than that. We have to work with that as best we can.

I’m interested in estimating the instantaneous velocity at t equals 2. So what I’m going to do is find the average velocity from 1.5 to 2. I actually just did that. And then I’ll find the average velocity from 2 to 2.5. I’ll average those average velocities. Between 1.5 and 2, I already know that the average velocity, I’ll use this symbol, v average, is 46 feet per second. Now what about between 2 and 2.5?

This is a new calculation. I’m going to need f(2.5) minus f(2); the change in position over the change in time. 2.5 minus 2. Now looking back at the table, f(2.5) is 80, f(2) is 50. So this is 80 minus 50 over 0.5. Again the units are feet per second. So 30 over 0.5, If I multiply the top and bottom by 2 I get 60 feet per second. That’s technically the average velocity from t equals 2 to t equals 2.5.

What I’m going to do, is average these two average velocities to get the instantaneous velocity. Add them together 46 plus 60 and divide by 2. I get 106 over 2, which is 53 feet per second. Now when you have a table of data and you want to find instantaneous velocity, the best you can do is approximate it using average velocities. What I always like to do is, again looking back at the table. To get instantaneous velocity at a point like this, find an average velocity to the left, an average velocity to the right and average them. Make sure you use the smallest increment of time possible. So use values that are adjacent to the value you’re interested in. Again averaging the two velocities we got 53 feet per second.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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