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Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2006

A honey bee makes several trips from the hive to a flower garden. The velocity graph is shown below. What is the total distance traveled by the bee? 200ft 100ft 700 feet

What is the displacement of the bee? 200ft -200ft 200ft -100ft 100 feet towards the hive

To find the displacement (position shift) from the velocity function, we just integrate the function. The negative areas below the x-axis subtract from the total displacement. To find distance traveled we have to use absolute value. Find the roots of the velocity equation and integrate in pieces, just like when we found the area between a curve and the x-axis. (Take the absolute value of each integral.) Or you can use your calculator to integrate the absolute value of the velocity function.

velocity graph position graph Displacement: Distance Traveled: Every AP exam I have seen has had at least one problem requiring students to interpret velocity and position graphs.

In the linear motion equation: V(t) is a function of time. For a very small change in time, V(t) can be considered a constant. We add up all the small changes in S to get the total distance.

As the number of subintervals becomes infinitely large (and the width becomes infinitely small), we have integration.

This same technique is used in many different real-life problems.

Example 5: National Potato Consumption The rate of potato consumption for a particular country was: where t is the number of years since 1970 and C is in millions of bushels per year. For a small, the rate of consumption is constant. The amount consumed during that short time is.

Example 5: National Potato Consumption The amount consumed during that short time is. We add up all these small amounts to get the total consumption: From the beginning of 1972 to the end of 1973: million bushels

Work: Calculating the work is easy when the force and distance are constant. When the amount of force varies, we get to use calculus!

Hooke’s law for springs: x = distance that the spring is extended beyond its natural length k = spring constant

Hooke’s law for springs: Example 7: It takes 10 Newtons to stretch a spring 2 meters beyond its natural length. F =10 N x =2 M How much work is done stretching the spring to 4 meters beyond its natural length?

F(x)F(x) x =4 M How much work is done stretching the spring to 4 meters beyond its natural length? For a very small change in x, the force is constant. newton-metersjoules 

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