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Consider the following generalized function of x(t) versus t: Two points P 1 and P 2 on the graph are shown, along with their generalized coordinates. Topic 2.1 Extended C – Velocity in function notation x t x(t)x(t) t1t1 x1x1 t2t2 x2x2 P1(t1,x1)P1(t1,x1) P2(t2,x2)P2(t2,x2) We call the line joining any two points on a graph a secant line. secant line The slope of the secant line is easily found by dividing the rise by the run: t x RISE RUN

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The average velocity is given by Topic 2.1 Extended C – Velocity in function notation x t x(t)x(t) t1t1 x1x1 t2t2 x2x2 P1(t1,x1)P1(t1,x1) P2(t2,x2)P2(t2,x2) secant line t x RISE RUN The slope of the secant line is the average velocity v = x t Average Velocity Now we're going to simplify our notation a bit, so we don't need subscripts: We begin by calling t 1 just plain "t." t Then t 2 becomes "t + t." t + t

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The average velocity is given by Topic 2.1 Extended C – Velocity in function notation x t x(t)x(t) x1x1 x2x2 secant line t x RISE RUN v = x t Average Velocity Now, using function notation we can express our two x- values in terms of the t-values: Thus x 1 is the same as x(t). t Then x 2 becomes x(t + t). t + t x(t)x(t) x(t + t) FYI: At first glance it appears that we are making things more complicated. But what we have actually done is we have taken four different values: x 1, x 2, t 1 and t 2, and condensed them into three different values: x, t, and t. And we have eliminated the subscripts.

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The average velocity is given by Topic 2.1 Extended C – Velocity in function notation x t x(t)x(t) secant line t x RISE RUN v = x t Average Velocity We can now express the average velocity in terms of the three new values: t t + t x(t)x(t) x(t + t) v = x t = x(t + t) - x(t) t Average Velocity function notation

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Suppose a particle has a position x which can be expressed as x = 2t 2, where x is in meters and t is in seconds. Topic 2.1 Extended C – Velocity in function notation (a) Find the average velocity between t = 1.000 s and t = 1.100 s using the original definition: v = x t Average Velocity x 1 = 2(1.000) 2 = 2.000 m t 1 = 1.000 s x 2 = 2(1.100) 2 = 2.420 m t 2 = 1.100 s t = 0.100 s x = 0.420 m v = x t = 0.420 m 0.100 s = 4.20 m/s

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Suppose a particle has a position x which can be expressed as x = 2t 2, where x is in meters and t is in seconds. Topic 2.1 Extended C – Velocity in function notation (b) Find the average velocity between t = 1.000 s and t = 1.100 s using the new definition of average velocity: t = 1.000 s v = x(t + t) - x(t) t Average Velocity function notation t + t = 1.100 s x(t) = 2(1.000) 2 = 2.000 m x(t + t) = 2(1.100) 2 = 2.420 m t = 0.100 s v = x(t + t) - x(t) t = 2.420 m - 2.000 m 0.100 s = 4.20 m/s

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