lecture25_1100_S12

lecture25_1100_S12 - PHYS 1100 Spring 2012 I. Graphical...

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I. Graphical Kinematics A. Slope problems B. Area under the curve problems PHYS 1100 Spring 2012 Reading for Wednesday: 2.10
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Derivative version: (at 1 point) Graphically: v = slope of tangent line of x v. t plot t x t 1 t 2 m 2 =v 2 m 1 =v 1 Average version: (between 2 points) Graphically: vavg = slope of secant line of x t plot t x t 1 t 2 m=v avg Position v. Time plots PHYS 1100 Spring 2012 v = d x dt v avg = x t
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CT1: The graph below shows the motion of an object in one dimension as a function of time. The positive direction points to the right. When is the object to the right of the origin? A. -3 s < t < -1 s B. -1 s < t < 1 s C. 1 s < t < 3 s D. 3 s < t < 5 s -3 -2 -1 0 1 2 3 4 5 -5 -3 -1 1 3 5 7 9 11 13 15 PHYS 1100 Spring 2012 t (s) x (m)
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On x v. t plot “located in positive direction” above t axis “has a positive position” “located in negative direction” below t axis “has a negative position” “located at the origin” on t axis PHYS 1100 Spring 2012
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CT2: The graph below shows the motion of an object in one dimension as a function of time.
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This note was uploaded on 03/29/2012 for the course PHYSICS 1100 taught by Professor Giammanco during the Spring '08 term at LSU.

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lecture25_1100_S12 - PHYS 1100 Spring 2012 I. Graphical...

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