lecture26_1100_S12

# lecture26_1100_S12 - I Graphical Kinematics A Slope...

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Unformatted text preview: I. Graphical Kinematics A. Slope problems B. Area under the curve problems PHYS 1100 Spring 2012 Reading for Friday: 2.9, 4.5 – 4.6 Derivative version: (at 1 point) Graphically: a = slope of tangent line of v v. t plot t v t 1 t 2 m 2 =a 2 m 1 =a 1 Average version: (between 2 points) Graphically: aavg = slope of secant line of v v. t plot t v t 1 t 2 m=a avg Velocity v. Time plots PHYS 1100 Spring 2012 a = d v dt a avg = ∆ v ∆ t CT9: The graph below shows the velocity of an object in one dimension as a function of time. When is the object traveling in the positive direction? A.-5 s < t < -1 s B.-1 s < t < 0 s C. 0 s < t < 3 s D. 3 s < t < 5 s-5-4-3-2-1 1 2 3 4 5 2 4 6 8 10 12 PHYS 1100 Spring 2012 t (s) v (m/s) On v v. t plot “moving in positive direction” above t axis “has a positive velocity” “has a displacement in positive direction” “moving in negative direction” below t axis “has a negative velocity” “has a displacement in negative direction” “not moving” on t axis “stopped” “turning around” “zero velocity” PHYS 1100 Spring 2012 1 2 3 4 5-10-5 5 10 15 CT10: The graph below shows the velocity of an object in one dimension as a function of...
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lecture26_1100_S12 - I Graphical Kinematics A Slope...

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