2c_lec2

2c_lec2 - Physics 2A! Olga Dudko UCSD Physics! Lecture 2!...

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Today: Variables of motion. Motion in a straight line. Kinematic equations. Objects in free fall. Feather and hammer drop” demo Physics 2A Olga Dudko UCSD Physics Lecture 2
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Motion in 1D Any description of motion involves three concepts: 1) Displacement 2) Velocity 3) Acceleration Studiodaily.com
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First, defne a coordinate system 0 x
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Displacement and Distance ! x x 1 x 2 x 0
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Velocity and Speed Average speed # 0 always v avg = ! x ! t = x 2 ± x 1 t 2 ± t 1 speed avg = d ! t = d t 2 ± t 1 A B t 1 = 3:00 pm x 1 = 0 km t 2 = 3:15 pm x 2 = 1 km t 3 = 4:00 pm x 3 = x 1 v avg = speed avg = 0 2 km/hr
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Average velocity from x(t) graph • Graphically, we fnd average velocity by examining the rise ( ! x) over the run ( ! t) in an x vs. t graph. Average velocity between points A and B is v avg = x 2 ± x 1 t 2 ± t 1 = 50 m ± 30 m 10 s ± 0 s = 20 m 10 s = 2 m s
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Instantaneous Velocity v = lim ! t ! 0 ! x ! t = dx dt infnitesimals v = dx dt
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Velocity from x(t) graph
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Velocity W E x 0 A B
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Velocity Similarly, Recall that displacement is v A = ! x A t ! x A = v A t ! x B = v B t ! x A = x A , fnal ± x A , initial x A , fnal = x A , initial + v A t x B , fnal = x B , initial + v B t => => => When the runners meet… x B , fnal = x A , fnal x B , initial + v B t = x A , initial + v A t their positions are equal: => We know every variable here except t. Solve for t. x 0 A B
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Velocity x B , fnal = x B , initial + v B t = ± 0 . 18 miles “-” means… = 3 . 0 miles + ± 5 . 0 mi
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2c_lec2 - Physics 2A! Olga Dudko UCSD Physics! Lecture 2!...

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