inst05.1 - Class 5.1 Analysis of Motion & Newton's Laws...

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Class 5.1 Analysis of Motion & Newton’s Laws
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Class Objective Learn and apply Newton’s First, Second, and Third Laws Unidirectional Multidirectional Learn the relationship between position, velocity, and acceleration
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RAT 1
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Some Definitions (1D) Position - location on a straight line -3 -2 -1 0 1 2 3 4 Displacement - change in location on a straight line -3 -2 -1 0 1 2 3 4 x x x = x 2 - x 1 = 4 - (-2) = 6
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Some Definitions (1D) Average Velocity - rate of position change with time (vector) Time Elapsed nt Displaceme t x v ave = = dt dx t x t v = = 0 lim Instantaneous Velocity (vector)
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Average and Instantaneous Velocity (1D) Position Time t 1 t 2 x 1 x 2 Slope = = Average Velocity x 2 - x 1 t 2 - t 1
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Some Definitions (2D) Position -- a location usually described by a graphic on a map or by a coordinate system 5 4 3 2 1 -3 -2 -1 4 3 2 1 0 -1 -2 -3 -4 (-2, -3) (4, 3) r A
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Some Definitions (2D) Displacement -- change in position, where r 4 -2 -3 -3 -4 (-2, -3) (4, 3) 5 1 r r Δ 2 r 1 2 r r r Δ - =
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Some Definitions (2D) Average velocity (vector) Instantaneous velocity (vector) Speed - the magnitude of instantaneous velocity (scalar) ( 29 ( 29 scalar Time Elapsed vector nt Displaceme t r v ave = = dt r d t r t v = = 0 lim v speed =
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Some Definitions (1D) Average Acceleration - rate of velocity change with time (vector) t v t t v v a 1 2 ave = - - = 1 2 Instantaneous Acceleration (vector) dt dv t v 0 t lim a = =
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Average and Instantaneous Acceleration (1D) Velocity Time t 1 t 2 v 1 v 2 Slope = = Average Acceleration v 2 - v 1 t 2 - t 1
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Some Definitions (2D) Average Acceleration (vector) Instantaneous Acceleration (vector) t v t t v v a 1 2 ave = - - = 1 2 dt v d t v 0 t lim a = =
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Paired Exercise 1 What is the distance traveled? What is the acceleration at 1.25 hours? Speed, miles per hour 3 2 1 0 0 10 20 Time, hours
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For constant acceleration. .. if acceleration is constant integrating both sides v 0 is the original value at the beginning of the time interval dt v d a = dt a v d o = ∫ ∫ = dt a v d o t a v v o o + = (Definition)
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Constant Acceleration substituting the velocity equation from the previous page integrating both sides yields dt x d v = dt v x d = ( 29 dt t a v x d o o + = ( 29 ∫ ∫ + = dt t a v x d o o 2 o o o t a 2 1 t v x x + + = (Definition)
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Equations of Motion (Constant Acceleration) Velocity Position (in terms of x) t a v v o o + = 2 o o o t a 2 1 t v x x + + =
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Multiple Directions
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inst05.1 - Class 5.1 Analysis of Motion & Newton's Laws...

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