PHY 102 1st EditionOutline of Last LectureI. Uniformly Accelerated Motiona. Exampleb. Y InterceptII. Equations of MotionIII. Steps of solving kinematic problemsIV. Graphs of Motiona. Constant motionb. Uniformly Accelerated MotionOutline of Current Lecture I. Positiona. Distanceb. DisplacementII. VectorIII. Relationships between velocitiesIV. Accelerationa. RelationshipsV. Example ProblemsCurrent LectureI. Position – magnitude and direction (vector) from some arbitrary locationa. Distance – track along which something travelsb. Displacement – straight line distance between points A & BThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.II. Vector – magnitude & directiona. 12 mi N – displacementb. 12 mi – distancec. Average speed – has to factor in time, NOT just total time/distancei. Ave speed: v = ∆ x / td. Speed = I velocity Ii. Vf = final velocityii. Vi = initial velocityiii. V0 = original velocitye. Tangents to a curved Position-Time Graph are instantaneous velocityIII. Relationships between vave and vinst a. V = ½ (v1 + vt)b. A = constantc.∆ x=vtd.i¿+ vtv¿∆ x=12¿IV. Accelerationa. Rate in change of velocity (speed or direction)b. Acceleration is a vector quantity that can be positive, negative, or zeroc. Negative acceleration does NOT necessarily mean slowing downd. Relationshipsi. If +v and +a, it goes faster in the positive directionii. If +v and –a, it goes slower in the positive directioniii. If –v and +a, it goes slower in the – directioniv. If –v and –a, it goes faster in the – directionv. Acceleration is in the same direction as the forceV. Examplesa. A vehicle travels 65s in 28 seconds.V = ?∆ x = 65 m t = 28 sv = ∆ xt65 m/28s2.1 m/sb. A vehicle accelerates from rest to 22 m/s in 4 s.A = ?V0 = 0 m/sV = 22 m/sT = 4sV = v0 + atv-v0 = ata = v-v0 /ta = (22 m/s -0)/4a = 5.5
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