PHY 101 1nd Edition Lecture 7 Outline of Last Lecture I. Free Falling ObjectsII. Problem Solving StrategiesIII. 3.1 Vectors and PropertiesIV. Adding VectorsOutline of Current Lecture V. 3.2 Components of a VectorVI. Vector Addiction AlgebraicallyVII. 3.3 Displacement, Velocity, Acceleration in 2 Dimensionsa. Average Velocityb. Instantaneous Velocityc. Average Accelerationd. Instantaneous AccelerationCurrent Lecture3.2: Components of a Vector- A vector can be decomposed into componentso Each component by itself is a vector- A two dimensional vector has 2 componentso A = Ax + Ayo 1D Vector = 1 componentso 3D Vector = 3 components- The magnitude of the componentso Ax = AcosΘo Ay = AsinΘo “A” is the magnitude of the vector A- Using the Pythagorean theorem we have:o A = (Ax2 + Ay2)1/2o The angle Θ can also be calculatedo tanΘ = Ay/Axo Θ = tan-1(Ay/Ax)o The calculated angle is correct only if the vector lies in the 1st or 4th quadrantThese 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.- Example:o Superman flies from the top of a building along a path. The magnitude of the displacement of superman is 100m.o A. Θ = -30.0° because it is clockwise from the axis. Find the x and y components. Ax = Acos (-30.0) Ax= 100(0.866) Ax = 86.6m Ay = Asin(-30.0) Ay = 100(-0.5) Ay = -50m- Vector Addition Algebraicallyo Graphically = tip to tail methodo Vector Addition can be done numericallyo Suppose R = A + Bo Then R = Rx + Ry, where Rx = Ax + Bx and Ry = Ay + Byo The x and y components are added separatelyo Graphically and algebraic methods should give the same answer- Example:o A hiker travevls 25 km 45° south of east on the first day then travels 40km 60° north of east on the second day.o A. Determine the components of the hikers displacement on the first and second day (day 1 = A, day 2 = B) Draw the coordinate vectors Ax = Acos (-45)** (Make sure the angle is negative!!) Ax = 25 (0.707) = 17.7 km Ay = Asin (-45) Ay = 25(-0.707) = -17.7km Bx = Bcos (60) Bx = 40(0.5) = 20 km By = Bsin (60) By = 40 (0.866) = 34.6 km Total displacement = R Rx = Ax + Bx Rx = 17.7 + 20 = 37.7km Ry = Ay + By Ry = -17.7 +34.6 = -16.9kmo B. what is the magnitude of the hikers total displacement for 2 days? R = (Rx2 + Ry2)1/2 R = [(37.7)2 + (-16.9)2]1/2 = 41.3 km Direction: Θ = tan -1 (Ry/Rx) Θ = tan-1(-16.9/37.7) Θ = 24.1°3.3 Displacement, Velocity, Acceleration in 2 Dimensions- For 2D motion, we must consider explicitly the vector nature of displacement, velocity and acceleration- Displacement: use graphs (and tip to tail method) to figure out the path of the object as vector quantities- Average velocity (during time interval Δt)o v = Δr/Δt SI Units: m/so direction of the average velocity is the same as r- instantaneous velocityo v = lim Δt 0 (Δr/ Δt) SI unit: m/so direction of instantaneous velocity along the line that is tangent to the object’s path- average accelerationo a = Δv/Δt SI Unit: m/s2o direction: same as Δv- instantaneous accelerationo a = lim Δt 0 (Δv/ Δt)- Important*** a nonzero acceleration does not necessarily lead to a change in speed. The object may simply change the direction of its velocity without changing its
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