PHYS 1443 – Section 003 Lecture #2AnnouncementsOne Dimensional MotionKinematic Equations of Motion on a Straight Line Under Constant AccelerationExample 2.8Free FallExample 2.12Coordinate SystemsExample 3.1Vector and ScalarProperties of VectorsVector OperationsExample 3.2Components and Unit VectorsExamples 3.3 & 3.4Displacement, Velocity, and Acceleration in 2-dimMonday, Sept. 9, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu1PHYS 1443 – Section 003Lecture #2Wednesday, Sept. 9, 2002Dr. Jaehoon Yu1. One dimensional motion w/ constant acceleration2. Kinetimatic equations of motion3. Free fall4. Coordinate systems 5. Vector; its properties, operations and components6. Two dimensional motion•Displacement, Velocity, and Speed•Projectile Motion•Uniform Circular MotionToday’s homework is homework #3, due 1am, next Monday!!Monday, Sept. 9, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu2Announcements•Your e-mail account is automatically assigned by the university, according to the rule: fml####@exchange.uta.edu. Just subscribe to the PHYS1443-003-FALL02. •e-mail:15 of you have subscribed so far. –This is the primary communication tool. So do it ASAP.–A test message will be sent this Wednesday.•Homework registration: 44 of you have registered (I have 56 of you)–Roster will be locked at the end of the day Wednesday, Sept. 11Monday, Sept. 9, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu3One Dimensional Motion•Let’s start with the simplest case: acceleration is constant (a=a0)•Using definitions of average acceleration and velocity, we can draw equations of motion (description of motion, velocity and position as a function of time)tvvttvvaxiixixxffxfIf tf=t and ti=0tavvxxixfFor constant acceleration, simple numeric averagetxxttxxviiixfff If tf=t and ti=0221tatvxtvxx xxiixif Resulting Equation of Motion becomestavtavvvvxxixxixfxix21222tvxxxifMonday, Sept. 9, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu4Kinematic Equations of Motion on a Straight Line Under Constant Acceleration tavtvxxixf221tatvxxxxiif tvvtvxxxixfxif2121 ifxfxxavvxxi 222Velocity as a function of timeDisplacement as a function of velocity and timeDisplacement as a function of time, velocity, and accelerationVelocity as a function of Displacement and accelerationYou may use different forms of Kinematic equations, depending on the information given to you for specific physical problems!!Monday, Sept. 9, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu5Example 2.8•Let’s call the time interval for police to catch the car; T•Set up an equation:Police catches the violator when his final position is the same as the violator’s.2200.32121TaTxPolicef )1(0.451 TTvxCarf)1(0.4500.321 ;2 TTxxCarfPolicefaacbbxcbxax24are 0for Solutions22A car traveling at constant speed of 45.0m/s (~162km/hr or ~100miles/hr), police starts chasing the car at the constant acceleration of 3.00m/s2, one second after the car passes him. How long does it take for police to catch the violator?Ex02-08.ip00.300.3000.12 TTpossible)(not 00.1or (possible) 0.31 TTMonday, Sept. 9, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu6Free Fall•Free fall is a motion under the influence of gravitational pull (gravity) only; Which direction is a freely falling object moving?•Gravitational acceleration is inversely proportional to the distance between the object and the center of the earth•The gravitational acceleration is g=9.80m/s2 on the surface of the earth, most of the time.•The direction of gravitational acceleration is ALWAYS toward the center of the earth, which we normally call (-y); where up and down direction are indicated as the variable “y”•Thus the correct denotation of gravitational acceleration on the surface of the earth is g=-9.80m/s2Monday, Sept. 9, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu7Example 2.12Stone was thrown straight upward at t=0 with +20.0m/s initial velocity on the roof of a 50.0m high building, 1. Find the time the stone reaches at maximum height (v=0)2. Find the maximum height3. Find the time the stone reaches its original height4. Find the velocity of the stone when it reaches its original height5. Find the velocity and position of the stone at t=5.00sg=-9.80m/s2stttavv yyif04.280.90.2000.080.90.20 1)(4.704.200.50)04.2()80.9(2104.2200.502122mtatvyy yyiif2st 08.4204.2 3 Other ways?)/(0.2008.4)80.9(0.20 smtavvyyiyf4)(5.27)00.5()80.9(2100.50.200.502122mtatvyyyyiif5-Position)/(0.2900.5)80.9(0.20smtavvyyiyf5-VelocityEx02-12.ipMonday, Sept. 9, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu8Coordinate Systems•Makes it easy to express locations or positions•Two commonly used systems, depending on convenience–Cartesian (Rectangular) Coordinate System•Coordinates are expressed in (x,y)–Polar Coordinate System •Coordinates are expressed in (r)•Vectors become a lot easier to express and computeO (0,0)(x1,y1)=(r)rsincosryrxHow are Cartesian and Polar coordinates related? 11tan2121xyyxry1x1+x+yMonday, Sept. 9, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu9Example 3.1 Cartesian Coordinate of a point in the xy plane are (x,y)= (-3.50,-2.50)m. Find the polar coordinates of this point.yx(-3.50,-2.50)mrs )(30.45.1850.250.32222myxrs1807550.350.2tan s5.3575tan1s2165.35180180 sMonday, Sept. 9, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu10Vector and ScalarVector quantities have both magnitude (size) and directionScalar quantities have magnitude onlyCan be completely specified with a value and its unitForce, gravitational pull, momentumNormally denoted in BOLD BOLD letters, FF, or a letter with arrow on topFTheir sizes or magnitudes are denoted with normal letters, F, or absolute values:For FEnergy, heat, mass, weightNormally denoted in normal letters, EBoth have units!!!Monday, Sept. 9, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu11Properties of Vectors•Two vectors are the same if their and the are the same, no matter where they
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