PHYS 1443 – Section 003 Lecture #3AnnouncementsSurvey ResutsDisplacement, Velocity and SpeedAccelerationMeanings of AccelerationExample 2.4Example 2.7One Dimensional MotionOne Dimensional Motion cont’dKinetic Equations of Motion on a Straight Line Under Constant AccelerationExample 2.11Free FallExample for Using 1D Kinetic Equations on a Falling objectExample of a Falling Object cnt’dCoordinate SystemsExampleWednesday, Sept. 3, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu1PHYS 1443 – Section 003Lecture #3Wednesday, Sept. 3, 2003Dr. Jaehoon Yu1. One Dimensional Motion AccelerationMotion under constant accelerationFree Fall2. Motion in Two DimensionsVector Properties and OperationsMotion under constant accelerationProjectile MotionToday’s homework is homework #2, due noon, next Wednesday!!Wednesday, Sept. 3, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu2Announcements•Homework: 33 of you have signed up (out of 36)–Roster will be locked at 5pm today–In order for you to obtain 100% on homework #1, you need to pickup the homework, attempt to solve it and submit it. Only about 27 of you have done this. –Homework system deducts points for failed attempts. •So be careful when you input the answers•Input the answers to as many significant digits as possible–All homework problems are equally weighted•e-mail distribution list:14 of you have subscribed so far. –This is the primary communication tool. So subscribe to it ASAP.–A test message will be sent after the class today for verification purpose•Physics Clinic (Supplementary Instructions): 11-6, M-F •Could I speak to Steven Smith after the class?Wednesday, Sept. 3, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu3Survey Resuts •25 of you have taken Physics before–18 in High School–10 of you had Classical Mechanics while 4 said E&M–13 of you are taking this course due to degree requirements–About 5 of you wanted to have better understanding•24 of you had some level of calculus before–10 in HS and 13 in college–You must have had it already or must be having it concurrently•31 of you said physics is mandatory•16 of you take this course because you like physicsLet’s have fun together this semester!!!Wednesday, Sept. 3, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu4Displacement, Velocity and Speeddtdxtxvxlim0ΔtDisplacementixxxfAverage velocitytxttxxviixffAverage speedSpent Time TotalTraveled Distance TotalvInstantaneous velocityInstantaneous speeddtdxtxvx lim0ΔtWednesday, Sept. 3, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu5Acceleration•In calculus terms: A slope (derivative) of velocity with respect to time or change of slopes of position as a function of timetvttvvaxfxfixixtxttxxviixffanalogous to22dtxddtdxdtddtdvtvaxxxlim0Δtdtdxtxvxlim0Δtanalogous toChange of velocity in time (what kind of quantity is this?)•Average acceleration:•Instantaneous acceleration:Wednesday, Sept. 3, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu6Meanings of Acceleration•When an object is moving in a constant velocity (v=v0), there is no acceleration (a=0)–Could there be acceleration when an object is not moving?•When an object is moving faster as time goes on, (v=v(t) ), acceleration is positive (a>0)•When an object is moving slower as time goes on, (v=v(t) ), acceleration is negative (a<0)•Is there acceleration if an object moves in a constant speed but changes direction?The answer is YES!!YES!Wednesday, Sept. 3, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu7Example 2.4tvttvvaxifxixfx)/(2.40.5210.50212smA car accelerates along a straight road from rest to 75km/h in 5.0s.What is the magnitude of its average acceleration?smsmvxf/ 21360075000smvxi/ 0 )/(105.4100036002.4242hkm)/( hkmWednesday, Sept. 3, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu8Example 2.7 ttdtddtdxtxvtx20.480.210.2lim20 smstvvxxi/6.1200.320.400.3 2/2.400.240.800.300.56.120.21smtvaxx(a) Compute the average acceleration during the time interval from t1=3.00s to t2=5.00s.A particle is moving on a straight line so that its position as a function of time is given by the equation x=(2.10m/s2)t2+2.8m. (b) Compute the particle’s instantaneous acceleration as a function of time. smstvvxxf/0.2100.520.400.5 2/20.420.4 smtdtddtdvtvaxxxlim0ΔtWhat does this mean?The acceleration of this particle is independent of time.This particle is moving under a constant acceleration.Wednesday, Sept. 3, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu9One 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)tvvttvvaaxiixixxxffxf If tf=t and ti=0tavvxxixfFor constant acceleration, simple numeric averagetxxttxxviiixfff If tf=t and ti=0221tatvxtvxxxxiiixfResulting Equation of Motion becomestavtavvvvxxixxixfxix21222tvxxxfi Wednesday, Sept. 3, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu10One Dimensional Motion cont’dxxixfixxixfxixfifavvxavvvvxx2222tvvaxixxf Average velocityxxiaxvtxf Sincetvvxtvxxxfxiixif2Substituting t in the above equation, we obtain2xfxivvvx ifxxixfxxavv 222Resulting inWednesday, Sept. 3, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu11Kinetic Equations of Motion on a Straight Line Under Constant Acceleration tavtvxxixf221tatvxxxxiif tvvtvxxxixfixf2121 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 Kinetic equations, depending on the information given to you for specific physical problems!!Wednesday, Sept. 3, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu12Example 2.11 ifxf xxavv xxi
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