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UT Arlington PHYS 1443 - One Dimensional Motion

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PHYS 1443 – Section 003 Lecture #3AnnouncementsDifference between Speed and VelocityExample 2.1Instantaneous Velocity and SpeedPosition vs Time PlotSlide 7Example 2.3Example 2.3 cont’dDisplacement, Velocity and SpeedAccelerationMonday, Aug. 30, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu1PHYS 1443 – Section 003Lecture #3Monday, Aug. 30, 2004Dr. Jaehoon Yu1. One Dimensional Motion Average VelocityAccelerationMotion under constant accelerationFree Fall2. Motion in Two DimensionsVector Properties and OperationsMotion under constant accelerationProjectile MotionMonday, Aug. 30, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu2Announcements•Homework: 38 of you have signed up (out of 43)–Roster will be locked at 5pm Wednesday–In order for you to obtain 100% on homework #1, you need to pickup the homework, attempt to solve it and submit it.  30 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:: 15 of you have subscribed so far. –This is the primary communication tool. So subscribe to it ASAP.–5 extra credit points if done by midnight tonight and 3 by Wednesday.–A test message will be sent after the class today for verification purpose•Physics Clinic (Supplementary Instructions, SH010): 12 – 6, M-F•Labs begin today!!!Monday, Aug. 30, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu3Difference between Speed and Velocity•Let’s take a simple one dimensional translation that has many steps:Let’s call this line as X-axisLet’s have a couple of motions in a total time interval of 20 sec.+10m+15m-15m-5m -10m+5mTotal Displacement:xDTotal Distance Traveled:D =Average Velocity:ffixix xvt t-�-Average Speed:Total Distance TraveledTotal Time Intervalv �xtD=D020=0( / )m s=fix x� -i ix x= -0( )m=10 15 5 15 10 5 60( )m+ + + + + =6020=3( / )m s=Monday, Aug. 30, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu4Example 2.12 1fix x x x xD � - = -•Displacement: •Average Velocity: ffixix xvt t-�-•Average Speed: Total Distance TraveledTotal Time Intervalv �The position of a runner as a function of time is plotted as moving along the x axis of a coordinate system. During a 3.00-s time interval, the runner’s position changes from x1=50.0m to x2=30.5 m, as shown in the figure. What was the runner’s average velocity? What was the average speed?30.5 50.0= -19.5( )m=-2 12 1x x xt t t- D= =- D19.56.50( / )3.00m s-= =-50.0 30.5 19.56.50( / )3.00 3.00m s- += = =+Monday, Aug. 30, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu5Instantaneous Velocity and Speed•Can average quantities tell you the detailed story of the whole motion?dtdxtxvxlim0Δt*Magnitude of Vectors are expressed in absolute values•Instantaneous speed is the size (magnitude) of the velocity vector:dtdxtxvxlim0Δt•Instantaneous velocity is defined as:–What does this mean?•Displacement in an infinitesimal time interval•Mathematically: Slope of the position variation as a function of timeMonday, Aug. 30, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu6Position vs Time Plottimet1t2t3t=0Positionx=0x112 31. Running at a constant velocity (go from x=0 to x=x1 in t1, Displacement is + x1 in t1 time interval)2. Velocity is 0 (go from x1 to x1 no matter how much time changes)3. Running at a constant velocity but in the reverse direction as 1. (go from x1 to x=0 in t3-t2 time interval, Displacement is - x1 in t3-t2 time interval)It is helpful to understand motions to draw them on position vs time plots.Does this motion physically make sense?Monday, Aug. 30, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu7Instantaneous VelocityTimeAverage VelocityInstantaneous VelocityMonday, Aug. 30, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu8Example 2.3mxxt3.5580.2)00.5(10.2200.522mxxt7.2180.2)00.3(10.2200.311 smtxvx/8.1600.26.3300.300.56.33 mxxx 6.337.213.5512(a) Determine the displacement of the engine during the interval from t1=3.00s to t2=5.00s.Displacement is, therefore:A jet engine moves along a track. Its position as a function of time is given by the equation x=At2+B where A=2.10m/s2 and B=2.80m.(b) Determine the average velocity during this time interval.Monday, Aug. 30, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu9Example 2.3 cont’d 1nnnCtCtdtd   smAstvx/0.210.1010.200.5200.5  BAtdtddtdxtxvtx20limAt2Calculus formula for derivative The derivative of the engine’s equation of motion is(c) Determine the instantaneous velocity at t=t2=5.00s.and  0CdtdThe instantaneous velocity at t=5.00s isMonday, Aug. 30, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu10Displacement, Velocity and Speeddtdxtxvxlim0ΔtDisplacementixxxfAverage velocitytxttxxviixffAverage speedSpent Time TotalTraveled Distance TotalvInstantaneous velocityInstantaneous speeddtdxtxvx lim0ΔtMonday, Aug. 30, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu11Acceleration•In calculus terms: A slope (derivative) of velocity with respect to time or change of slopes of position as a function of timexa �xv �analogs toxa �dtdxtxvxlim0Δtanalogs toChange of velocity in time (what kind of quantity is this?)•Average acceleration:•Instantaneous acceleration:xffxiiv vt t-=-xvtDDffiix xt t-=-xtDDxvt�D=DΔt 0limxdvdt=d dxdt dt� �=� �� �22d


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UT Arlington PHYS 1443 - One Dimensional Motion

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