Wednesday, Feb. 4, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu1PHYS 1441 – Section 004Lecture #5Wednesday, Feb. 4, 2004Dr. Jaehoon Yu• Chapter two: Motion in one dimension– One dimensional motion at constant acceleration• Falling Motion– Coordinate systems• Chapter three: Motion in two dimension– Vector and Scalar– Properties of vectors– Vector operations– Components and unit vectorsToday’s homework is homework #3, due 1pm, next Wednesday!!Wednesday, Feb. 4, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu2Announcements• There will be a quiz next Wednesday, Feb. 11– Will cover•Sections A5 –A9• Chapter 2• Homework Registration: 60/61– Roster will be locked again at 2:40pm today– Come see me if you haven’t registered.• E-mail distribution list (phys1441-004-spring04)– 45 of you subscribed as of 10am this morning– Test message will be issued today, Feb. 4Wednesday, Feb. 4, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu3Displacement, Velocity and Speeddtdxtxvx =∆∆=→lim0ΔtDisplacementixxx f−≡∆Average velocitytxttxxviixff∆∆=−−≡Average speedSpent Time TotalTraveled Distance Total≡vInstantaneous velocityInstantaneous speeddtdxtxvx =∆∆=→lim0ΔtWednesday, Feb. 4, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu4Acceleration• 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:xffxiivvtt−=−xvt∆∆ffiixxtt−=−xt∆∆xvt→∆=∆Δt0limxdvdt=ddxdt dt⎛⎞=⎜⎟⎝⎠22dxdtWednesday, Feb. 4, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu5One Dimensional Motion• Let’s start with the simplest case: acceleration is a constant (a=a0)• Using definitions of average acceleration and velocity, we can derive equations of motion (description of motion, velocity and position as a function of time)xa =(If tf=t and ti=0)xfv=For constant acceleration, average velocity is a simple numeric averagexv =fx=Resulting Equation of Motion becomesxv=fx=xi xvat+xa =xffxiivvtt−−xfxixvvat−=2xi xfvv+=22xi xvat+=12xixvat+ffiixxtt−−f ixxxvt−=(If tf=t and ti=0)xixvt+xixvt+ =212ixi xxvt at+ +Wednesday, Feb. 4, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu6One Dimensional Motion cont’dfx=xa =Average velocityt=Sincefx =Substituting t in the above equation, xv=2xfv=Solving for tResulting in 2xixfvv+xixvt+=2xi xfivvxt+⎛⎞+⎜⎟⎝⎠xf xivvt−xf xixvxa−2xf xi xf xiixvv vvxa+−⎛⎞⎛⎞+=⎜⎟⎜⎟⎝⎠⎝⎠222xf xiixvvxa−+()22xi xfivaxx+−Wednesday, Feb. 4, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu7Kinematic Equations of Motion on a Straight Line Under Constant Acceleration()tavtv xxixf +=221tatvxx xxiif += +()tvvtvxx xixfixf+==−2121()ifxf xxavv xxi−+= 222Velocity as a function of timeDisplacement as a function of velocities 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, Feb. 4, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu8How do we solve a problem using a kinematicformula for constant acceleration?• Identify what information is given in the problem.– Initial and final velocity?– Acceleration?–Distance?–Time?• Identify what the problem wants.• Identify which formula is appropriate and easiest to solve for what the problem wants.– Frequently multiple formula can give you the answer for the quantity you are looking for. Î Do not just use any formula but use the one that can be easiest to solve.• Solve the equation for the quantity wantedWednesday, Feb. 4, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu9Example 2.102xfv=fixx−=xiv=xfv=Suppose you want to design an air-bag system that can protect the driver in a head-on collision at a speed 100km/s (~60miles/hr). Estimate how fast the air-bag must inflate to effectively protect the driver. Assume the car crumples upon impact over a distance of about 1m. How does the use of a seat belt help the driver? t=How long does it take for the car to come to a full stop? As long as it takes for it to crumple. We also know thatandUsing the kinematic formulaThe acceleration isxa=Thus the time for air-bag to deploy isThe initial speed of the car is0/ms100 /km h =10000028 /3600mmss=1m()22fixi xvaxx+−()222xffixivvxx−=−()2028/21msm−=×2390 /ms−xf xivva−=2028/390 /msms−=−0.07s=Wednesday, Feb. 4, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu10Falling Motion• Falling motion is a motion under the influence of gravitational pull (gravity) only; Which direction is a freely falling object moving?– A motion under constant acceleration– All kinematic formula we learned can be used to solve for falling motions. • Gravitational acceleration is inversely proportional to the distance between the object and the center of the earth• The gravitational acceleration is g=9.80m/s2on 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/s2Wednesday, Feb. 4, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu11Example for Using 1D KinematicEquations on a Falling object Stone was thrown straight upward at t=0 with +20.0m/s initial velocity on the roof of a 50.0m high building,g=-9.80m/s2fv =fy =(a) Find the time the stone reaches at the maximum height.What is the acceleration in this motion?What is so special about the maximum height? V=0t=(b) Find the maximum height.yi yvat+ =20.0 9.80t+− =0.00 /msSolve for t20.09.80=2.04s2150.0 20 2.04 ( 9.80) (2.04)2+× +×− ×50.0 20.4 70.4( )m=+=212iyi yyvt at++ =Wednesday, Feb. 4, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu12Example of a Falling Object cnt’dt =yfv =fy =Positionyfv =Velocity(c) Find the time the stone reaches back to its original height.(d) Find the velocity of the stone when it reaches its original height.(e) Find the velocity and position of the stone at t=5.00s.2.04 2×= 4.08syi yvat+=20.0 ( 9.80) 4.08+−×=20.0( / )ms−yi yvat+=20.0 ( 9.80) 5.00 29.0( /
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