Monday, Feb. 2, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu1PHYS 1441 – Section 004Lecture #4Monday, Feb. 2, 2004Dr. Jaehoon Yu• Chapter two: Motion in one dimension– Acceleration (Average and instantaneous)– One dimensional motion at constant acceleration• Free Fall– Coordinate systemsMonday, Feb. 2, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu2Announcements• Homework Registration: 58/61– Roster has been locked– Come see me if you haven’t registered.• E-mail distribution list (phys1441-004-spring04)– 43 of you subscribed as of 10am this morning– Send an e-mail to [email protected]• “subscribe phys1441-004-spring04 fn ln” • Without a subject. Put the above in the body ONLY!––1 point1 pointextra credit if done by 6pm Monday, Feb. 2– Test message will be issued Wednesday, Feb. 4• Lab begins today, Monday, Feb. 2• Use the Physics Clinic– When: MWF: 12-7pm, T,Th: 12-7:30pm– Where: SH010•Quiz– Average 49– Top score: 90Monday, Feb. 2, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu3Instantaneous 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, Feb. 2, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu4Position vs Time Plottimet1t2t3t=0Positionx=0x112 31. Running at a constant velocity (go from x=0 to x=x1in t1, Displacement is + x1in t1time interval)2. Velocity is 0 (go from x1to x1no matter how much time changes)3. Running at a constant velocity but in the reverse direction as 1. (go from x1to x=0 in t3-t2time interval, Displacement is - x1in t3-t2time interval)It is useful to understand motions to draw them on position vs time plots.Does this motion physically make sense?Monday, Feb. 2, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu5Instantaneous VelocityTimeAverage VelocityInstantaneous VelocityMonday, Feb. 2, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu6Displacement, Velocity and Speeddtdxtxvx =∆∆=→lim0ΔtDisplacementixxx f−≡∆Average velocitytxttxxviixff∆∆=−−≡Average speedSpent Time TotalTraveled Distance Total≡vInstantaneous velocityInstantaneous speeddtdxtxvx =∆∆=→lim0ΔtMonday, Feb. 2, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu7Acceleration• 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⎛⎞=⎜⎟⎝⎠22dxdtMonday, Feb. 2, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu8Example 2.4xa=)/(2.40.5210.50212sm==−=A car accelerates along a straight road from rest to 75km/h in 5.0s.What is the magnitude of its average acceleration?xfv =xiv =())/(105.4100036002.4242hkm×=×=)/( hkm0 /ms750003600ms=21 /msxf xifivvtt−=−xvt∆∆Monday, Feb. 2, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu9Example for Acceleration• Velocity, vx, is express in: • Find average acceleration in time interval, t=0 to t=2.0s()xvt•Find instantaneous acceleration at any time t and t=2.0s(0)xi ivt=()xat(2.0)xat=Instantaneous Acceleration at any timeInstantaneous Acceleration at any time t=2.0s()240 5 /tms=−(2.0)xf fvt=xa40( / )ms=()240 5 2.0=−× 20( / )ms=xfxifivvtt−=−xvt∆=∆20 402.0 0−=−210( / )ms=−xdvdt≡()240 5dtdt=−10t=−10 (2.0)=−×220( / )ms=−Monday, Feb. 2, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu10Meanings of Acceleration• When an object is moving in a constant velocity (v=v0), there is no acceleration (a=0)– Is there any 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) – Incorrect since an object might be moving in negative direction initially• When an object is moving slower as time goes on, (v=v(t) ), acceleration is negative (a<0)– Incorrect since an object might be moving in negative direction initially• In all cases, velocity is positive, unless the direction of the movement changes. – Incorrect since an object might be moving in negative direction initially• Is there acceleration if an object moves in a constant speed butchanges direction?The answer is
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