Page 1Physics 207 – Lecture 3Physics 207: Lecture 3, Pg 1Physics 207, Lecture 3l Today (Finish Ch. 2 & start Ch. 3) v Examine systems with non-zero acceleration (often constant)v Solve 1D problems with zero and constant acceleration (including free-fall and motion on an incline)v Use Cartesian and polar coordinate systemsv Perform vector algebraPhysics 207: Lecture 3, Pg 2““2D2D””Position, DisplacementPosition, Displacementtime (sec) 1 2 3 4 5 6 position -2,2 -1,2 0,2 1,2 2,2 3,2 (x,y meters)originposition vectorsxyPage 2Physics 207 – Lecture 3Physics 207: Lecture 3, Pg 3““2D2D””Position, DisplacementPosition, Displacementtime (sec) 1 2 3 4 5 6 position -2,2 -1,2 0,2 1,2 2,2 3,2 (x,y meters)originposition vectorsxydisplacement vectorsPhysics 207: Lecture 3, Pg 4Position, Displacement, VelocityPosition, Displacement, Velocitytime (sec) 1 2 3 4 5 6 displacement vectorstxttxxvififx∆∆=−−=(avg) velocity vectors1 2 2 3 3 4 4 5 5 6Velocity always has same magnitude & length CONSTANT2xinitial final m/s 0/a 0 )(⇒∆∆≡=−=∆∆+=∆tvvvvtvxtxxxxxxixyPage 3Physics 207 – Lecture 3Physics 207: Lecture 3, Pg 5AccelerationAccelerationl The average acceleration of a particle as it moves is defined as the change in the instantaneous velocity vectordivided by the time interval in which that change occurs. l Bold fonts are vectors al The average acceleration is a vector quantity directed along ∆vPhysics 207: Lecture 3, Pg 6AccelerationAccelerationl Particle motion often involves non-zero accelerationv The magnitude of the velocity vector may changev The direction of the velocity vector may change (true even if the magnitude remains constant)v Both may change simultaneouslyl E.g., a “particle” with smoothly decreasing speedv0=∆+vrv1v101v vrr−=∆vtva∆∆≡rravgv1v3v5v2v4vr∆vr∆vr∆vr∆vr∆v0Page 4Physics 207 – Lecture 3Physics 207: Lecture 3, Pg 7Average AccelerationAverage Accelerationl Question: A sprinter is running around a track. 10 seconds after he starts he is running north at 10 m/s. At 20 seconds he is running at 10 m/s eastwards.l What was his average acceleration in this time (10 to 20 secs)?tvvtttvtvaififif∆−=−−≡rrrrrinitialfinalavg)()(Physics 207: Lecture 3, Pg 8Average AccelerationAverage Accelerationl Question: A sprinter is running around a track. 10 seconds after he starts he is running north at 10 m/s. At 20 seconds he is running at 10 m/s eastwards.l What was his average acceleration in this time (10 to 20 secs)?tvvtttvtvaififif∆−=−−≡rrrrrinitialfinalavg)()(vvvifrrr∆+=14 m/s /10 s= 1.4 m/s2 to the SENPage 5Physics 207 – Lecture 3Physics 207: Lecture 3, Pg 9Instantaneous Accelerationl Average accelerationl The instantaneous acceleration is the limit of the average acceleration as ∆v/∆t approaches zerotvvaif∆−≡rrravgPhysics 207: Lecture 3, Pg 10Position, velocity & acceleration for motion along a linel If the position x is known as a function of time, then we can find both the instantaneous velocity vxand instantaneous acceleration axas a function of time!tvxtxaxt22vadtxddtdxx==dtdxx=v] offunction a is [ )(txtxx∆∆=Page 6Physics 207 – Lecture 3Physics 207: Lecture 3, Pg 11Position, velocity & acceleration for motion along a linel If the position x is known as a function of time, then we can find both the instantaneous velocity vxand instantaneous acceleration axas a function of time!tvxtxaxt22vadtxddtdxx==dtdxx=v] offunction a is [ )(txtxx∆∆=Physics 207: Lecture 3, Pg 12Going the other way….l Particle motion with constant accelerationv The magnitude of the velocity vector changesxadtdxxva =xxddtva=ixxxxtv-vva=∆=∆ati∆t0ttf∆v = area under curve = a ∆tPage 7Physics 207 – Lecture 3Physics 207: Lecture 3, Pg 13Going the other way….l Particle motion with constant accelerationv The magnitude of the velocity vector changesl A particle with smoothly increasing speed:vv11vv00vv33vv55vv22vv44aavt0vf= vi+ a ∆t= vi+ a (tf- ti)∆v = area under curve = a ∆txaati∆t0ttftxxxi∆+= avvPhysics 207: Lecture 3, Pg 14So if constant acceleration we can integrate twicetxxxi∆+= avv221 a v ttxxxxii∆+∆+=consta=xtaxxtxivxtviPage 8Physics 207 – Lecture 3Physics 207: Lecture 3, Pg 15Two other relationshipsl If constant acceleration then we also get:)( 2avv22ixxxxxi−=−)vv(v21(avg) xxxi+=Physics 207: Lecture 3, Pg 16Example probleml A particle moves to the right first for 2 seconds at 1 m/sand then 4 seconds at 2 m/s.l What was the average velocity?l Two legs with constant velocity but ….vxt2vvv21Avg+≠Page 9Physics 207 – Lecture 3Physics 207: Lecture 3, Pg 17Example probleml A particle moves to the right first for 2 seconds at 1 m/sand then 4 seconds at 2 m/s.l What was the average velocity?l Two legs with constant velocity but ….l We must find the displacement (x2–x0)l And x1= x0+ v0(t1-t0) x2= x1+ v1(t2-t1)l Displacement is (x2- x1) + (x1– x0) = v1(t2-t1) + v0(t1-t0) l x2–x0= 1 m/s (2 s) + 2 m/s (4 s) = 10 m in 6 seconds or 5/3 m/svxt2vvv21Avg+≠Physics 207: Lecture 3, Pg 18A particle starting at rest & moving along a line with constant acceleration has a displacement whose magnitude is proportional to t2221 a)( txxxi∆+=−221 a v ttxxxxii∆+∆+=1. This can be tested2. This is a potentially useful result Displacement with constant accelerationPage 10Physics 207 – Lecture 3Physics 207: Lecture 3, Pg 19Free Falll When any object is let go it falls toward the ground !! The force that causes the objects to fall is called gravity.l This acceleration on the Earth’s surface, caused by gravity, is typically written as “little” gl Any object, be it a baseball or an elephant, experiences the same acceleration (g) when it is dropped, thrown, spit, or hurled, i.e. g is a constant. 2210 g v)(0ttytyy∆−∆+=∆ga-y=Physics 207: Lecture 3, Pg 20Gravity facts:l g does not depend on the nature of the material !v Galileo (1564-1642) figured this out without fancy clocks & rulers!l Feather & penny behave just the same in vacuuml Nominally, g = 9.81 m/s2v At the equator g = 9.78 m/s2v At the North pole g = 9.83 m/s2Page 11Physics 207 – Lecture 3Physics 207: Lecture 3, Pg 21When throwing a ball straight up, which of the following is true about its velocity v and its acceleration a at the highest point in its path?A. Both v = 0 and a = 0B. v ≠ 0, but a = 0C. v = 0, but a ≠ 0D. None of the aboveyExercise 1Motion in One
View Full Document
Unlocking...