Physics 207, Lecture 3AccelerationAverage AccelerationSlide 8Instantaneous AccelerationPosition, velocity & acceleration for motion along a lineSlide 11Going the other way….Slide 13So if constant acceleration we can integrate twiceTwo other relationshipsExample problemSlide 17A particle starting at rest & moving along a line with constant acceleration has a displacement whose magnitude is proportional to t2Free FallGravity facts:When 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?Slide 22Exercise 2 1D FreefallSlide 26Problem Solution Method:A science projectA “science” projectSlide 32See you WednesdayPhysics 207: Lecture 3, Pg 1Physics 207, Lecture 3Today (Finish Ch. 2 & start Ch. 3) Examine systems with non-zero acceleration (often constant) Solve 1D problems with zero and constant acceleration (including free-fall and motion on an incline) Use Cartesian and polar coordinate systems Perform vector algebraPhysics 207: Lecture 3, Pg 5AccelerationAccelerationThe average acceleration of a particle as it moves is defined as the change in the instantaneous velocity vector divided by the time interval in which that change occurs. Bold fonts are vectors aThe average acceleration is a vector quantity directed along ∆vPhysics 207: Lecture 3, Pg 7Average AccelerationAverage AccelerationQuestion: 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.What was his average acceleration in this time (10 to 20 secs)?tvvtttvtvaifififinitialfinalavg)()(Physics 207: Lecture 3, Pg 8Average AccelerationAverage AccelerationQuestion: 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.What was his average acceleration in this time (10 to 20 secs)?tvvtttvtvaifififinitialfinalavg)()(vvvif14 m/s /10 s= 1.4 m/s2 to the SENPhysics 207: Lecture 3, Pg 9Instantaneous AccelerationAverage accelerationThe instantaneous acceleration is the limit of the average acceleration as ∆v/∆t approaches zerotvvaifavgPhysics 207: Lecture 3, Pg 10Position, velocity & acceleration for motion along a lineIf the position x is known as a function of time, then we can find both the instantaneous velocity vx and instantaneous acceleration ax as a function of time!tvxtxaxt22vadtxddtdxxdtdxxv] offunction a is [ )( txtxx Physics 207: Lecture 3, Pg 11Position, velocity & acceleration for motion along a lineIf the position x is known as a function of time, then we can find both the instantaneous velocity vx and instantaneous acceleration ax as a function of time!tvxtxaxt22vadtxddtdxxdtdxxv] offunction a is [ )( txtxx Physics 207: Lecture 3, Pg 12Going the other way….Particle motion with constant acceleration The magnitude of the velocity vector changesxadtdxxva xxddt va ixxxxt v-vva atit0ttfv = area under curve = a tPhysics 207: Lecture 3, Pg 13Going the other way….Particle motion with constant acceleration The magnitude of the velocity vector changesA particle with smoothly increasing speed:vv11vv00vv33vv55vv22vv44aa vt0vf = vi + a t = vi + a (tf - ti )v = area under curve = a txaatit0ttftxxxi avvPhysics 207: Lecture 3, Pg 14So if constant acceleration we can integrate twicetxxxi avv221 a v ttxxxxiiconsta xtaxxtxivxtviPhysics 207: Lecture 3, Pg 15Two other relationshipsIf constant acceleration then we also get:)( 2avv22ixxxxxi)vv(v21(avg) xxxiPhysics 207: Lecture 3, Pg 16Example problemA particle moves to the right first for 2 seconds at 1 m/s and then 4 seconds at 2 m/s.What was the average velocity?Two legs with constant velocity but ….vxt2vvv21AvgPhysics 207: Lecture 3, Pg 17Example problemA particle moves to the right first for 2 seconds at 1 m/s and then 4 seconds at 2 m/s.What was the average velocity?Two legs with constant velocity but ….We must find the displacement (x2 –x0)And x1 = x0 + v0 (t1-t0) x2 = x1 + v1 (t2-t1)Displacement is (x2 - x1) + (x1 – x0) = v1 (t2-t1) + v0 (t1-t0) x2 –x0 = 1 m/s (2 s) + 2 m/s (4 s) = 10 m in 6 seconds or 5/3 m/svxt2vvv21AvgPhysics 207: Lecture 3, Pg 18A particle starting at rest & moving along a line with constant acceleration has a displacement whose magnitude is proportional to t2 221 a)( txxxi221 a v ttxxxxii1. This can be tested2. This is a potentially useful result Displacement with constant accelerationPhysics 207: Lecture 3, Pg 19Free FallWhen any object is let go it falls toward the ground !! The force that causes the objects to fall is called gravity.This acceleration on the Earth’s surface, caused by gravity, is typically written as “little” gAny 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)(0ttytyyga -yPhysics 207: Lecture 3, Pg 20Gravity facts:g does not depend on the nature of the material ! Galileo (1564-1642) figured this out without fancy clocks & rulers!Feather & penny behave just the same in vacuumNominally, g = 9.81 m/s2 At the equator g = 9.78 m/s2At the North pole g = 9.83 m/s2Physics 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 DimensionPhysics 207: Lecture 3, Pg 22When throwing a ball straight up, which of the following is When throwing a ball straight up, which of the following is true about its velocity true about its velocity vv and its acceleration and its acceleration aa at the highest at the highest point in its path?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 DimensionPhysics 207: Lecture 3, Pg 25Exercise 2 1D FreefallA. vA < vB B. vA = vB C. vA > vBAlice and Bill are standing at the top of a cliff of height
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