KinematicsPage3ofthetextbookTimePositionSpeedVelocityDistanceDirectionAcceleration012-2-1Howcanwedescribethechangeofposition?0.80.20.40.60.8DISPLACEMENT=newposition-oldposition-1catdisplacement=-1-0.8=-1.8origincoordinate-anumberlinedisplacement=3DISTANCE

-Magnitudeofdisplacementbetweentwopositions012-2-1displacement=-2displacementa=-2displacementb=3totaldisplacement=distancetravelled=2+3=51Howmuchofthedistancedoesthisathleterun?Whatistherunner’sdisplacement?Question:Totaldistancetravelled:400mTotaldisplacement:0mAVERAGEVELOCITYDISPLACEMENTTIMEOFTRAVEL=SIunit:m/susingcalculus,visdefinedas:VELOCITY=DISPLACEMENT/TIMEInstantaneousvelocitySpeed–themagnitude(absolutevalue)ofvelocityACCELERATIONSIunit:m/s2AVERAGEACCELERATIONchangeinvelocitytimeelapsed=“changeinvelocity”canbeachangeinspeed,butalsoachangeindirectionInstantaneousaccelerationisexpressedas:finalvelocityforaccelerationaafteratimeΔt:positionx

time

Δttime

Δt//velocityvdisplacement

Δx

changeaccelerationa

ΔvchangeUSINGGRAPHS

TODESCRIBEMOTIONCartesiancoordinatesystem(rectangularcoordinatesystem)RenéDescartes(1596

–1650)Latinized:RenatusCartesiusRunningataconstantspeed3km/15minutes3km/15minutesChangingvelocity(constantacceleration)Cargoesfrom0to100km/hour(negativedirection)in4.5secondsWhatisitsaverageacceleration?v0=0vf=-100km/hrΔt=4.5sa=(-27.8-0)/4.5=-6.17m/s2=-100000m/3600s=-27.8m/sexampleMotionequationsConstantaccelerationinonedimensionSolvingfordisplacementandfinalpositionaveragevelocity

finalposition:Airplanelandingat250km/hr,deceleratingat-1.5m/s2Whatdistancetostopcompletely?Solvingforfinalvelocitywithconstantaccelerationknowinginitialandfinalposition

(ifwedon’tknowtimet)SummaryofKinematicEquations(constanta)GRAVITYAllobjectsfalltowardsthegroundwiththesameaccelerationg=9.80m/s2……neglectingairresistanceandfriction!a=-g=-9.80m/s2KINEMATICEQUATIONSWITHGRAVITY(upwardasthepositivedirectionofy)WhathappenswhenIthrowaballupintheair?positionandvelocityattimest=1s,2s,3st=1sUsethesamemethodfor

y3,v3t=2spositionvstimeparabolavelocityvstimestraightlinequadraticlinearIfIthrowaballupintheair,whatheightwillitreachbeforeitstartstofall?velocityiszeroaroundhereWhatarewelookingfor?yatatimewhenv=0Whatdoweknow?Step1:findthetimewhenv=0Step2:findyatthetimewhenv=0Orusethisformulaproportionaltothesquareofthevelocity!Maximumheightreachedbyanobjectthrownupwithavelocityv0

question6stepsforsolvingakinematicsproblemStep1:ExaminethesituationtodeterminewhichphysicalprinciplesareinvolvedStep6:Checktheanswertoseeifitisreasonable.Doesitmakesense?Step5:Substitutetheknownsalongwiththeirunitsintotheappropriateequation,andobtainnumericalsolutionscompletwithunitsStep4:FindanequationofsetofequationsthatcanhelpyousolvetheproblemStep3:Identifyexactlywhatneedstobedeterminedintheproblem(identifytheunknowns)Step2:Makealistofwhatisgivenorcanbeinferredfromtheproblemasstated(identifytheknowns)physicsVelocityjustbeforetheballstrikesthefloor1.5m1.1mUse:1.5m1.1mVelocityjustaftertheballleavesthefloorUse:accelerationduringcontactwiththefloor-5.42m/s4.64m/st=0.0035stv=(4.64m/s-(-5.42m/s))/0.0035scontactwiththefloorfallingdownmovingupv=v0+atfallingdown=2876m/s2v0=-5.42m/svf=0m/sy=0yf

=?a=2876m/s2y-y0=(-5.42m/s)2/(2*2876m/s2)=0.005m=5mmHowmuchdidtheballcompress?Use:straightlineKinematicsin2dimensions:usingvectorsVECTORSSCALARSmagnitude+directiondisplacementmagnitudeonlytemperatureaccelerationspeeddistancevelocityvsmassenergyforcesumofvectors“headtotail”methodindependentoftheordersubtractionofvectorsaddavectorofoppositedirectionvectormultipliedbyascalarsamedirection"3"timeslongervectordividedbyascalar–thesame!

θAxθAvectorscanberesolvedintoperpendicularcomponentsAAAyAyAxAxAyA2=Ax2

+Ay2Ax=AcosθAy=Asinθθ=tan-1(Ay/Ax)A=vectorA=magnitudeθsumofvectors:sumbycomponentsABRAxAyBxByRx=Ax+Bx

Ry=Ay+By

RxRyMotionin2dimensionsThemotionsindirectionsxandyareindependentDisplacement:Velocity:Acceleration:ProjectilemotionconstantvelocityconstantaccelerationProcedureforanalyzingprojectilemotionin2dimensions1.Separatedisplacement,velocityandaccelerationintoxandycomponents2.Analyzethex-motionandy-motionseparately3.RecombinebysummingthecomponentsRangeofaprojectile:initialspeedv0,angleθvy0=v0sinθv0θRvx0=v0cos