KinematicsinOneDimension1.Introduction

1.DifferentTypesofMotionWe'lllookat:2.Dimensionalityinphysics3.Onedimensionalkinematics4.Particlemodel

2.DisplacementVector1.Displacementin1-D2.DistanceTraveled

3.SpeedandVelocity1....withadirection

4.Changeinvelocity.1.Acceleration2.Acceleration,themath.3.Slowingdown4.Accelerationinthenegative5.Summaryofaccelerationsignage.

5.Kinematicequations1.EquationsofMotion(1-D)

6.SolvingProblems7.Plotting8.FreeFall

1.Dropawrench2.Howhighwasthis?3.Everypointonalinehasatangent

IntroductionMotion:changeinpositionororientationwithrespecttotime.

Vectorshavegivenussomebasicideasabouthowtodescribethepositionofobjectsintheuniverse/Now,we'llcontinuebyextendingthoseideastoaccountforchangesinthatposition.Ofcoursetheworldwouldbeawfullyboringifthepositionofeverythingwasconstant.

DifferentTypesofMotionWe'lllookat:Linearmotioninvolvesthechangeinpositionofanobjectinonedirectiononly.Anexamplewouldbeatrainonastraightsectionofthetrack.Thechangeinpositionisonlyinthehorizontaldirection.

Projectilemotionoccurswhenobjectsarelaunchedinthegravitationalfieldneartheearthssurface.Theyexperiencemotioninboththehorizontalandtheverticaldirections.

Circularmotionoccursinafewspecificcaseswhenanobjecttravelsinaperfectcircle.Somespecialmathcanbeusedinthesecases.

Rotationalmotionimpliesthatthebodyinquestionisrotatingaroundanaxis.Theaxisdoesn'tnecessaryneedtopassthroughtheobject.

...oracombinationofthem.

Dimensionalityinphysics

PHY 20700 - 1 Dimensional Kinematics - J. Hedberg - 2021

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Fig.1

Preludetoadvancedphysicsandengineering:Lateron,you'llhavetoexpandyournotionofdimensionsabit.Itwon'tsimplymeanstraightorcurvy,butwillinsteadbeusedtodescribethedegreesoffreedominasystem.Forexample,anorbitingbody,thoughitmovesinacirclewhichrequiresxandyvaluestodescribe,canalsobedescribedbyconsideringtheradiusandtheangleofrotationinstead.Thisisjustanothercoordinatesystem:polarcoordinates(usually: and ).Ifwedescribetheorbitingplanetinthissystem,andsay,it'sgoingaroundinaperfectcircle,thenthe valuedoesn'tchangeandthe valuebecometheonlydimensionofinterest.Let'sholdoffonthisapproachfornow,butwhenitcomesbacklateron,welcomeitwithopenarmsbecauseitallowsformuchmorepowerfulandsimpleanalysisofsystems.

OnedimensionalkinematicsForthecaseof1-dimensionalmotion,we'llonlyconsiderachangeofpositioninonedirection.

Itcouldbeanyofthethreecoordinateaxes.

Justadescriptionofthemotion,withoutattemptingtoanalyzethecause.Todescribemotionweneed:

1.CoordinateSystem(origin,orientation,scale)2.theobjectwhichismoving

1dkinematicswillbeourstartingpoint.Itisthemoststraightforwardandeasiestmathematicallytodealwithsinceonlyonepositionvariablewillbechangingwithrespecttotime.

ParticlemodelWe'llneedtouseanabstraction:

Allrealworldobjectstakeupspace.We'llassumethattheydon't.Inotherwords,thingslikecars,cats,andducksarejustpoint-likeparticles.

r θ

r θ

y

x

z

Fig.2Acoordinatesystem

0 20-20 40 60 80 100

+x (m)-x (m)

Fig.3A1-dcoordinatesystem

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Thisisourfirstrealabstraction.Again,sincewearetryingtopredicteverything,wewouldliketofigureouttherulesthatdescribehowanyobjectwouldmove.Takeatrainforexample.Ifweaskedaquestionlike"whendoestheCtrainenter59thstreetstation?",anatural

follow-upwouldbe"well,doyoumeanthefrontofthetrain,orthemiddleofthetrain,ortheendofthetrain?Eachoftheseanswersmightbedifferentbyafewseconds.

Howdowedealwiththis?Byconsideringthetraintobea'point',wecanneglecttheactuallengthofthetrainandfocusonwhat'smoreinteresting:howthetrainmoves.

Thegoalistofindtheunderlyingphysicsthatdescribesalltrains.Oncewedothat,thenwecanimproveourmodelbyincludinginformationaboutthelengthoftheindividualtrainweareinterestedin.

DisplacementVectorToquantifythemotion,we'llstartbydefiningthedisplacementvector.

Inthecaseofourwanderingbug,thiswouldbethedifferencebetweenthefinalpositionandtheinitialposition.

Thisfigureshowsthedisplacementvector .Thismightbedifferentthanthedistancetraveledbythebug(showninthedottedline).

NoteonNotation!isthesamethingas

isthesamethingas

Whendescribingmotions,weusuallyhaveaninitialpositionandafinalposition.Wecancallthese and respectively,whenwedoouralgebra.

Oranotherwayofwritingthesequantitiesistosayourinitialpositionis andourfinalpositionisjust .Thisisaslightlymoregeneralwayofwritingthings.

Displacementin1-D

0 20-20 40 60 80 100

+x (m)-x (m)

Here'sacarthatmovesfrom to creatingadisplacementvectorof:

Thecarthenreversesto .

0 20-20 40 60 80 100

+x (m)-x (m)

Theleadstoadisplacementvectorof .

Fig.4Theparticlemodel

x∆ x

Δx = x − x0

Δx

xf x

xi x0

xi xf

x0 x

x0 x

Δx = x − = 60m − 0m = 60mx0

x = −20

Δx = −80m

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Aboutnotation. ("deltax")referstothechangein .Thatis,differencebetweenafinalandinitialvalue:

Or,inwords,thefinalxpositionminustheoriginalxpositionisequaltothechangeinx.

DistanceTraveledTogetthedistancetraveled,wejustneettotakethemagnitudeofthedisplacementduringacertainmotion.

Thisequationwillonlybetrueifthedisplacementisalwaysinthesamedirection.Ifhowever,thedisplacementvectorweretochangedirectionduringatrip,thethedistancetraveledmightnotbeequaltothetotaldisplacement.Forexample,ifyouwalk100feetforward,thenturnaroundandwalk50backwards.Youdisplacementfromtheinitialtofinalpositionwillonlybe50feet,butyouwillhavewalkedatotalof150feet.

SpeedandVelocity

The'elapsedtime'isdeterminedinthesamewayasthedistance:

Again, isthestartingtime,and isthefinaltime.

TakingtheAtrainbetween59thand125thtakesabout8minutes.TheC,whichisalocal,takes12minutes(onagoodday).Findtheaveragespeedforbothofthesetrips.

...withadirectionCalculatingtheaveragespeeddidn'ttellusanythingaboutthedirectionoftravel.Forthis,we'llneedaveragevelocity.

Inmathematicalterms:

Δx x

Δx = x− x0

|Δx| = DistanceTraveled

AverageSpeed ≡Distanceinagiventime

Elapsedtime

Δt = t− .t0

t0 t

Example Problem#1:

AverageVelocity ≡DisplacementElapsedtime

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(SIunitsofaveragevelocityarem/s)Inone-dimension,velocitycaneitherbeinthepositiveornegativedirection.

ThinkingabouttheAtrain,it'sclearthatitsspeedandvelocitystayedessentiallyconstantbetween59thand125thideally).However,theCtrainhadtostartandstopat7stations.Toquantify,thisdifferenceinmotion,we'llneedtointroducetheconceptofinstantaneousvelocity.

Ifweimaginemakingmanymeasurementsofthevelocityoverthecourseofthetravel,byreducingthe weareconsidering,thenwecanbegintoseehowwecanmoreaccuratelyassessthemotionofthetrain.

Theconceptofinstantaneousvelocityinvolvesconsideringaninfinitesimallysmallsectionofthemotion:

Thiswillenableustotalkaboutthevelocityataparticle'sspecificpositionortimeratherthanforanentiretrip.

Ingeneral,thisiswhatwe'llmeanwhenwesay'velocity'or'speed'.

≡ =vx − x0t− t0

ΔxΔt

Δx

v = =limΔt→0

ΔxΔt

dx

dt

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Changeinvelocity.Naturally,inordertobeginmoving,anobjectmustchangeitsvelocity.

Here'sagraphofabicyclistridingataconstantvelocity.(Inthiscaseit's10m/s)

+x (m)100 20 30 40

0s 1s 2s

Now,here'sagraphofthesamebicyclistridingandchanginghisvelocityduringthemotion

+x (m)100 20 30 40

0s 1s 2s 3s

Intheuppermotiongraph,noticehowthelengthofthedisplacementvector isthesameateachintervalintime.Meaning,thatafter1secondhaspassed,thedisplacementis10m,afteranothersecondpasses,another10metersdisplacementhasoccurred,makingthetotaldisplacementequalto20m.Thisismotionataconstantvelocity.Thisalsoapparentinthelengthofthevelocityvectorsateachpoint.Theyarealwaysthesame.

Inthebottomgraph,thedisplacement,andvelocityvectors,changeeachtimetheyaremeasured.Thisisrepresentativeofmotionwithnon-constantvelocity.Thevelocityischangingastimemoveson.

AccelerationThischangeinvelocitywe'llcallacceleration,andwecandefineitinaverysimilarwaytoourdefinitionofvelocity:

Again,inthiscasewe'retalkingaboutaverageacceleration.

At ,theAtrainisatrestat59thstreet.5secondslater,it'stravelingnorthat19meterspersecond.Whatistheaverageaccelerationduringthistimeinterval?

Ifweconsideredthesameverysmallchangeintime,theinfinitesimalchange,thenwecouldtalkaboutinstantaneousacceleration

TheSIunitsofaccelerationaremeterspersecondpersecond,or .That'sprobablyalittlebitofaweirdunit,but,itmakessensetothinkaboutlikethis:

Acceleration,themath.Toquantifytotheaccelerationofamovingbody,saythiscar,we'llneedtoknowitsinitialandfinalvelocities

Thecarhasabuildinspeedometer,sowecanlookatthattogetthespeed,andifwedon'tchangedirection,thenthevelocitywillbealwayspointedinthesamedirection.

d→

= =av − v0t− t0

ΔvΔt

Example Problem#2:

t = 0

a = =limΔt→0

ΔvΔt

dv

dt

ms−2

or( )m

s

s

vel

s

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Forthiscaseofacarstartingfromrest,andthenincreasingvelocity,theaccelerationwillbeapositivequantity.

SlowingdownWhatifweaskaboutacarslowingdown.Now,our while .

Nowthemathlookslikethis:

Wenoticethattheaccelerationisnegative.

AccelerationinthenegativeWhatifthecarstartsacceleratinginthenegativedirection?

Now,eventhespeedisincreasing,thevelocityisgettingmorenegative.

Ifwedothemath,we'llseethattheaccelerationvectorpointsinthenegativedirection.

Summaryofaccelerationsignage.Whenthesignsofanobject’svelocityandaccelerationarethesame(insamedirection),theobjectisspeedingupWhenthesignsofanobject’svelocityandaccelerationareopposite(inoppositedirections),theobjectisslowingdownandspeeddecreases

+vx

t

+vx

t

+vx

t

+vx

t

Kinematicequations

Wecandoalotbyrearrangingtheseequations.

Putting from(1)into(2)willgiveus:

= = =av − v0t− t0

20mph − 0mph2s − 0s

20mph2s

= = +4.5ma9m/s2s

s−2

= +9m/sv0 v = 0

= = = − = −4.5m/av − v0t− t0

0m/s − 9m/s2s − 0s

9m/s2s

s2

1. = a = ⇒ v = + atav− v0

tv0

2. = ⇒ x− = t = ( + v)tvx− x0

t− t0x0 v

12

v0

v

3.x− = t+ ax0 v012

t2

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or,solving(1)for ,theninsertingthatinto(2)willgiveus:

1.2.3.4.

Herewehaveanequationforvelocitywhichischangingduetoanacceleration, .

Ittellsushowfastsomethingwillbegoing(andthedirection)ifhasbeenacceleratedforatime, .

Itcandetermineanobject’svelocityatanytimetwhenweknowitsinitialvelocityanditsaccelerationDoesnotrequireorgiveanyinformationaboutpositionEx:“Howfastwasthecargoingafter10secondswhileacceleratingfromrestat10m/s ”Ex:“Howlongdidittaketoreach20milesperhour”

Thisequationwilltellusthepositionofanobjectbasedontheinitialandfinalvelocities,andthetimeelapsed.

Itdoesnotrequireknowing,norwillitgiveyou,theaccelerationoftheobject.

Ex:Howfardidtheduckwalkifittook10secondstoreach50milesperhourunderconstantacceleration.

Givespositionattimetintermsofinitialvelocityandacceleration

Doesn’trequireorgivefinalvelocity.Ex:“Howfarupdidtherocketgo?”

Givesvelocityattimetintermsofaccelerationandposition

Doesnotrequireorgiveanyinformationaboutthetime.Ex:“Howfastwaspennygoingwhenitreachedthebottomofthewell?”

EquationsofMotion(1-D)Thingstobeawareof:

1.Theyareonlyforsituationswheretheaccelerationisconstant.2.Thewaywehavewrittenthemisreallyjustfor1-Dmotion.

t

4. = + 2a(x− )v2 v20 x0

v = + atv0x = t = ( + v)tv 1

2v0

x = + t+ ax0 v012

t2

= + 2axv2 v20

v = + atv0

a

t

2

x = t =v(v+ )tv0

2

x = + t+x0 v0at2

2

= + 2axv2 v20

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Equation MissingVariable Goodforfinding

, ,

, ,

, ,

, ,

SolvingProblems1.Diagram:drawapicture2.Characters:Considertheproblemastory.Whoarethecharacters?3.Find:clearlylistsymbolicallywhatwe'relookingfor.4.Solve:statethebasicideabehindsolution,inafewwords(physicalprinciplesused,etc.)5.Assess:doesanswermakesense?

Ataxiissittingataredlight.Thelightturnsgreenandthetaxiacceleratesat2.5m/s for3seconds.Howfardoesittravelduringthistime?

Aparticleisatrest.Whataccelerationvalueshouldwegiveitsothatitwillbe2metersawayfromitsstartingpositionafter0.4seconds?

Asubwaytrainacceleratesstartingatx=200muniformlyuntilitreachesx=350m,atauniformaccelerationvalueof0.5m/s .

a.Ifithadaninitialvelocityof0m/s,whatwillthedurationofthisaccelerationbe?b.Ifithadaninitialvelocityof8m/s,whatwillthedurationofthisaccelerationbe?

If ,find and .Also,findthetimewhenthevelocityiszero.

TraianVuia,aRomanianInventor,wantedtoreach17m/sinordertotakeoffinhisflyingmachine.Hisplanecouldaccelerateat2m/s .Theonlyrunwayhehadaccesstowas80meterslong.Willhereachthenecessaryspeed?

v = + atv0 x a t v

x = (v+ )tv02

a x t v

x = + t+x0 v0at2

2v x a t

= +2axv2 v20 t a x v

Example Problem#3:

2

Example Problem#4:

Example Problem#5:

2

Example Problem#6:

x(t) = 4 − 27t+ t3 v(t) a(t)

Example Problem#7:

2

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Plotting

+x (ft)0 10 20 30 40 50

1s0s 2s 3s 4s 5s 6s 7s 8s 9s

Let'slookatthemotionofahoneybadger.

Aftereachsecond,wenotewherethehoneybadgerisalongthexaxis.

t[s] x[ft]

0 0.00

1 5.00

2 10.0

3 15.0

4 17.5

5 20.0

6 22.5

7 25.0

8 35.0

9 50.0

dist

ance

[ft]

0

10

20

30

40

50

time [s]1 2 3 4 5 6 7 8 9 10

Derivekinematicsusingcalculus.

Wecanderivenearlyallofkinematics(forcasseswithconstantacceleration)byconsideringtherelationshipsbetweenderivativesandintegrals.Let'sbeginwiththedefinitionofacceleration:

Ifwemakethe and infinitesimallysmall, and ,thenwecanrewritethisas:

a =Δv

Δt

Δv Δv dv dt

a = ⇒ dv = a dtdv

dt

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Now,wecantaketheindefiniteintegralofbothsides:

Since isassumedtobeconstant,wecanremovefromtheintegrand.Performingtheindefiniteintegrals:

where istheconstantofintegration.Todeterminetheconstant ,considertheequationwhen .Thisisthe'initialcondition',thusthevelocityatthispointwillbetheinitialvelocity: .Wethereforeobtain:

byconsideringjustthedefinitionofaccelerationandtheconceptofintegration.

Wecanlikewiseconsiderthedefinitionofinstantaneousvelocity:

Asimilaroperationleadsto:

Now,wecannotremove fromthisintegrandsinceitisnotaconstantvalue.However,wejustfiguredoutarelationbetweenvelocityandtimeabove,so:

Inthiscase, and arebothconstants.Sotheindefiniteintegralcanbesolved:

Again,wehaveaconstantofintegrationtosolvefor: .Let'sagainconsider ,i.e.theinitialcondition.When ,theobjectwillbelocatedattheinitial position, .Thus .Finally,wehaveanequationfor asafunctionoftimegivenalltheinitialconditionsofpositionandvelocity:

Thisisourfundamentalquadraticequationthatdescribesthemotionofaparticleundergoingtranslationwithconstantacceleration.

v(t)

t0

x(t)

t0

x(t)

t0

velocityasafunctionoftime: Accelerationisconstant

positionasafunctionoftime .(vel.constant,accel=0)

positionasafunctionoftime

∫ dv = ∫ a dt

a

v = at+ C1

C1 C t = 0v0

v = + atv0

v =dx

dt

∫ dx = ∫ v dt

v

∫ dx = ∫ ( + at) dtv0

v0 a

x = t+ a +v012

t2 C2

C2 t = 0 t = 0x x0 =C2 x0 x

x = + t+ ax0 v012

t2

v = + atv0 x = vtx = t+v0

at2

2

v(t) x(t)

Example Problem#8:

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Aturtleandarabbitaretohavearace.Theturtle’saveragespeedis0.9m/s.Therabbit’saveragespeedis9m/s.Thedistancefromthestartinglinetothefinishlineis1500m.Therabbitdecidestolettheturtlerunbeforehestartsrunningtogivetheturtleaheadstart.What,approximately,isthemaximumtimetherabbitcanwaitbeforestartingtorunandstillwintherace?

Acarandamotorcycleareat at .Thecarmovesataconstantvelocity .Themotorcyclestartsatrestandaccelerateswithconstantaccelerationa.

a.Findthe wheretheymeet.b.Findtheposition wheretheymeet.c.Findthevelocityofthemotorcyclewhentheymeet.

Thisproblemisaskingustodescribethekinematicsofthesituationinthemostgeneraltermspossible.Therearenonumbersgiven,sowemustdoeverythingusingsymbolicalgebra.First,let'smakesureweunderstandthesetup.Therearetwovehicles:acarandamotorcycle.Theycanbeconsideredparticlesmeaningtheyarepointlike.Theactionstartsatt=0.Atthistime,bothvehiclesarelocatedattheorigin.Themotorcycleisstationary,butthecarhasavelocity, .(* isjustasymbolthatcouldbeanumber,like10m/sor34.3mph.Butweleaveitasasymbolsothatwecansolvethisprobleminageneralway,applicabletoanycar!)Nowthecarwillmovefartherthanthemotorcycleatfirst.However,themotorcyclewillcatchupandovertakethecarbecauseitisaccelerating.

a)Findoutwhen,i.e.atwhattime,theyareatthesameposition.So,weneedfunctionsthattelluswhereeachvehicleislocatedatagiventime.Wecanstartwiththebasickinematicequationofmotion:

Forthecar,sincethereisnoacceleration, ,and ,thisequationsimplifiesto:

Forthemotorcycle,ithasnointitialvelocity, ,butitdoeshasanacceleration .Italsostartsfromtheorigin:

Thequestionaskwhentheobjectsmeet?Thatis,whenarethexvaluesthesame.So,wecanjustsetthetwoequationsequaltoeachother.

andsolvethisfor .

.Nowwehaveanequationfor thatwecanusegivenanyaccelerationandinitialvelocity.

b)Wheredoesthisoccur?Wecanusethetimeexpressioninoneofthepreviouspositionequations.

Itshouldalsobethesameifweputinthetimeinthemotorcycle'spositionequation:

c)Whatisthespeedofthemotorcycle?Wefirstneedtofindanequationforspeedofthemotorcycles.Let'stherelationshipbetween

Example Problem#9:

= 0x0 t = 0 v0

t

x

v0 v0

x = + t+ ax0 v012

t2

a = 0 = 0x0

= txcar v0

= 0v0 a

= axmoto12

t2

=xcar xmoto

t = av012

t2

t

t =2v0

a

t

= = 2xcar v02v0

a

v20

a

= a = a = 2xmoto12

t212

( )2v0

a

2 v20

a

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positionandvelocity:

So,whentimeis ,thespeedofthemotorcyclewillbe:

Noticehowtheaccelerationtermisgone.Thespeedofthemotorcyclewhenthetwoobjectmeetisindependentofitsacceleration.That'saninterestingbitofinformationthatwouldhavebeenlostifwedidthisproblemusingnumbersinsteadofletters.

x

t(s)

x

t(s)

carmeeting time

motorcycle

Thisplotshowsgraphicallythesituation.Wecancomparetheslopesthattheintersectionandseethattheslopeofthemotorcycleisroughlytwicethatofthecar.

FreeFallAfreelyfallingobjectisanyobjectmovingfreelyundertheinfluenceofgravityalone.

Objectcouldbe:

1.Dropped=releasedfromrest2.Throwndownward3.Thrownupward

Itdoesnotdependupontheinitialmotionoftheobject.

1.Theaccelerationofanobjectinfreefallisdirecteddownward(negativedirection),regardlessoftheinitialmotion.

2.Themagnitudeoffreefallaccelerationis .

3.Wecanneglectairresistance.4.We'llchooseouryaxistobepositiveupward.5.ConsidermotionnearEarth’ssurfacefornow.

v = = atdx

dt

t = 2v0a

v = at = a ( ) = 22v0

av0

9.8m/ = gs2

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Kinematicequationinthecaseoffreefall:

1.2.3.4.

Theyarethesame.Wejustreplaced and .

Anobjectisthrownupwardat20m/s:

a.Howlongwillittaketoreachthetopb.Howhighisthetop?c.Howlongtoreachthebottom?d.Howfastwillitbegoingwhenitreachesthebottom?

Ifanobjectisthrownupwardfromaheight withaspeed ,whenwillithittheground?

DropawrenchAworkerdropsawrenchdowntheelevatorshaftofatallbuilding.

a.Whereisthewrench1.5secondslater?b.Howfastisthewrenchfallingatthattime?

Arockisthrownupwardwithavelocityof49m/sfromapoint15mabovetheground.

a.Whendoestherockreachitsmaximumheight?b.Whatisthemaximumheightreached?c.Whendoestherockhittheground?

Howhighwasthis?

Drawposition,velocity,andaccelerationgraphsasafunctionsoftime,foranobjectthatisletgofromrestoffthesideofacliff.

v = − gtv0y = t = ( + v)tv 1

2v0

y = + t− gy0 v012

t2

= − 2gyv2 v20

x → y a → −g

Example Problem#10:

Example Problem#11:

y0 v0

Example Problem#12:

Example Problem#13:

Example Problem#14:

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