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