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Kinematics…
• Branchofphysicsthatdealswithdescribingthemotionofobjects
• Motioncanbedescribedin3ways:a) Words/sentencesb) Mathematicalequationsc) graphically
Distance
• Theseparationbetween2points
• Expressesaquantityofmeasure(scalarquantity)
• Distancestraveledin1direction
Displacement• Theseparationbetweenanobjectandareferencepoint.
• Indicatesdistanceanddirection(vectorquantity)
• Usedwhendistancetraveledisnotequaltodistancefromstart
• Referencepoint– zeropointusedtodescribemotioninaframeofreference
Speed• Thedistanceanobjecttravelsinaunitoftime
• Instantaneousspeedishowfastyourgoingataninstantoftime(i.e.,speedometerreadingincar)
• Average speedisthetotaldistancetraveledoverthetotaltime.
v = ∆d∆t
v=averagevelocity(m/s)∆d =displacement(m)∆t =timetakenforchange(s)
Example:Acartravels25kmin0.70h,thentravels35kminthenext1.50h.Whatistheaveragevelocityoftheentiretrip?
Velocity
• indicatesspeed anddirection ofamovingobject.
• hasmagnitudeanddirection(vectorquantity)
• equaltotheslope ofadisplacement-timegraph.
ConstantVelocity• Achievedwhentheaveragevelocityofanobjectisthesameforallintervals(uniformmotion)
• Constantvelocitiesproducestraightlinesondisplacementtimegraphs.
d
t
• Steeperslope= greatervelocity
• Horizontalslope= zerovelocity(standingstill)
• Positivevelocity= indicatesmotionawayfromthestart
• Negativevelocity= indicatesmotiontowardsstartingpoint
VA =positivevelocity
VB =zerovelocity
VC =negativevelocity
VD =fastestvelocity
VE =slowest(non-zero)velocity t
d
A
B
C
D
E
SlopeofChangingVelocity
• Whenvelocitychangestheobjectisaccelerating
• Thelineonadisplacement-timegraphwillbecurved.
d
t
Tangent– InstantaneousVelocity
Tangent – straightlinethathasthesameslopeasapointonacurve
• Drawalinethroughasinglepoint,butdonotcrossthegraph– justtouchthepoint.
• Thevelocityatthatpoint isequaltotheslopeofthetangent.
d
t
tangent
VA =5.4m– 1.2m = 4.2m =0.66m/s10.9s– 4.5s 6.4s
VB =8.5m– 8.5m = 0m =0m/s17.0s– 7.0s 10s
VC =4.0m– 7.2m = -3.2m =-0.68m/s16.7s– 12.0s 4.7s
Acceleration
• therate atwhichvelocity changes(increasesordecreases).
• changingvelocity=acceleration
• constantvelocity=zeroacceleration
a= ∆vt
a=averageacceleration(m/s2)∆v = changeinvelocity(m/s)t=timetakenforchange(s)vf =finalvelocityvi =initialvelocity
=vf - vit
Example:Acaracceleratesfrom20km/hto80km/hin10seconds.Calculatetheaverageacceleration.(Assume2sig.figs.forallmeasurements)
Example4:Aturtlewantstoacceleratefrom2mm/s to8mm/s.Howlongdoesittake,ifitsmaximumaccelerationis3mm/s2?
PositiveAcceleration• Constantaccelerationproducesastraightline(increaseofvelocityisthesameforeachunitoftime).
• Changingacceleration producesacurvedline(increaseinvelocityisnotthesameforeachunit).
v v
tt
NegativeAcceleration• Slowingdownproducesnegativeacceleration
• Straightlinewhendecreaseinvelocityisthesameforeachunitoftime.
• Curvedlinewhendecreaseinvelocitychangeswithtime.
v v
tt
ChangeinDirection
• Ifdirectionchanges,thevelocitychangesandtherefore,thereisacceleration.
i.e.,Ferriswheelrotatesataconstantspeedinacircularmotion,butthedirectionchanges,resultingin
acceleration.
ChangingAcceleration
• CurveonaV-Tgraph
• Averageacceleration=slopeofthestraightlinejoiningtwopointsonacurveofaV-Tgraph
v
t
3.Whatistheaverageaccelerationforeachofthefollowingtimeintervals?
a) 5.0to15.0s?b) 9.0to13.0s?c) 15.0to20.0s?
(7.8-3.9m/s)/10.0s=0.39m/s2
(5.0-6.3m/s)/4.0s=-0.33m/s2
(11.0-7.2m/s)/5.0s=0.76m/s2
4.Whatistheaccelerationforeachofthefollowingtimes?
a) 15.0s?b) 11.0s?c) 17.0s?
(10.0-3.0m/s)/(16.0-13.0s)=2.3m/s2
(3.0-8.8m/s)/(15.0-6.0s)=-0.64m/s2
VelocitywithUniformAcceleration
• Velocity-timegraphisastraightline• Slopeoftheline=acceleration
a= vf – vit
vf =vi +at
vf =finalvelocity(m/s)vi =initialvelocity(m/s)a=acceleration(m/s2)t=time(s)
Ex#2.Asupersonicjetthatisflyingat145m/sisaccelerateduniformlyattherateof23.1m/s2 for20.0s.Whatisitsfinalvelocity?
Displacement:GivenVelocityandTime
• displacementisequaltothetotalarea underthelineofavelocity-timegraph.
• Duringconstantvelocityfromrest:
vave = dt
d=vave t
d=½(vf +vi)t
vf
vit
Displacement:GivenAccelerationandTime
If vf =vi +at
and d=½(vf+vi)t
then d=½(vi +at+vi)t
d=½(2vi +at)t
d=½(2vit +at2)
d=vit +½at2
Ex#2.Whatistheaccelerationofanobjectthatacceleratessteadilyfromrest,travelingadistanceof150mover10.0s?
Displacement:GivenVelocityandAcceleration
If vf =vi +at
t =v - via
and
d= vf - via
æ
è ç
ö
ø ÷ vi +
12a vf - vi
a
æ
è ç
ö
ø ÷
é
ë ê
ù
û ú
d= vit+ 12at2
= (t)(vi +12at)
ad= vf - vi( ) vi + 12vf - vi( )
é
ë ê
ù
û ú
ad= vf - vi( ) vi +12vf -
12vi
æ
è ç
ö
ø ÷
ad= vf - vi( ) 12vf + 1
2vi
æ
è ç
ö
ø ÷
ad= 12vf - vi( ) vf + vi( )
2ad= vf - vi( ) vf + vi( )
2ad= vf2- vi
2
vf
2= vi
2+ 2ad
Ex#1.Abulletacceleratesat6.8x104 m/s2 fromrestasittravelsthe0.80m oftheriflebarrel.Whatvelocitydoesthebullethaveasitleavesthebarrel?
Ex#2.Adrivertravelingat95km/hseesadeerstandingontheroad150mahead.Heslamsonhisbreaksanddeceleratesatarateof-2.0m/s2.Willhestopintime?
AccelerationduetoGravity
• Galileoshowedthatallobjectsfalltoearthwithaconstantacceleration,ifairresistancecanbeignored.
• g isthesymbolforaccelerationduetogravity.
• Onthesurfaceoftheearth,g =9.80m/s2
(variesslightly,dependingondistancefromcentreofearth)
Assumingnoairresistance,allaccelerationformulasapplytofallobjects(substituteg fora)
Ex#1:The“Hellevator”rideatPlayland fallsfreelyforatimeof1.8s.
a) Whatisthevelocityattheendofthedrop?