ChapterChapter 6A.6A. AccelerationAcceleration AAA PowerPointPowerPointPowerPoint PresentationPresentationPresentation bybyby PaulPaulPaul E.E.E. Tippens,Tippens,Tippens, ProfessorProfessorProfessor ofofof PhysicsPhysicsPhysics SouthernSouthernSouthern PolytechnicPolytechnicPolytechnic StateStateState UniversityUniversityUniversity

© 2007 The Cheetah: A cat that is built for speed. Its strength and agility allow it to sustain a top speed of over 100 km/h. Such speeds can only be maintained for about ten seconds. Photo © Vol. 44 Photo Disk/Getty Objectives:Objectives: AfterAfter completingcompleting thisthis module,module, youyou shouldshould bebe ableable to:to: •• DefineDefine andand applyapply conceptsconcepts ofof averageaverage andand instantaneousinstantaneous velocityvelocity andand acceleration.acceleration. •• SolveSolve problemsproblems involvinginvolving initialinitial andand finalfinal velocityvelocity,, accelerationacceleration,, displacementdisplacement,, andand timetime.. •• DemonstrateDemonstrate youryour understandingunderstanding ofof directionsdirections andand signssigns forfor velocity,velocity, displacement,displacement, andand acceleration.acceleration. •• SolveSolve problemsproblems involvinginvolving aa freefree--fallingfalling bodybody inin aa gravitationalgravitational fieldfield.. UniformUniform AccelerationAcceleration inin OneOne Dimension:Dimension: •• MotionMotion isis alongalong aa straightstraight lineline (horizontal,(horizontal, verticalvertical oror slanted).slanted). •• ChangesChanges inin motionmotion resultresult fromfrom aa CONSTANTCONSTANT forceforce producingproducing uniformuniform acceleration.acceleration. •• TheThe causecause ofof motionmotion willwill bebe discusseddiscussed later.later. HereHere wewe onlyonly treattreat thethe changes.changes. •• TheThe movingmoving objectobject isis treatedtreated asas thoughthough itit werewere aa pointpoint particle.particle. DistanceDistance andand DisplacementDisplacement

DistanceDistance isis thethe lengthlength ofof thethe actualactual pathpath takentaken byby anan object.object. ConsiderConsider traveltravel fromfrom pointpoint AA toto pointpoint BB inin diagramdiagram below:below:

DistanceDistance ss isis aa scalarscalar quantity (no direction): B quantity (no direction): s = 20 m ContainsContains magnitudemagnitude onlyonly A andand consistsconsists ofof aa numbernumber andand aa unit.unit. (20(20 m,m, 4040 mi/h,mi/h, 1010 gal)gal) DistanceDistance andand DisplacementDisplacement

DisplacementDisplacementDisplacement isisis thethethe straightstraightstraight-line--lineline separationseparationseparation ofofof twotwotwo pointspointspoints ininin aaa specifiedspecifiedspecified direction.direction.direction.

A vector quantity: D = 12 m, 20o B Contains magnitude A AND direction, a  number, unit & angle. (12 m, 300; 8 km/h, N) DistanceDistance andand DisplacementDisplacement

••• ForFor For motionmotionmotion alongalongalong xxx ororor yyy axis,axis,axis, thethethe displacementdisplacementdisplacement isisis determineddetermineddetermined bybyby thethethe xxx ororor yyy coordinatecoordinatecoordinate ofofof itsitsits finalfinalfinal position.position.position. Example:Example:Example: ConsiderConsiderConsider aaa carcarcar thatthatthat travelstravelstravels 888 m,m,m, EEE thenthenthen 121212 m,m,m, W.W.W.

NetNet displacementdisplacement DD isis from the origin to the from the origin to the D finalfinal position:position: 8 m,E x DD == 44 m,m, WW x = -4 x = +8 WhatWhat isis thethe distancedistance 12 m,W traveled?traveled? 20 m !! TheThe SignsSigns ofof DisplacementDisplacement •• DisplacementDisplacement isis positivepositive (+)(+) oror negativenegative ((--)) basedbased onon LOCATIONLOCATION.. Examples: 2 m The displacement is the y-coordinate. Whether motion is -1 m up or down, + or - is based on LOCATION. -2 m

TheThe directiondirection ofof motionmotion doesdoes notnot matter!matter! DefinitionDefinition ofof SpeedSpeed

••• SpeedSpeedSpeed isisis thethethe distancedistancedistance traveledtraveledtraveled perperper unitunitunit ofofof timetimetime (a(a(a scalarscalarscalar quantity).quantity).quantity).

s 20 m s = 20 m B v = = t 4 s A vv== 55 m/sm/s

Not direction dependent! Time t = 4 s DefinitionDefinition ofof VelocityVelocity ••• VelocityVelocityVelocity isisis thethethe displacementdisplacementdisplacement perperper unitunitunit ofofof time.time.time. (A(A(A vectorvectorvector quantity.)quantity.)quantity.) s = 20 m D 12 m B v  D=12 m t 4 s A o 20 vv == 33 m/sm/s atat 202000 NN ofof EE

Time t = 4 s Direction required! ExampleExample 1.1. AA runnerrunner runsruns 200200 m,m, east,east, thenthen changeschanges directiondirection andand runsruns 300300 m,m, westwest.. IfIf thethe entireentire triptrip takestakes 6060 ss,, whatwhat isis thethe averageaverage speedspeed andand whatwhat isis thethe averageaverage velocity?velocity?

RecallRecall thatthat averageaverage s = 300 m s = 200 m speedspeed isis aa functionfunction 2 1 onlyonly ofof totaltotal distancedistance andand totaltotal timetime:: start TotalTotal distance:distance: ss == 200200 mm ++ 300300 mm == 500500 mm total path 500 m Average speed  Avg. speed time 60 s 8.33 m/s DirectionDirection doesdoes notnot matter!matter! ExampleExample 11 (Cont.)(Cont.) NowNow wewe findfind thethe averageaverage velocity,velocity, whichwhich isis thethe netnet displacementdisplacement divideddivided byby timetime.. InIn thisthis case,case, thethe directiondirection matters.matters. x  x tt == 6060 ss f 0 x = -100 m x = +200 m v  f 1 t

xo = 0 xx0 == 00 m;m; xxf == --100100 mm Direction of final 100 m 0 Direction of final v 1.67 m/s displacementdisplacement isis toto 60 s thethe leftleft asas shown.shown. Average velocity: v 1.67 m/s, West

Note:Note: AverageAverage velocityvelocity isis directeddirected toto thethe west.west. ExampleExample 2.2. AA skysky diverdiver jumpsjumps andand fallsfalls forfor 600600 mm inin 1414 s.s. AfterAfter chutechute opens,opens, hehe fallsfalls anotheranother 400400 mm inin 150150 s.s. WhatWhat isis averageaverage speedspeed forfor entireentire fall?fall? 14 s TotalTotal distance/distance/ totaltotal time:time:

xxAB 600 m + 400 m A v  625 m ttAB 14 s + 150 s 1000 m v  v  6.10 m/s 164 s B

AverageAverageAverage speedspeedspeed isisis aaa functionfunctionfunction 356 m onlyonlyonly ofofof totaltotaltotal distancedistancedistance traveledtraveledtraveled andandand thethethe totaltotaltotal timetimetime required.required.required. 142 s ExamplesExamples ofof SpeedSpeed

Orbit 2 x 104 m/s Light = 3 x 108 m/s

Jets = 300 m/s Car = 25 m/s SpeedSpeed ExamplesExamples (Cont.)(Cont.)

Runner = 10 m/s

Glacier = 1 x 10-5 m/s

Snail = 0.001 m/s AverageAverage SpeedSpeed andand InstantaneousInstantaneous VelocityVelocity

... TheTheThe averageaverageaverage speedspeedspeed dependsdependsdepends ONLYONLYONLY ononon thethethe distancedistancedistance traveledtraveledtraveled andandand thethethe timetimetime required.required.required.

s = 20 m B TheThe instantaneousinstantaneous C velocityvelocity isis thethe magn-magn- A itudeitude andand directiondirection ofof thethe speedspeed atat aa par-par- ticularticular instant.instant. (v(v atat Time t = 4 s pointpoint C)C) TheThe SignsSigns ofof VelocityVelocity

... VelocityVelocityVelocity isisis positivepositivepositive (+)(+)(+) ororor negativenegativenegative (((-)--)) basedbasedbased ononon directiondirectiondirection ofofof motion.motion.motion.

+ - FirstFirst choosechoose ++ direction;direction; + thenthen vvisis positivepositive ifif motionmotion isis withwith thatthat direction,direction, andand - negativenegative ifif itit isis againstagainst thatthat + direction.direction. AverageAverage andand InstantaneousInstantaneous vv

AverageAverage Velocity:Velocity: InstantaneousInstantaneous Velocity:Velocity:

xxx21 x vavg  vtinst  (0) ttt21 t

slope x xx 2 x x xx 1 t tt Displacement,

tt 1 tt 2 Time DefinitionDefinition ofof AccelerationAcceleration

.. AnAn accelerationacceleration isis thethe changechange inin velocityvelocity perper unitunit ofof time.time. (A(A vectorvector quantity.)quantity.) .. AA changechange inin velocityvelocity requiresrequires thethe applicationapplication ofof aa pushpush oror pullpull ((forceforce).).

AA formalformal treatmenttreatment ofof forceforce andand accelerationacceleration willwill bebe givengiven later.later. ForFor now,now, youyou shouldshould knowknow that:that:

• The direction of accel- • The acceleration is eration is same as proportional to the direction of force. magnitude of the force. a 2a 2F F Acceleration and Force Acceleration and Force

Pulling the wagon with twice force produces twice the acceleration and acceleration is in direction of force. Pulling the wagon with twice force Pulling the wagon with twice force produces twice the acceleration and produces twice the acceleration and acceleration is in direction of force. acceleration is in direction of force. ExampleExample ofof AccelerationAcceleration

+ Force t = 3 s

v0 = +2 m/s vf = +8 m/s

The wind changes the speed of a boat from 2 m/s to 8 m/s in 3 s. Each second the speed changes by 2 m/s.

WindWind forceforce isis constant,constant, thusthus accelerationacceleration isis constant.constant. TheThe SignsSigns ofof AccelerationAcceleration ••• AccelerationAccelerationAcceleration isisis positivepositivepositive (((+)++)) ororor negativenegativenegative (((-)--)) basedbasedbased ononon thethethe directiondirectiondirection ofofof forceforceforce...

+ ChooseChooseChoose +++ directiondirectiondirection first.first.first. F a (-) ThenThenThen accelerationaccelerationacceleration aaa willwillwill havehavehave thethethe samesamesame signsignsign asasas thatthatthat ofofof thethethe forceforceforce FFF ——— a(+) regardlessregardlessregardless ofofof thethethe F directiondirectiondirection ofofof velocity.velocity.velocity. AverageAverage andand InstantaneousInstantaneous aa

vvv21 v aavg  atinst  (0) ttt21 t

slope v2 v v v 1 t t

t1 t2 time ExampleExample 33 (No(No changechange inin direction):direction): AA constantconstant forceforce changeschanges thethe speedspeed ofof aa carcar fromfrom 88 m/sm/s toto 2020 m/sm/s inin 44 ss.. WhatWhat isis averageaverage acceleration?acceleration?

+ Force t = 4 s

v1 = +8 m/s v2 = +20 m/s

StepStep 1.1. DrawDraw aa roughrough sketch.sketch. StepStep 2.2. ChooseChoose aa positivepositive directiondirection (right).(right). StepStep 3.3. LabelLabel givengiven infoinfo withwith ++ andand -- signs.signs. StepStep 4.4. IndicateIndicate directiondirection ofof forceforce F.F. ExampleExample 33 (Continued):(Continued): WhatWhat isis averageaverage accelerationacceleration ofof car?car? + Force t = 4 s

v1 = +8 m/s v2 = +20 m/s

StepStep 5.5. RecallRecall definitiondefinition 20 m/s - 8 m/s a 3 m/s ofof averageaverage acceleration.acceleration. 4 s

vvv21 aavg  a  3 m/s, rightward ttt21 ExampleExample 4:4: AA wagonwagon movingmoving easteast atat 2020 m/sm/s encountersencounters aa veryvery strongstrong headhead--wind,wind, causingcausing itit toto changechange directions.directions. AfterAfter 55 ss,, itit isis travelingtraveling westwest atat 55 m/sm/s.. WhatWhat isis thethe averageaverage acceleration?acceleration? (Be(Be carefulcareful ofof signs.)signs.)

+ Force E

vf = -5 m/s vo = +20 m/s

StepStep 1.1. DrawDraw aa roughrough sketch.sketch. StepStep 2.2. ChooseChoose thethe eastwardeastward directiondirection asas positive.positive. StepStep 3.3. LabelLabel givengiven infoinfo withwith ++ andand -- signs.signs. ExampleExample 44 (Cont.):(Cont.): WagonWagon movingmoving easteast atat 2020 m/sm/s encountersencounters aa headhead--wind,wind, causingcausing itit toto changechange directions.directions. FiveFive secondsseconds later,later, itit isis travelingtraveling westwest atat 55 m/sm/s.. WhatWhat isis thethe averageaverage acceleration?acceleration?

ChooseChooseChoose thethethe eastwardeastwardeastward directiondirectiondirection asasas positive.positive.positive. InitialInitial velocity,velocity, vv == +20+20 m/s,m/s, easteast (+)(+) Initial velocity, voo = +20 m/s, east (+) FinalFinal velocity,velocity, vv == --55 m/s,m/s, westwest ((--)) Final velocity, vff = -5 m/s, west (-) TheThe changechange inin velocity,velocity, vv == vv -- vv The change in velocity, v = vff - v00 vvv=== (((-5--55 m/s)m/s)m/s) --- (+20(+20(+20 m/s)m/s)m/s) === ---252525 m/sm/sm/s ExampleExample 4:4: (Continued)(Continued) + Force E vo = +20 m/s vf = -5 m/s v = (-5 m/s) - (+20 m/s) = -25 m/s

vv vv f -- vv o --2525 m/sm/s aaavg === aa == tt tt f -- tt o 55 ss Acceleration is directed to a = - 5 m/s22 a = - 5 m/s left, west (same as F). SignsSigns forfor DisplacementDisplacement

+ C Force D E A B 2 vf = -5 m/s vo = +20 m/s a = - 5 m/s

TimeTime tt == 00 atat pointpoint AA.. WhatWhat areare thethe sisiggnsns (+(+ oror --)) ofof displacementdisplacement atat BB,, CC,, andand DD?? At B, x is positive, right of origin At C, x is positive, right of origin At D, x is negative, left of origin SignsSigns forfor VelocityVelocity x = 0 + C Force D E A B 2 vf = -5 m/s vo = +20 m/s a = - 5 m/s

WhatWhat areare thethe signssigns (+(+ oror --)) ofof velocityvelocity atat pointspoints B,B, C,C, andand D?D? .. AtAt B,B, vv isis zerozero -- nono signsign needed.needed. .. AtAt CC,, vv isis positivepositive onon wayway outout andand negativenegative onon thethe wayway back.back. .. AtAt DD,, vv isis negativenegative,, movingmoving toto left.left.

SignsSigns forfor AccelerationAcceleration

+ C Force D E A B 2 vf = -5 m/s vo = +20 m/s a = - 5 m/s

What are the signs (+ or -) of acceleration at points B, C, and D? .. AtAt B,B, C,C, andand DD,, aa == --55 m/s,m/s, negativenegative atat allall points.points. . The force is constant and always directed to left, so acceleration does not change. DefinitionsDefinitions

Average velocity:

xxx21 vavg  ttt21

Average acceleration:

vvv21 aavg  ttt21 VelocityVelocity forfor constantconstant aa Average velocity: Average velocity:

x xxf  0 vv v  0 f avg ttt vavg  f 0 2

Setting to = 0 and combining we have: vv xx0 f t 0 2 ExampleExample 5:5: AA ballball 55 mm fromfrom thethe bottombottom ofof anan inclineincline isis travelingtraveling initiallyinitially atat 88 m/sm/s.. FourFour secondsseconds later,later, itit isis travelingtraveling downdown thethe inclineincline atat 22 m/sm/s.. HowHow farfar isis itit fromfrom thethe bottombottom atat thatthat instant?instant?

+ x F vf vo 5 m -2 m/s 8 m/s t = 4 s CarefulCareful

vo + vf 8 m/s + (-2 m/s) x = xo + t = 5 m + (4 s) 2 2 = 17 m = 17 m x x F t = 4 s -2 m/s

f 2 v

o 2

x v 8 m/s + (-2 m/s) 8 m/s 8 m/s - 2 + 5 m = 5 m + (4 s) (Continued) (Continued) x = 5 m + (4 s)

x ConstantConstant AccelerationAcceleration

v vvf  0 Acceleration:Acceleration: a avg  tttf  0

SettingSetting tto == 00 andand solvingsolving forfor v,v, wewe have:have:

vvatf  0 

FinalFinal velocityvelocity == initialinitial velocityvelocity ++ changechange inin velocityvelocity AccelerationAcceleration inin ourour ExampleExample

+ F vvatf 0 x v vo vv a  f 0 5 m -2 m/s t 88 m/sm/s t = 4 s

( 2 m/s) ( 8 m/s) a 2 m/s2 4 s WhatTheWhatThe forceforce isis thethe changingchanging meaningmeaning 2 aa == -2.50-2.50 m/sm/s2 speedofspeedof negativenegative isis downdown signsign plane!plane! forfor aa?? FormulasFormulas basedbased onon definitions:definitions:

vv0  f xx t vvatf  0  0 2 DerivedDerived formulas:formulas

1 2 1 2 xx00 vtat 2 xx0 vtatf 2

22 2(ax x00 ) vf v ForFor constantconstant accelerationacceleration onlyonly UseUse ofof InitialInitial PositionPosition xx 0 inin Problems.Problems.

00 vv0  f IfIf youyou choosechoose thethe xx t origin of your x,y 0 2 origin of your x,y axesaxes atat thethe pointpoint ofof 00 xx vtat 1 2 thethe initialinitial position,position, 002 you can set x = 0, you can set x00 = 0, 00 simplifying these xx vtat 1 2 simplifying these 0 f 2 equations.equations. 00 22The x term is very 2(ax x ) v v The xo term is very 00f usefuluseful forfor studyingstudying problemsproblems involvinginvolving vvatf 0 motionmotion ofof twotwo bodies.bodies. ReviewReview ofof SymbolsSymbols andand UnitsUnits

• Displacement ((x,x, xx );); metersmeters ((mm)) •Displacement(x, xoo ); meters (m) • Velocity ((v,v, vv );); metersmeters perper secondsecond ((m/sm/s)) •Velocity(v, voo ); meters per second (m/s) •• Acceleration Acceleration (((aaa););); metersmetersmeters perperper sss22 (((m/sm/sm/s22))) •••Tim TimeTimee (((t);tt);); secondssecondsseconds (((sss)))

ReviewReview signsign conventionconvention forfor eacheach symbolsymbol TheThe SignsSigns ofof DisplacementDisplacement

••• DisplacementDisplacementDisplacement isisis positivepositivepositive (+)(+)(+) ororor negativenegativenegative (((-)--)) basedbasedbased ononon LOCATIONLOCATIONLOCATION...

22 mm The displacement is the y-coordinate. Whether motion is --11 mm up or down, + or - is --22 mm based on LOCATION. TheThe SignsSigns ofof VelocityVelocity

••• VelocityVelocityVelocity isisis positivepositivepositive (+)(+)(+) ororor negativenegativenegative (((-)--)) basedbasedbased ononon directiondirectiondirection ofofof motionmotionmotion... + + - First choose + direction; then velocity v is positive if motion is with that + direction, and negative if + - it is against that positive direction. AccelerationAcceleration ProducedProduced byby ForceForce

••• AccelerationAccelerationAcceleration isisis (((+)++)) ororor (((-)--)) basedbasedbased ononon directiondirectiondirection ofofof forceforceforce (((NOTNOTNOT basedbasedbased ononon vvv).).).

A push or pull (force) is necessary to change FF aa((--)) velocity, thus the sign of a is same as sign of F.

FF aa(+)(+) More will be said later on the relationship between F and a. ProblemProblem SolvingSolving Strategy:Strategy:

.. DrawDraw andand labellabel sketchsketch ofof problem.problem.

.. IndicateIndicate ++ directiondirection andand forceforce direction.direction.

.. ListList givensgivens andand statestate whatwhat isis toto bebe found.found.

Given: ____, _____, _____ (x,v,vo ,a,t) Find: ____, _____ . Select equation containing one and not the other of the unknown quantities, and solve for the unknown. ExampleExample 6:6: AA airplaneairplane flyingflying initiallyinitially atat 400400 ft/sft/s landslands onon aa carriercarrier deckdeck andand stopsstops inin aa distancedistance ofof 300300 ft.ft. WhatWhat isis thethe acceleration?acceleration?

+400 ft/s v = 0 300 ft v + F o

XX0 == 00

Step 1. Draw and label sketch. Step 2. Indicate + direction and FF direction. Example:Example: (Cont.)(Cont.) +400 ft/s v = 0 300 ft vo + F XX0 == 00

StepStep 3.3. List given; find Given:Given: vv o == +400+400 ft/sft/s information with signs. vv == 00 xx == +300+300 ftft ListList tt == ?,?, eveneven thoughthough timetime waswas notnot askedasked for.for. Find:Find: aa == ?;?; tt == ?? ContinuedContinued ...... x +400 ft/s v = 0 300 ft vo + F

X0 = 0 0 0 StepStep 4.4. Select equation 2 2 that contains aa and not tt. 2a(x -xo) = v -vo -v 2 -(400 ft/s)2 InitialInitial positionposition andand o finalfinal velocityvelocity areare zero.zero. a = = a = - 267 ft/s22 2x 2(300 ft) aa == -- 267267 ft/sft/s

BecauseBecauseWhyWhy isis ForceForce thethe accelerationacceleration isis inin aa negativenegative negative?negative? direction!direction! AccelerationAcceleration DueDue toto GravityGravity

••• EveryEvery Every objectobjectobject ononon thethethe earthearthearth experiencesexperiencesexperiences aaa commoncommoncommon force:force:force: thethethe forceforceforce dueduedue tototo gravity.gravity.gravity. ••• ThisThis This forceforceforce isisis alwaysalwaysalways directeddirecteddirected towardtowardtoward thethethe centercentercenter ofofof thethethe earthearthearth g W (downward).(downward).(downward). ••• TheThe The accelerationaccelerationacceleration dueduedue tototo gravitygravitygravity isisis relativelyrelativelyrelatively constantconstantconstant nearnearnear thethethe EarthEarthEarth’s’’ss surface.surface.surface. Earth GravitationalGravitational AccelerationAcceleration

••• InIn In aaa vacuum,vacuum,vacuum, allallall objectsobjectsobjects fallfallfall withwithwith samesamesame acceleration.acceleration.acceleration. ••• EquationsEquations Equations forforfor constantconstantconstant accelerationaccelerationacceleration applyapplyapply asasas usual.usual.usual. ••• NearNear Near thethethe EarthEarthEarth’s’’ss surface:surface:surface:

aa == gg == 9.809.80 m/sm/s2 oror 3232 ft/sft/s2 DirectedDirected downwarddownward (usually(usually negative).negative). ExperimentalExperimental DeterminationDetermination ofof GravitationalGravitational Acceleration.Acceleration. tt TheThe apparatusapparatus consistsconsists ofof aa devicedevice whichwhich measuresmeasures thethe timetime requiredrequired forfor aa ballball toto fallfall aa givengiven distance.distance. yy

SupposeSuppose thethe heightheight isis 1.201.20 mm andand thethe dropdrop timetime isis recordedrecorded asas 0.6500.650 s.s. WhatWhat isis thethe accelerationacceleration duedue toto gravity?gravity? ExperimentalExperimental DeterminationDetermination ofof GravityGravity (y(y0 == 0;0; yy == --1.201.20 m)m) tt yy == --1.201.20 m;m; tt == 0.4950.495 ss

1 2 yvtatv002 ;  0 22(1.20y  m) yy a 22 t (0.495 s) ++

Acceleration 2 of Gravity: a 9.79 m/s WW AccelerationAcceleration aa isis negativenegative becausebecause forceforce WW isis negative.negative. SignSign Convention:Convention: av = = - 0 AA BallBall ThrownThrown y = + VerticallyVertically UpwardUpward

av y== = -+ + yva = == + -- ••• DisplacementDisplacement isis positivepositive (+)(+) oror negativenegative ((-)-) basedbased UP = + onon LOCATIONLOCATION.. Release Point vya = == - 0-

••• VelocityVelocity isis positivepositive (+)(+) oror negativenegative ((-)-) basedbased onon directiondirection ofof motionmotion.. yv= = -- aa == -- NegativeNegative •• Acceleration is (+) or (-) based on direction of force (weight). Tippens SameSame ProblemProblem SolvingSolving StrategyStrategy ExceptExcept aa == gg:: .. DrawDraw andand labellabel sketchsketch ofof problem.problem.

.. IndicateIndicate ++ directiondirection andand forceforce direction.direction.

.. ListList givensgivens andand statestate whatwhat isis toto bebe found.found. Given: ____, _____, a = -9.8 m/s2 Find: ____, _____ . Select equation containing one and not the other of the unknown quantities, and solve for the unknown. ExampleExample 7:7: AA ballball isis thrownthrown verticallyvertically upwardupward withwith anan initialinitial velocityvelocity ofof 3030 m/sm/s.. WhatWhat areare itsits positionposition andand velocityvelocity afterafter 22 ss,, 44 ss,, andand 77 ss??

Step 1. Draw and label a sketch. + Step 2. Indicate + direction a = g and force direction. Step 3. Given/find info.

a = -9.8 ft/s2 t = 2, 4, 7 s vo = +30 m/s vo = + 30 m/s y = ? v = ? FindingFinding Displacement:Displacement: Step 4. Select equation that contains y and not v. + 0 a = g 1 2 yy00 vtat2 yy == (30(30 m/s)m/s)tt ++ ½((--9.89.8 m/sm/s2))tt2

SubstitutionSubstitution ofof tt == 2,2, 4,4, andand 77 ss vo = 30 m/s willwill givegive thethe followingfollowing values:values:

yy == 40.440.4 m;m; yy== 41.641.6 m;m; yy == -30.1-30.1 mm FindingFinding Velocity:Velocity: StepStep 5.5. FindFind vv fromfrom equationequation + thatthat containscontains vvandand notnot xx:: a = g vvatf 0

2 vtf 30 m/s ( 9.8 m/s ) vo = 30 m/s SubstituteSubstitute tt == 2,2, 4,4, andand 77 s:s: vv == +10.4+10.4 m/s;m/s; vv == -9.20-9.20 m/s;m/s; vv == -38.6-38.6 m/sm/s ExampleExample 7:7: (Cont.)(Cont.) NowNow findfind thethe maximummaximum heightheight attained:attained: DisplacementDisplacement isis aa maximummaximum + whenwhen thethe velocityvelocity vvf isis zero.zero. 2 a = g vtf 30 m/s ( 9.8 m/s ) 0 30 m/s tt; 3.06 s 9.8 m/s2 vo = +96 ft/s ToTo findfind yymax wewe substitutesubstitute tt == 3.063.06 ss intointo thethe generalgeneral equationequation forfor displacement.displacement.

yy == (30(30 m/s)m/s)tt ++ ½((--9.89.8 m/sm/s2))tt2 ExampleExample 7:7: (Cont.)(Cont.) FindingFinding thethe maximummaximum height:height:

yy == (30(30 m/s)m/s)tt ++ ½((--9.89.8 m/sm/s2))tt2 tt == 3.063.06 ss

OmittingOmitting units,units, wewe obtain:obtain: + 1 2 a = g y (30)(3.06)2 ( 9.8)(3.06)

yy == 91.891.8 mm -- 45.945.9 mm vo =+30 m/s ymax = 45.9 m SummarySummary ofof FormulasFormulas

vv0  f xx t vvatf  0  0 2 DerivedDerived Formulas:Formulas

1 2 1 2 xx00 vtat 2 xx0 vtatf 2

22 2(ax x00 ) vf v For Constant Acceleration Only Summary:Summary: ProcedureProcedure

.. DrawDraw andand labellabel sketchsketch ofof problem.problem.

.. IndicateIndicate ++ directiondirection andand forceforce direction.direction.

.. ListList givensgivens andand statestate whatwhat isis toto bebe found.found. Given: ____, _____, ______Find: ____, _____ . Select equation containing one and not the other of the unknown quantities, and solve for the unknown. CONCLUSIONCONCLUSION OFOF ChapterChapter 66 -- AccelerationAcceleration