Kinematic Equations: Sample Problems and Solutions - The ...

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EarlierinLesson6,fourkinematicequationswereintroducedanddiscussed.Ausefulproblem-solvingstrategywaspresentedforusewiththeseequationsandtwoexamplesweregiventhatillustratedtheuseofthestrategy.Then,theapplicationofthekinematicequationsandtheproblem-solvingstrategytofree-fallmotionwasdiscussedandillustrated.InthispartofLesson6,severalsampleproblemswillbepresented.Theseproblemsallowanystudentofphysicstotesttheirunderstandingoftheuseofthefourkinematicequationstosolveproblemsinvolvingtheone-dimensionalmotionofobjects.Youareencouragedtoreadeachproblemandpracticetheuseofthestrategyinthesolutionoftheproblem.Thenclickthebuttontochecktheanswerorusethelinktoviewthesolution.   CheckYourUnderstanding Anairplaneacceleratesdownarunwayat3.20m/s2for32.8suntilisfinallyliftsofftheground.Determinethedistancetraveledbeforetakeoff. SeeAnswer   Answer:d=1720m Seesolutionbelow.   Acarstartsfromrestandacceleratesuniformlyoveratimeof5.21secondsforadistanceof110m.Determinetheaccelerationofthecar. SeeAnswer   Answer:a=8.10m/s/s Seesolutionbelow.   UptonChuckisridingtheGiantDropatGreatAmerica.IfUptonfreefallsfor2.60seconds,whatwillbehisfinalvelocityandhowfarwillhefall? SeeAnswer   Answers:d=33.1mandvf=25.5m/s Seesolutionbelow.   Aracecaracceleratesuniformlyfrom18.5m/sto46.1m/sin2.47seconds.Determinetheaccelerationofthecarandthedistancetraveled. SeeAnswer   Answers:a=11.2m/s/sandd=79.8m Seesolutionbelow.   Afeatherisdroppedonthemoonfromaheightof1.40meters.Theaccelerationofgravityonthemoonis1.67m/s2.Determinethetimeforthefeathertofalltothesurfaceofthemoon. SeeAnswer   Answer:t=1.29s Seesolutionbelow.   Rocket-poweredsledsareusedtotestthehumanresponsetoacceleration.Ifarocket-poweredsledisacceleratedtoaspeedof444m/sin1.83seconds,thenwhatistheaccelerationandwhatisthedistancethatthesledtravels? SeeAnswer   Answers:a=243m/s/s d=406m Seesolutionbelow.   Abikeacceleratesuniformlyfromresttoaspeedof7.10m/soveradistanceof35.4m.Determinetheaccelerationofthebike. SeeAnswer   Answer:a=0.712m/s/s Seesolutionbelow.   Anengineerisdesigningtherunwayforanairport.Oftheplanesthatwillusetheairport,thelowestaccelerationrateislikelytobe3m/s2.Thetakeoffspeedforthisplanewillbe65m/s.Assumingthisminimumacceleration,whatistheminimumallowedlengthfortherunway? SeeAnswer   Answer:d=704m Seesolutionbelow.   Acartravelingat22.4m/sskidstoastopin2.55s.Determinetheskiddingdistanceofthecar(assumeuniformacceleration). SeeAnswer   Answer:d=28.6m Seesolutionbelow.   Akangarooiscapableofjumpingtoaheightof2.62m.Determinethetakeoffspeedofthekangaroo. SeeAnswer   Answer:vi=7.17m/s Seesolutionbelow.   IfMichaelJordanhasaverticalleapof1.29m,thenwhatishistakeoffspeedandhishangtime(totaltimetomoveupwardstothepeakandthenreturntotheground)? SeeAnswer   Answer:vi=5.03m/sandhangtime=1.03s(exceptforinsportscommericals) Seesolutionbelow.   Abulletleavesariflewithamuzzlevelocityof521m/s.Whileacceleratingthroughthebarreloftherifle,thebulletmovesadistanceof0.840m.Determinetheaccelerationofthebullet(assumeauniformacceleration). SeeAnswer   Answer:a=1.62*105m/s/s Seesolutionbelow.   Abaseballispoppedstraightupintotheairandhasahang-timeof6.25s.Determinetheheighttowhichtheballrisesbeforeitreachesitspeak.(Hint:thetimetorisetothepeakisone-halfthetotalhang-time.) SeeAnswer   Answer:d=48.0m {^cosymantecnisbfw^} Seesolutionbelow. Theobservationdeckoftallskyscraper370mabovethestreet.Determinethetimerequiredforapennytofreefallfromthedecktothestreetbelow. SeeAnswer   Answer:t=8.69s Seesolutionbelow.   Abulletismovingataspeedof367m/swhenitembedsintoalumpofmoistclay.Thebulletpenetratesforadistanceof0.0621m.Determinetheaccelerationofthebulletwhilemovingintotheclay.(Assumeauniformacceleration.) SeeAnswer   Answer:a=-1.08*10^6m/s/s Seesolutionbelow.   Astoneisdroppedintoadeepwellandisheardtohitthewater3.41safterbeingdropped.Determinethedepthofthewell. SeeAnswer   Answer:d=-57.0m(57.0metersdeep)  Seesolutionbelow.   ItwasoncerecordedthataJaguarleftskidmarksthatwere290minlength.AssumingthattheJaguarskiddedtoastopwithaconstantaccelerationof-3.90m/s2,determinethespeedoftheJaguarbeforeitbegantoskid. SeeAnswer   Answer:vi=47.6m/s Seesolutionbelow.   Aplanehasatakeoffspeedof88.3m/sandrequires1365mtoreachthatspeed.Determinetheaccelerationoftheplaneandthetimerequiredtoreachthisspeed. SeeAnswer   Answer:a=2.86m/s/sandt=30.8s Seesolutionbelow.   Adragsteracceleratestoaspeedof112m/soveradistanceof398m.Determinetheacceleration(assumeuniform)ofthedragster. SeeAnswer   Answer:a=15.8m/s/s Seesolutionbelow.   Withwhatspeedinmiles/hr(1m/s=2.23mi/hr)mustanobjectbethrowntoreachaheightof91.5m(equivalenttoonefootballfield)?Assumenegligibleairresistance. SeeAnswer   Answer:vi=94.4mi/hr Seesolutionbelow.   SolutionstoAboveProblems Given: a=+3.2m/s2 t=32.8s vi=0m/s Find: d=?? d=vi*t+0.5*a*t2 d=(0m/s)*(32.8s)+0.5*(3.20m/s2)*(32.8s)2 d=1720m ReturntoProblem1   Given: d=110m t=5.21s vi=0m/s Find: a=?? d=vi*t+0.5*a*t2 110m=(0m/s)*(5.21s)+0.5*(a)*(5.21s)2 110m=(13.57s2)*a a=(110m)/(13.57s2) a=8.10m/s2 ReturntoProblem2   Given: a=-9.8m t=2.6s vi=0m/s Find: d=?? vf=?? d=vi*t+0.5*a*t2 d=(0m/s)*(2.60s)+0.5*(-9.8m/s2)*(2.60s)2 d=-33.1m(-indicatesdirection) vf=vi+a*t vf=0+(-9.8m/s2)*(2.60s) vf=-25.5m/s(-indicatesdirection) ReturntoProblem3   Given: vi=18.5m/s vf=46.1m/s t=2.47s Find: d=?? a=?? a=(Deltav)/t a=(46.1m/s-18.5m/s)/(2.47s) a=11.2m/s2 d=vi*t+0.5*a*t2 d=(18.5m/s)*(2.47s)+0.5*(11.2m/s2)*(2.47s)2 d=45.7m+34.1m d=79.8m (Note:thedcanalsobecalculatedusingtheequationvf2=vi2+2*a*d) ReturntoProblem4   Given: vi=0m/s d=-1.40m a=-1.67m/s2 Find: t=?? d=vi*t+0.5*a*t2 -1.40m=(0m/s)*(t)+0.5*(-1.67m/s2)*(t)2 -1.40m=0+(-0.835m/s2)*(t)2 (-1.40m)/(-0.835m/s2)=t2 1.68s2=t2 t=1.29s ReturntoProblem5   Given: vi=0m/s vf=444m/s t=1.83s Find: a=?? d=?? a=(Deltav)/t a=(444m/s-0m/s)/(1.83s) a=243m/s2 d=vi*t+0.5*a*t2 d=(0m/s)*(1.83s)+0.5*(243m/s2)*(1.83s)2 d=0m+406m d=406m (Note:thedcanalsobecalculatedusingtheequationvf2=vi2+2*a*d) ReturntoProblem6       Given: vi=0m/s vf=7.10m/s d=35.4m Find: a=?? vf2=vi2+2*a*d (7.10m/s)2=(0m/s)2+2*(a)*(35.4m) 50.4m2/s2=(0m/s)2+(70.8m)*a (50.4m2/s2)/(70.8m)=a a=0.712m/s2 ReturntoProblem7   Given: vi=0m/s vf=65m/s a=3m/s2 Find: d=?? vf2=vi2+2*a*d (65m/s)2=(0m/s)2+2*(3m/s2)*d 4225m2/s2=(0m/s)2+(6m/s2)*d (4225m2/s2)/(6m/s2)=d d=704m ReturntoProblem8   Given: vi=22.4m/s vf=0m/s t=2.55s Find: d=?? d=(vi+vf)/2*t d=(22.4m/s+0m/s)/2*2.55s d=(11.2m/s)*2.55s d=28.6m ReturntoProblem9   Given: a=-9.8m/s2 vf=0m/s d=2.62m Find: vi=?? vf2=vi2+2*a*d (0m/s)2=vi2+2*(-9.8m/s2)*(2.62m) 0m2/s2=vi2-51.35m2/s2 51.35m2/s2=vi2 vi=7.17m/s ReturntoProblem10   Given: a=-9.8m/s2 vf=0m/s d=1.29m Find: vi=?? t=?? vf2=vi2+2*a*d (0m/s)2=vi2+2*(-9.8m/s2)*(1.29m) 0m2/s2=vi2-25.28m2/s2 25.28m2/s2=vi2 vi=5.03m/s Tofindhangtime,findthetimetothepeakandthendoubleit. vf=vi+a*t 0m/s=5.03m/s+(-9.8m/s2)*tup -5.03m/s=(-9.8m/s2)*tup (-5.03m/s)/(-9.8m/s2)=tup tup=0.513s hangtime=1.03s ReturntoProblem11   Given: vi=0m/s vf=521m/s d=0.840m Find: a=?? vf2=vi2+2*a*d (521m/s)2=(0m/s)2+2*(a)*(0.840m) 271441m2/s2=(0m/s)2+(1.68m)*a (271441m2/s2)/(1.68m)=a a=1.62*105m/s2 ReturntoProblem12   Given: a=-9.8m/s2 vf=0m/s t=3.13s Find: d=?? (NOTE:thetimerequiredtomovetothepeakofthetrajectoryisone-halfthetotalhangtime-3.125s.)   Firstuse: vf =vi +a*t 0m/s=vi +(-9.8 m/s2)*(3.13s) 0m/s=vi -30.7m/s vi =30.7m/s (30.674m/s) Nowuse: vf2 =vi2 +2*a*d (0m/s)2 =(30.7m/s)2 +2*(-9.8 m/s2)*(d) 0m2/s2 =(940m2/s2)+(-19.6 m/s2)*d -940 m2/s2 =(-19.6 m/s2)*d (-940 m2/s2)/(-19.6 m/s2)=d d=48.0m ReturntoProblem13   Given: vi=0m/s d=-370m a=-9.8m/s2 Find: t=?? d=vi*t+0.5*a*t2 -370m=(0m/s)*(t)+0.5*(-9.8m/s2)*(t)2 -370m=0+(-4.9m/s2)*(t)2 (-370m)/(-4.9m/s2)=t2 75.5s2=t2 t=8.69s ReturntoProblem14     Given: vi=367m/s vf=0m/s d=0.0621m Find: a=?? vf2=vi2+2*a*d (0m/s)2=(367m/s)2+2*(a)*(0.0621m) 0m2/s2=(134689m2/s2)+(0.1242m)*a -134689m2/s2=(0.1242m)*a (-134689m2/s2)/(0.1242m)=a a=-1.08*106m/s2 (The-signindicatesthatthebulletsloweddown.) ReturntoProblem15   Given: a=-9.8m/s2 t=3.41s vi=0m/s Find: d=?? d=vi*t+0.5*a*t2 d=(0m/s)*(3.41s)+0.5*(-9.8m/s2)*(3.41s)2 d=0m+0.5*(-9.8m/s2)*(11.63s2) d=-57.0m (NOTE:the-signindicatesdirection) ReturntoProblem16   Given: a=-3.90m/s2 vf=0m/s d=290m Find: vi=?? vf2=vi2+2*a*d (0m/s)2=vi2+2*(-3.90m/s2)*(290m) 0m2/s2=vi2-2262m2/s2 2262m2/s2=vi2 vi=47.6m/s ReturntoProblem17   Given: vi=0m/s vf=88.3m/s d=1365m Find: a=?? t=?? vf2=vi2+2*a*d (88.3m/s)2=(0m/s)2+2*(a)*(1365m) 7797m2/s2=(0m2/s2)+(2730m)*a 7797m2/s2=(2730m)*a (7797m2/s2)/(2730m)=a a=2.86m/s2 vf=vi+a*t 88.3m/s=0m/s+(2.86m/s2)*t (88.3m/s)/(2.86m/s2)=t t=30.8s ReturntoProblem18   Given: vi=0m/s vf=112m/s d=398m Find: a=?? vf2=vi2+2*a*d (112m/s)2=(0m/s)2+2*(a)*(398m) 12544m2/s2=0m2/s2+(796m)*a 12544m2/s2=(796m)*a (12544m2/s2)/(796m)=a a=15.8m/s2 ReturntoProblem19   Given: a=-9.8m/s2 vf=0m/s d=91.5m Find: vi=?? t=?? First,findspeedinunitsofm/s: vf2=vi2+2*a*d (0m/s)2=vi2+2*(-9.8m/s2)*(91.5m) 0m2/s2=vi2-1793m2/s2 1793m2/s2=vi2 vi=42.3m/s Nowconvertfromm/stomi/hr: vi=42.3m/s*(2.23mi/hr)/(1m/s) vi=94.4mi/hr ReturntoProblem20   NextSection: KinematicEquationsandKinematicGraphs FollowUs PhysicsTutorial»1-DKinematics»Lesson6-DescribingMotionwithEquations»SampleProblemsandSolutions