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CBSEClass9Science

ImportantQuestions

Chapter8

Motion

3MarksQuestions

1.Anobjecthasmovedthroughadistance.Canithavezerodisplacement?Ifyes,

supportyouranswerwithanexample.

Ans.Yes,ifanobjecthasmovedthroughadistanceitcanhavezerodisplacementbecause

displacementofanobjectistheactualchangeinitspositionwhenitmovesfromone

positiontotheother.SoifanobjecttravelsfrompointAtoBandthenreturnsbacktopoint

Aagain,thetotaldisplacementiszero.

2.Afarmermovesalongtheboundaryofasquarefieldofside10min40s.Whatwillbe

themagnitudeofdisplacementofthefarmerattheendof2minutes20seconds?

Ans.Distancecoveredbyfarmerin40seconds=

Speedofthefarmer=distance/time=40m/40s=1m/s.

Totaltimegiveninthequestion=2min20seconds=60+60+20=140seconds

Sincehecompletes1roundofthefieldin40secondssoinhewillcomplete3roundsin120

seconds(2mins)or120mdistanceiscoveredin2minutes.Inanother20secondswillcover

another20msototaldistancecoveredin2min20sec=120+20=140m.

Displacement = 200= (asperdiagram) =14.14m.

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3.Atrainstartingfromarailwaystationandmovingwithuniformaccelerationattains

aspeed40km in10minutes.Finditsacceleration.

Ans.Sincethetrainstartsfromrest(railwaystation)=u=zero

Finalvelocityoftrain=v=

=

=

time(t)=10min= =600seconds

Sincea=(v–u)/t

4.Whatcanyousayaboutthemotionofanobjectwhosedistance-timegraphisa

straight-lineparalleltothetimeaxis?

Ans.Iftheobject’sdistancetimegraphisastraightlineparalleltothetimeaxisindicates

thatwithincreasingtimethedistanceofthatobjectisnotincreasinghencetheobjectisat

resti.e.notmoving.

5.Whatcanyousayaboutthemotionofanobjectifitsspeedtimegraphisastraight

lineparalleltothetimeaxis?

Ans.Suchagraphindicatesthattheobjectistravellingwithuniformvelocity.



6.Atrainistravellingataspeedof .Brakesareappliedsoastoproducea

uniformaccelerationof .Findhowfarthetrainwillgobeforeitisbrought

torest.

Ans.u= =

,v=0(trainisbroughttorest)

v=u+at=25+(-0.5)xt

0=25–0.5x

0.5t=25,ort=25/0.5=50seconds

=

=1250–625=625m

7.Astoneisthrowninaverticallyupwarddirectionwithavelocityof .Ifthe

accelerationofthestoneduringitsmotionis inthedownwarddirection,

whatwillbetheheightattainedbythestoneandhowmuchtimewillittaketoreach

there?



Ans.

v=0(sinceatmaximumheightitsvelocitywillbezero)

v=u+at=

0=5–10t

10t=5,or,t=5/10=0.5second.

=

=2.5–1.25=1.25m

8.Derivethesecondequationofmotion graphically?

Ans.letattimeT=0bodymoveswithinitialvelocityuandattime‘t’bodyhasfinalvelocity

‘v’anduntime‘t’itcoversadistance’s.

AC=v,AB=u,OA=t,DB=OA=t,BC=AC-AB=V-u

Areaunderav-tcurvegivesdisplacementso,

S=Areaof DBC+AreaofrectangleOABD (i)

Areaof DBC= Base Height DB BC

= t (v-u) (ii)

AreaofrectangleOABD=length Breadth



=OA BA

=t u (iii)

S=ut+ t (v-u)

S=ut+ t at( useV-u=at)

S=ut+ at2

9.Acarmovingwithacertainvelocitycomestoahaltiftheretardationwas ,

findtheinitialvelocityofthecar?

Ans.V=0(comestorest)V=finalvelocity

S=62.5m

(retardation)

U=?

From3rdequationofmotion,

=2 (-5) 62.5

=-10 62.5

,

u= [u=25m/s]

10.TwocarsAandBaremovingalonginastraightline.CarAismovingataspeedof

80KMphwhilecarBismovingataspeed50KMphinthesamedirection,findthe



magnitudeanddirectionof

(a)tivevtherelativeofcarAwithrespecttoB

TherelativevelocityofcarBwithrespecttoA.

Ans.(a)VelocityofcarA=80KMph

VelocityofCarB=-50kmph

(-vesignindicatesthatCarBismovinginoppositedirectiontoCarA)

RelativevelocityofcarAwithrespecttoB

=velocityofcarA+(-velocityofcarB)

=80+(-(-50))

=80+50

=+130KMph

+130KMphshowsthatforapersonincarB,carAwillappeartomoveinthesamedirection

withspeedofsumoftheirindividualspeed.

(b)RelativevelocityofcarBwithrespecttoA

=velocityofcarB+(-velocityofcarA)

=-50+(-80)

=-130kmph

ItshowsthatcarBwillappeartomovewith130kmphinoppositedirectiontocarA

11.Aballstartsfromrestandrollsdown16mdownaninclinedplanein4s.

(a)Whatistheaccelerationoftheball?

(b)Whatisthevelocityoftheballatthebottomoftheincline?



Ans.u=initialvelocity=0(bodystartsfromrest)

S=distance=16m

T=time=4s

(i)From,

16= a 16

(ii)From,v=u+at

v=0+2 4

[v=8m/s]

12.TwoboysAandB,travelalongthesamepath.Thedisplacement–timegraphfor

theirjourneyisgiveninthefollowingfigure.

(a)HowfardowntheroadhasBtravelledwhenAstartsthejourney?



(b)Withoutcalculation,thespeed,statewhoistravelingfasterAorB?

(c)WhatisthespeedofA?

(d)WhatisthespeedofB?

(e)ArethespeedofAandBuniform?

(f)WhatdosepointXonthegraphrepresent?

(g)WhatisthespeedofapproachofAtowardsB?

WhatisthespeedofseparationofAfromB?

Ans.(a)WhenAstartshisjourneyat4sec,Bhasalreadycoveredadistanceof857m

(b)AtravelsfasterthanBbecauseAstartshisjourneylatebutcrossesBandcoversmore

distancethenBinthesametimeasB

(c)SpeedofA=

Letatt=12min,distancecovered=3500m

(d)SpeedofB=

(e)SpeedofapproachofAtowardsB=375m/min-214m/min

=161m/min

(f)SpeedofseparationofAfromB=161m/min.

13.Abodyisdroppedfromaheightof320m.Theaccelerationduetothegravityis



?

(a)Howlongdoesittaketoreachtheground?

(b)Whatisthevelocitywithwhichitwillstriketheground?

Ans.Height=h

Distance=s=320m

Accelerationduetogravity=

Initialvelocity=u=0

(a)froms=ut+

(b)

14.Derivethirdequationofmotion numerically?

Ans.Weknow;



………(i)

…….(ii)

Where,v=finalvelocity

u=initialvelocity

a=acceleration

t=time

s=distance

Fromequation(i)t=

Putthevalueoftinequation(ii)



15.Thevelocitytimegraphofrunnerisgiveninthegraph.

(a)Whatisthetotaldistancecoveredbytherunnerin16s?

(b)Whatistheaccelerationoftherunneratt=11s?

Ans.(a)Weknowthatareaunderv-tgraphgivesdisplacement:

So,Area=distance=s=areaoftriangle+areaofrectangle

Areaoftriangle=

=

=30m

Areaofrectangle=length breadth

=(16-6) 10

=10 10

=100m

Totalarea=180m

Totaldistance=180m

(b)Sinceatt=11sec,particlestravelswithuniformvelocityso,thereisnochangeinvelocity

henceacceleration=zero.



16.Aboythrowsastoneupwardwithavelocityof60m/s.

(a)Howlongwillittaketoreachthemaximumheight ?

(b)Whatisthemaximumheightreachedbytheball?

(c)Howlongwillittaketoreachtheground?

Ans.u=60m/s; ;v=0

(a)Thetimetoreachmaximumheightis;

(b)Themaximumheightis;

(c)Thetimetoreachtopisequaltotimetakentoreachbacktoground.Thus,timetoreach

thegroundafterreachingtopis6sOrthetimetoreachthegroundafterthrowingis6+6

=12s.

17.Thedisplacementxofaparticleinmetersalongthex-axiswithtime‘t’inseconds

accordingtotheequation-

(a)drawagraphifxversustfort=0andt=5sec

(b)Whatisthedisplacementcomeoutoftheparticlesinitially?

(c)Whatisslopeofthegraphobtained?



Ans.X=20m+(12)t

(i)Att=0

X=20+12 0=12m

(ii)Att=1

X=20+12=32m

(iii)Att=2

X=20+24=44m

(iv)Att=5

X=20+12 5=72m

(a)

(b)AtT=0(initially)

Displacement=20m.

(c)Slope=

18.Thevelocityofabodyinmotionisrecordedeverysecondasshown-



calculatethe–

(a)Acceleration

(b)distancetravelledanddrawthegraph.

Ans.(a)Acceleration=slopeofthevelocitytimegraph

a=

(b)Distance

=600-300=300m

(c)

19.Drawthegraphforuniformretardation–

(a)position–timegraph

(b)velocity–time

(c)Acceleration-time

Ans.(1)Position–time



(2)Velocity–time

(3)Acceleration-time

20.Thedisplacement–timegraphforabodyisgiven.Statewhetherthevelocityand

accelerationofthebodyintheregionBC,CD,DEandEFarepositive,negativeorZero.



Ans.(i)ForAB,thecurveisupwardstoppingi.e.slopeisincreasingsovelocityispositive

andremainssameso,V=+vebuta=0

(ii)ForBC,curvehasstillhas+veslopeso,V=+vebutvelocityisdecreasingwrttimeso,

a=negative

(iii)ForCD,bothvelocityandaccelerationareZerobecauseslopeisZero.

(iv)ForDE,velocityisthe(visincreasingwrttime)andsoisaccelerationis+ve.

(v)ForEF,velocityis+ve(positiveslopeofx-tgraph)butaccelerationisZerobecause

velocityremainssomewithtime.

21.Derivethethirdequationofmotion asgraphically?

Ans.Letattimet=0,bodymoveswithinitialvelocityuandtimeat‘t’hasfinalvelocity‘v’

andintime‘t’coversadistance‘s’



Areaunderv-tgraphgivesdisplacement

S=Areaof DBC+AreaofrectangleOABD

S=

Now,v-u=at

Putthevalueof‘t’inequation(i)

thirdequationofmotion


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