2 t” The displacement of a body S = a where g is constant, The velocity of the body at any time ‘tis 2gt 2et a @ fr Fos wr Fre = BE wet g feria 2 fret eer 'e ac ag a aT — 2gt 2gt yn = 2) =8t a) 3 Q) 5 Ans. 2 Sol. y= ds _ 8, M dt 3 a) A body can't have (1) A constant speed and varying velocity (3) A constant velocity and varying speed wag fataatwa sate (1) Pagar a oat ar (3) fara a vada art Ans. 3 Sol. Velocity has magnitude and direction but speed = [Velocity] a acceleration = “7 GB) gv3 (4) gts (3) gu3 (4) gus (2) Anacceleration and a constant speed (4) Non-zero speed and zero acceleration Q) wera Faas 4 saa yar The velocity of a body depends on time according to equation, v= 0.1 1? + 10t +20, the body has motion — (1). Uniform acceleration 3) Non-uniform acceleration ‘Fret fos 3 ar Ht attr ara a (2). Uniform retardation (4) Zero acceleration V=0.1 t+ L0t+ 20% sagan steattia Stet a at fos rahe e — () Ga aoT Q) Sear aT Ans. 3 Sol. .-W 2+10 dt (2) (Ran Aer ace5. A car travels from A to B ata speed of 10 km/h and then B to A at a speed of 20 km/h and then again ‘Ato Bata speed of 40 km/h The average speed of the car for the whole journey is WHKARB l0kmhAMATTABSA 20 kv A AAT ATG: A AB 40 keh Hl aT ae a, a Sa 120 aw =] koh (2) 20 km/h (3) 35 km/h @ Bam Ans. 1 Sol ya. ___3d 3x40 Auge eg tigi eta or 20 * 0 = 2h 7 A stone is thrown vertically upward with an initial speed U from the top of a tower and it reaches the ground with a speed 4U then the height of the tower is = fedftoen ar tian a Seater oe A oi U orf 1a thas ae a AU Ta IAT, aH A Seas es 2 o 155 @ " 2 32 (3) os a4 Ans. 1 Sol. V>=U? +2as w+ 2(+g)(h) _ 15U? = ‘Two particles held at different height 8 m and 12 m above the ground are allowed to fall from rest. The ratio of their speeds on reaching the ground is : alia & erste Sarg 8 ft. 12 oH, oe fear at ait at cada Be @ fares er Si Sa aT aT h 2 4 a) 2 4) = Os OsA patticle stats from rest and moves along a straight line with constant acceleration. The graph of velocity v with displacement s is - a) vy @Q vy @B) vw ae @) S o a aon Feeney B rer Shane Prat caer A ee ear 3 sear afer ase V1 Ser v ar Ree s 3 eT AT B — y y v y w 2) Oe @ os s os s Ans. 2 Sol. V>=U? +2as; V?=0+2as Vas In the given velocity-time graph, magnitude of average acceleration of body is— 3040 time Ges) (1) 0.5 mis? (2) Amis? (3) 2m/s* (4) 15 mvs? aay areas site aeq & sik career ar aheaoT B— (2) mis? (1) 0.5 mis® (3) 2mis* (4) 1.5 mis? Ans. 1% 10. Position-time graph of a body is given. Average velocity of body for time interval t = 0 to t= 35 second is position time (ee) () sms ) 2ms 6 3) Says @) pms (4) Imis ‘evel eg ar Referer a ear rar &1 err siarTet t = 0B t= 35 awe ae athe a B— x elt a0. 10} 0 10 pS () Sous Q) =ms @) 2ms (4) Imis A balloon is moving upward with velocity 20 m/s. It release a stone which comes down to the ground in 15 sec. The height of the balloon from the ground at the moment when the stone was dropped is Uae Oa 20 Hiv, TB Seite ST Soa A GATT Se eT 1 eA HH Te Frc SHAT 8, ST eit Ie 15 das vara 8 ee fr Ss oy, eat ot aot a Sag ae ah? (1) 620m (2) 825m @) 1035 m (4) 1215 m Ams. 2 Sol. s = (_20)(15) + 7 caoysy? S=~300 + 1125 = 825 m11, The displacement of a moving particle is proportional to square of time taken, for this particle (v > velocity, a acceleration) (A) v=constant — (B) v = variable (©)a=constant — (D) a= variable Which one is correct : () A&C @) B&D @) A&D (@) BEC em fer oT TRY a Pereears ere eee a TT ae TATA Vl eT HHT eT, Wo F,aS aM) (A) v= aa (B) v= Shadt (Qa=fat (D)a=ahadt wel ae ee () AwiC @ BwD @) A@D (@) BC Ans. 4 Sol. s oc? k— Racias s=kt va2kt Svat a=2k = a~constant 12, The coordinates of a particle is x = at® and y = Bt° then the rate of change of acceleration with respect to time is (a. and Bare constant) ) 6 fab +p 2) ore +p? @) Jo2+p? (3 fab +p? ‘ret aor Réatiss x = ata y = Beal a area ater caer Ht afters al ax &- (a. Us B Paci 8) 2) 6 ye+p? @ 3 erp aay) _ 6B13. 14, A car moving with a speed of 40 km/h can be stopped by brakes after at least 4m. If it has a speed of 100 kawh, the minimum stopping distance is- 40 keonfh ret Sere are a hes TAK TA 4 ahee A Rast ST AT BI ake ae we 100 knv/h A are Bet ater at ea A age ate (1) 16m 2) 20m @) 25m (4) 35m Ans. 3 ee ae Si. (vi) S. Vz oe oy 046 S2 100 1 4 _ 400 s,-—t = 100 Reig eciet Atul iseleased from height 2h. it takes T seconds to reach ground. Whats its hight from ground at seconds vara an viene air ara tae oa amagaAH deed Tas eA Te q) wa Q) 3W4 @) 1sw8 (4) 17W8 Ans.3 Sol. t=0 u=o015. If a particle is moving with uniform velocn,, then INSHWAPage-43] I. Average speed = |Average velocity! Il. Instantaneous speed = Instantaneous velocity! IIL. Instantaneous speed = Average speed 1) Only Tis correct 2) Only His correct 3) Only I and Il are correct 4) All are correct Sol. (4): Conceptual Velocity-time graph for a particle is shown in| 16. figure. Starting from f = 0, at what instant t, average acceleration is zero between 0 to f? (mis) UNSERE Page- o1esas5e Te reed) 1s 2)35s 3) 63s 4) 738 Sol. (3): Conceptual If velocity v = 62 + 2t + 3 m/s, then find the, 17. average velocity from f= 0 to t= 15. INGEN Page-47 1) U1 m/s 2) 8 m/s 3) 6 m/s 4) 7 m/s Sol. (3): 18. A train starts to move from rest at t = 0. Its acceleration is a = at + Bt?, here a and B are constants. The velocity of the train at t=1sis (INGER Page-45] @ | Dare ao | 8)0 4a 1 2p 4}So1. «2: ¥= fea (obs05) ie 19. A particle moves a distance xin time t according to equation x = (f+ 5). The accelerationof the particle is proportional to [SNSHR Page-45] 1) (velocity)? 2) (velocity)#2 3) (distance)? 4) (distancey? 1 ace Soh Qh xe Fa Acceleration, wo = = a= (velocity)? 7p ow | 20. Aconstantaccelerationactsfor20sona 1 particle, starting from rest. If the particle covers distance d, in the first 10 s and distance d,inthe next 10s,then —(SN@HMHPage-50] 4 | Daz 2) d, 3) dy =3d, 4) d, Sol. (3): d= facior= ; xax 100 +a?) ayo aca 5 a(ior= $ax300= 34, 21, A particlemoves along ; x-axis in such a way that its a-coordinate varies with time according to the equation x = 8 - 4¢ + 6F. The distance covered by particle between t=0 to t= 2 sec is 8 2) $m 4) $mSol. (4): ven sccooidinate varies as x= velocity» = & = ~4+12¢ and when velocity become 1 zero v $ second Which ~4412t we get ‘means particle tums att = } seconds —_ Att = 0 x, = B~ 4(0)+ 6(0)? = 8m sarod) - jeain8-a(3)re(3) -o0 2 ete 22. Atx = 0, velocity of the particle is 2 m/s. Find the velocity of the particle when acceleration becomes zero. INGE Page-t6" ols 1) V248 m/s 2) 282 m/s 3) 32 m/s 4) V184 m/s Sol. (1): Acceleration becomes zero at 2+1.2=3.2m % 4amis? omit Slope = tan = x 3-x 9 12-4r=6r = 12= 10r x= 1.2m Now, a= ae 2 ‘Area of graph from 0 to 3.2 m =: pm ing t 72248 2 v=V248 ms -4t +6t? so its |. 123. Displacement-time graph of an object moving; along x-axis is given. Select the correct statement regarding it. (In OA region, both v and a are positive. (1) In AB region, both v and a are zero. (111)In region BCD, both » and a are negative. (IV)In region DE, v is negative and acceleration is positive. Page-44, 45] 1) @ and (iil) 2) @ and @V) 3) (M1) and (IV) 4) @ and (1) Jol. (2): In region OA, slope is positive and increasing. In region DE, slope is negative. 24. Assertion: : Magnitude of average velocity is equal to average speed if velocity is uniform. Reason : If velocity is uniform, then there is no change in the direction of motion. (INGHRHPage-12] 1) If both assertion and reason are true and the reasonis the correctexplanation of the assertion. 2) Ifboth assertion and reason are true, but reason is not the correct explanation of the assertion. 3) If assertion is true, but reason is false. 4) Ifassertion is false, but reason is true. Sol. (1): Conceptual 25, Initially, busA is 112.5 m ahead of bus B. Both start moving at time t= 0 in the same direction along a straight line. The velocity-time graph of the two cars is shown in the figure. The time when the bus B will catch the bus A, will be (NHRD Page-14} v bus B 20 ms| bus A 5" > 1) 45s 2) 2255 3) 325 4) 18sSol. (1): Given; x, = 207 fuse Bost asad * 120 a “4 Bus 20's} Bus A vol Tome xt 25 = O+dxix? 201+ 112.5= 0.57 0.57 —201- 1125=0 P 401-225 =0 1245s 3. The velocity of a particle is given by v= ¥i80—16x mis. Its acceleration will be (NHRD Page-48] 1) -8 m/s? 2) 5 m/s* 3) -9 m/s? 4) 10 m/s? a1. (1): Velocity of the particle is given as v= (180-16x m/s ‘Squaring on both sides v? = 180-16x Now compare with v? =u? + 2as = 2a=-16 = a=-8m/s?27. An elevator is ascending at a constant speed 10m/s. A passenger drops a coin. What will be the acceleration of coin as seen by the passenger. (g=10mvs") () 10m downward 2) Zero @) 10m/s* Upward @) Sis? downward oh ferve Faere ret 10m/s sea a Bite fetter Bl a fae et A Ae Perse Frere 8 at Se ae TT faeharame- (g= 10mis’) (1) 10 mis? Qw 3) 10m/s* sat 4) Sms? Ans. 1 Sol. dooinpusenger = Bec — @pasenger =CH)-0= 83) 28. A body A is moving with 10mvs and B is moving with 5 m/s in same direction, A is 200 m behind B. Time taken by Ato meet B is 10m/s Sm/s 200 m— FEA, 10m/s & Sk B, 5 m/s Baars fee H fete V1 PTET A, BS 200 + HS a ah A arr Ba eet arate (1) 40sec: (2) 20sec @) B sco (4) 15sec Ans. 1 Sol. ,_ Sie _ _200 29. Four particles situated at the comers of square of side 10 m move at a constant speed of 2m/s. Each particle maintains a direction towards the next particle in. succession. Calculate the time when they meet : 10 m ar a art Se aaa ae Pera a OT a AA eT 2m/s B a WER ferret & fee Meas aT seerhh Fee ATT Som area or A Site CaaT AI aT soit Bt TG Hers Hee aa tT; (1) Ssec 2) 10sec @) 5v3 see ( ~%& sec v2 Ans. 1 Sol. ,_ 10 a t=Ssec30. Graph shown in figure are the position time graph for two children going home from school, their relative velocity - : Cy CQ ol t (1) First increases and then decreases (2) Fist decreases and then increases (3) is zero (4) is non zero constant we Reger ord geared feat rer are er wafer, at greater () vest ae are Haar 2) vee sear are Farge @) erent (4) seer Feri eet Ans. 4 Sol. V,=constant (of Cy) V2 =constant (of C;) (v = =) Vi>V> Relative velocity = Vj — V2 = constant (Non zero)31.A bird flying witha speed of 10 m/salonga _ 1 vector (-31+4j). If initial position of bird is (22 m, 10 m, 5 m), then find the x-coordinate (in m) after the 2s. [MGB Page-47) yi 2)0 3)2 4) None of these ( x10 m/s 3 6t-48))(2)+L(022 61 +8))(2)+4(0)2 \isplacement after 2 secoordinate after 2s 32, TwoRajdhanitrains,each50mlongare _ > travelling in opposite directions with a velocity of 10 m/s and 15 m/s. The time of crossing is UN@HRA Page-51] 12s 2)4s 3) Bs 4) Bs Sol, (2): ,| 38. A boy standing at the top of a20mhigh tower drops a stone. Assuming g = 10 m/s’, the velocity with which it hits the ground is (NSH Page-s8] 1) 10.0 m/s 2) 20.0 m/s 3) 40.0 m/s 4) 5.0 m/s Sol. (2): v=figh y= V2xT0%20 = 20 mis 34, Abodyisfallingfromheight t h. It takes t seconds to reach the ground. Calculate the time taken by it to cover the first /16 height, (ING Page-50} 1) ts 2) t/2 3)t/4 4)t/8 Sol. (3): Conceptual 35, Water drops come out at regular interval from 1 the water tank at the roof. When the first drop hits ground, third drop comes out of the tank. The height of the second drop from the ground is [Height of tank is Hy]. (IN@HIH Page-49] 4 H, 1) 5Mo ao 3H, A, 3) a 4) 7Sol. (3): Time taken by first drop to come to ground, Drop falls at regular interval, therefore, 1 distance covered by second drop in > time is Ho = 3Ho 404 «height from ground = Hy ~- au36. A particle starts from rest with uniform acceleration a. Its velocity after n seconds is v. The displacement of the particle in the last three seconds is— frat 3 Rad a7 a Beales ge aor at n Sarvs Gee AT y a a Safer 3 Baas F aor aT Pea eg — a) av -% @ @ <= q@ 300 Ans. 1 Sol. V=0+an 37. OT ane 4 Baaws Ha ae (1) 50m Q) 45m 3) 40m (4) 80m Ans. 4 Sol. Distance = |Aj|+|Ao| velocity preva Distance = [4 «25+ 1)=20] + [5 15«<0] =35+45 = 80 meter38. 39. Water drops fall at regular intervals from a tap which is 10 m above the ground. The fifth drop is leaving the tap at the instant, the first drop touches the ground. How far above the ground is the third drop at that instant. (g = 10 m/s?) a Bm 2 Bm @) 5m (4) 6m art ot oa sete & 10 tex Sang we Reera et B area Seeeret F firech V1 via fe sa wes ge wt eet 8, er weet fe sata ag wah Bich Ui et aT Gla eg Goes a Peer Sars SrA? (g = 10 m/s) @ Bm ® 2m @ 5m @) 6m Ans. 1 Sol. x+3x+5x+7x=10 nal [4 Lye height of 3rd drop To ae = 12x sos | =12x2=2-75m m] 8 2 ser A body freely falling from rest has a velocity v after it falls through distance 4 m. The distance it has to fall down further in meter for its velocity to become triple is () 16m Q) 322m @) 48m (4) 64m ip ae ia 4 oh, ah ee ada wT tee 8 a ge Shere eae aT Es A v A aT I AT ata TAT ATS fa ‘att gt site rear eo — @) 16m Q) 32m G) 48m (4) 64m 2 i : Sya36m= re Ve 32m The position of a particle is given by x = (t — 2) x in meter t in second. The total distance travelled by the particle between t = 0 to t= 1 sec is : fret aor aft Rafe x = C) aweias (xtieta dans 8 t=08 t= 1 dave h che aor ar ach as Ga Ae: (1) Om @ 2m @) Im @ $mat 41, Ifa body loses 4° of its velocity after penetrating 3 cm in a wooden block, then how much will it penetrate more before coming to rest. ) Sem @ ten @) Sem (@) 6em aft eg ewer nH 3 cm Se a wc meer 3 eet RA ya ee err a eo aw Bom 2 Fem @) Yom (4) 6om 2a(3) 2a(s) 3-48 5 5- bem 3 16 a Further distance = 48-3 = Lem 42, From a tower of height H, a particle is thrown up with speed 5 m/s. The time taken to hit the ground is 3 times that of time to reach by it to highest point. Height of the tower H is : () H=105m (Q) H=5.25m 3) H=3.75m (4) H=7.25m H Soar St Hae Gy, eam aor aT 5S Ha, B Soe oA hear are ar aT Seta ae Prey er ae, Sea as ae eh 3s eer a 3 aT Bae AY See Let () H=105m Q) H=5.25m @) H=3.75m (4) H=7.25m Ans.343. A stone is dropped from a certain height which can reach the ground in 10 sec. It is stopped momentarily after 6 sec of its fall and then it is again released. The total time taken by stone to reach the ground will be - (1) 8 see, Q) 14sec. 3) 12sec. (4) 20sec. ifs Sead vee Pre Tee ae 10 5 HAT 1 AAS ee aE wa 6 5 Tea flrs BUS Fe | aT ete a a Here AT wee TA get (1) 85sec. (2) 14sec. (3) 12sec. (4) 20sec.45. Ans, 2 Sol. oe H= + g(o) =1 osteo H= 78 +580) > t=8 sec. =6+8=14 sec. From the top ofa tower, a stone is dropped. Ifit covers 25 min the last second of its motion, then height of tower is (g= 10 m/s") ( 45m (2) 15m (3) 50m @ 40m eR AS aA Oe TAT HT TAT, Ae ae ae aS H 25 He LTT ATS a HAR AT STS (= 10 mis?) () 45m Q) 15m @) 50m (4) 40m Ans. 1 Sol. H= 3 gt=s? (i) H-25= + g(t-1=sP+5—10t ...(ii) SP -25=5P+5—10t 10t= 30 sec. H=5%9=45 m Match the column — Cofumn-t ‘Colum-tt (@ instantaneous velocity | (®) | vector quantity (b) | Instantaneous speed (Q)_| Scalar quantity (© | Avene velocity (| depends only on initial and nel position (© | Average speed ()_| ts magnitude ean increase with time () @P,S 2) @P,R @) @P,S (4) @P.R )QS )QR O@R (b)P,S ()P,R,S ©P,R ©P,QS (©P,QR @MQs @s @MQR @s feat gata srg — eH SPE (a) | areafirs a P| aeerofr (b) | areerftres ater Q | sferakr © | seat (| Rab sara ota Raat Paice @ | steerer (8) | seer orn aaa arerarg were (@ @P,s 2) @P.R GB) @P,S @) @P,R 0)Q8 ()QR ()QR @)P,S OPRS PR OP,.QS ©P,QR @@s @Qs @MQR @Qs Ans. 1 Sol. As per theroy.