motion in straight line lecture notes, Lecture notes of Physics

this is a explained notes on motion in straight line, class 11 cbse ,taken by rajwant, also in hindi ,but fully in english

Typology: Lecture notes

2024/2025

Available from 06/14/2025

parikshith-narayan
parikshith-narayan 🇮🇳

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Motion in a Straight line
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Motion in a Straight line

& FOR JEE ASPIRANTS Motion in a Straight line @ Topics ,, .../ Plane (ay) Distance , Displacement, speed , velocity Vectors => aud A/Sub. Pro i (2) Acceleration ject q Relettve -mIb220, ) Kinematics equation (4) Motion under gravity — FOR NOTES & DPP CHECK DESCRIPTION — aaa Y ia Rosh ‘ a) — Pahoa ten wie fosttton vam rt Fine A Observer. is Vang yt time ut lot Obcerver, Obrewer., we Position, = 2. = Constant Ra 3) SS |x23 Bee to xed X= Vint \ mh Mubio, Note: Ye content bilkul sufficient hai ( MAINS/ Advanced) For class 11,12,13 Definition TRad Poth Covered, awe age genta! =PChrargs Py, Postttan . Scalar / Vector Scag Vector. Unit / Dimension an Li ait be Positive/Negative / Xero ,(+) ENecO. C+ve) Decrease/ Increase / Sher Cunstant or Inevanne. O Mer, deer. Jab chabe ht nah? Atanderd Result = Arel Radtans / eR Find distance and displacement from A to B. fstan ce. Afstance = Are le ng th +RO., Rsinl Oy) RSin(O/, re a. Se Atstlocerert =lQrce7a lA = .’B Distance = Bao t 400 i 2RSin(Q.) “-\26D = (eee) Mblacemert = (Sedmn, 3 Alec: YRStH/ lap z SAR Sib = RB. Find displacement of A after half A person moves 20m South 20m West and rotation A 20V2m towards North East. Find distance ‘ee2eo2®® 20m fs} = 2oxred Bla ‘ diss = 0. Ye Posttfon =m = PC) om RD HSH HESS XA , Time Ke Saath Sets We Ms wr pies 420 *=2 0 ° * K pesitfo +=| = am |. 9> OA sblacemert fn aSec = ~lbm Ca yecdty met \h-m), See \é Average Sheed = Bab atshone remap. Veloaty = Utah dp 2 en ee eS | TWekting Tel bi S Gm = 8m] — 2 Average Speed = (Rak dfstonce Average Speed = lita. dfstance Tatar tte. Wha, ttn. 7 VAT ae > RAT = ee mir Rit oe * 4 qaty V9 Sheed = HAG, Jer TEE Mains a @ For equal distance average speed Person travel with u for half the distance. Then for remaining half distance he travels gcse eens Pe with v for half the_time and w for next_half — KX 1 time. We =2% wage Ft * verage veloci 2 : —t + t+ XH. Avg Speed = Teta dtst NT 9 a ab \Pud Teak that es ee = 3% 2%, 7 Ki5 Ae Avg Velodty = Thats = oat, =¥ 255 Mea r Ta erty eee = \ po SES Seal ee YW \ty Particle move with v, for x distance and then take 120° turn then travel another x with v, Average speed = Wakes . 2%. - 20d, tWeaktim, 2A” Vay Vie = Th ap . = oe Average velocity = SP iN Trt TSX 9 Ve x = 6t?-¢? Time when velocity is zero. Posit en = POtY Particle —> Meter . Pete 4 “+3) = ae - %> eS at When FE O= REA WE 0: It(F-t) Position x = 6t* — 12t + 10 Cmikton) Find distance in 2 sec. t=0 Ku «lod — &m— > =6 xe RSH Z=Atd t+ 2 K¥= 6x4 -2%24'D * = 24-24 +'D = |v —- _— +—€m——> WShlacement = 0. Rs F-l24lo rede 2 Rei " at &) Z=t!O. h>S a wen oo" Particle move with 15 m/s for 2 sec in East then 5 = for 8 sec in North. Average speed. (La pe tie= tom f Tétal dtstonee = 26440 Tha tiny lo = | = Trl. eS Note: Ye content bilkul sufficient hai ( MAINS/ Advanced) For class 11,12,13 ‘om 4, )\/Acceleration Acceleration: | 8,.4¢ - \re Por) | Porttcle Sheed. Qee = ae = [LT ; tele Speed up or cs Ge | ait sete] Vedor Oty, aaa Instantaneous acceleration Q = | tsb dh