Motion in a Straight Line (Seedhi Line mein Motion): Class 11 Physics Chapter 2 Hinglish mein

Physics Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

70 min
27 practice questions
17 MCQs

Is chapter mein aap kya seekhenge

Line ke along motion ke liye position, displacement, path length, average velocity aur average speed ko ek consistent sign convention ke saath describe karna
Instantaneous velocity aur acceleration ko limits ke roop mein, x–t aur v–t graphs ke slopes se, aur differentiation se nikaalna
Graphs se motion ki direction padhna aur yeh ki speed badh rahi hai ya ghat rahi hai, aur v–t graph ke neeche ke area se displacement nikaalna
Constant acceleration ke liye kinematic equations ko area method aur integration, dono se derive karna
Vertical throws, free fall, stopping distance aur reaction time par problems solve karna
Relative velocity se yeh pata karna ki ek line par chal rahe do objects kab aur kahan milte hain ya ek doosre ko overtake karte hain

NCERT Physics Class 11, Chapter 2, pp. 13–26 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

2.1 Introduction

Bowler ke haath se nikalti cricket ball, tower mein upar jaati lift, red light par brake lagata autorickshaw: har ek mein koi cheez lagatar apni position badal rahi hai. Yeh chapter woh language banata hai jisse physicists theek-theek bata sakein ki position kaise badalti hai. Yeh kaam teen ideas karte hain, position, velocity aur acceleration, aur hum har ek ko itna precise banayenge ki unse calculation kar sakein.

Plan ke teen stages hain. Pehle hum velocity aur acceleration ko graphs aur calculus ke saath exactly pakka karenge. Phir fixed acceleration ke common special case ke liye hum teen chhoti equations derive karenge jo batati hain ki body kahan hogi aur kitni tezi se chal rahi hogi. Aakhir mein hum poochenge ki koi motion us observer ko kaisi dikhti hai jo khud bhi chal raha hai. Inmein se kuch bhi yeh nahi poochta ki body ki speed kyun badhti hai; forces ke liye Chapter 4, Laws of Motion, tak rukna hoga. Motion ke sirf description ko kinematics (शुद्धगतिकी) kehte hain.

Poore chapter mein do simplifications chalti hain. Motion ek hi seedhi line par rehta hai, jaise track ke seedhe hisse par train ya kuen mein giraya gaya patthar. Ise rectilinear motion (सरल रेखीय गति) kehte hain, aur ek akela number $x$ position fix kar deta hai. Aur har chalti body ko hum ek point object maan lete hain: hum uska size aur shape bhool kar us par ek point ko follow karte hain, jo tab theek hai jab body us zameen ke muqaable bahut chhoti ho jo woh cover karti hai.

Position, displacement aur average velocity: ek recap

Yeh ideas tum Class 9 mein padh chuke ho; ab inhe signs aur symbols milte hain. Position batane ke liye hum pehle ek reference point fix karte hain, origin $O$ , aur line ke along ek positive direction. Positive side ke points ka $x$ positive hota hai, doosri side ke points ka $x$ negative. Choice hamari hai, lekin koi bhi sign likhne se pehle use batana zaroori hai, aur poore problem mein wahi rakhna hai.

Maan lo koi object time $t_{1}$ par $x_{1}$ par hai aur baad ke time $t_{2}$ par $x_{2}$ par. Is interval mein uska displacement (विस्थापन) $Δ x = x_{2} - x_{1}$ hai: positive direction mein net shift ke liye positive, doosri taraf net shift ke liye negative. Route ke along asal mein cover ki gayi total distance ko path length kehte hain. Jaise hi object wapas mudta hai, dono alag ho jaate hain (Fig. 2.1).

3/3Displacement $= x_{R} - x_{P} = + 4 k m$ , lekin path length $8 k m$ hai.

Fig. 2.1Path length chalaaye gaye har kilometre ko ginti hai; displacement sirf yeh compare karta hai ki trip kahan khatam hui aur kahan shuru hui thi.

Kilometres mein ek horizontal number line, minus 2 se 6 tak, jismein origin O tea stall par hai aur positive direction right ki taraf hai. Step 1 axis aur starting point P ko minus 1 kilometre par dikhata hai. Step 2 mein P se plus 5 par Q tak ek arrow judta hai, jis par 6 km out likha hai, aur Q se wapas plus 3 par R tak doosra arrow, jis par 2 km back likha hai. Step 3 mein P se seedha R tak displacement arrow banta hai, jis par displacement plus 4 kilometre likha hai, aur note hai ki path length 8 kilometre hai.

Lagne wale time se divide karne par do alag averages milte hain. Ek hai average velocity (औसत वेग), jo displacement use karti hai, aur doosri average speed (औसत चाल), jo path length use karti hai:

Average velocity

\overset{ˉ}{v} = \frac{x_{2} - x_{1}}{t_{2} - t_{1}} = \frac{Δ x}{Δ t}

Average speed

average speed = \frac{path length}{Δ t}

Path length kabhi bhi displacement ke size se kam nahi ho sakti, isliye average speed kabhi bhi average velocity ke magnitude se kam nahi hoti. Dono tabhi barabar hote hain jab object interval ke dauraan wapas na mude. Position–time ( $x$ – $t$ ) graph par $\overset{ˉ}{v}$ do points ko jodne wali seedhi chord ka slope hai. Uniform motion mein graph khud ek seedhi line hota hai, isliye har chord ka slope same hota hai: har interval mein velocity same rehti hai.

Example

E-rickshaw ka ek chakkar

Tea stall ko origin lo aur school ki taraf jaane wali road ko positive. Ek e-rickshaw $P$ se shuru hota hai, jo stall ki doosri taraf $1.0 k m$ par hai, 15 min mein $6.0 k m$ chal kar $Q$ tak jaata hai, phir mud kar 5 min mein $2.0 k m$ wapas $R$ tak aata hai. Poori trip ke liye uska displacement, path length, average velocity aur average speed nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Kis question ke liye chalti body ko point object maana ja sakta hai?

Ghoomte fan blade ki tip kitni tezi se chalti haiSomersault mein gymnast ki body kaise ghoomti haiSatellite ko Earth ka ek chakkar lagaane mein kitna time lagta haiBand hote waqt darwaaze ka kinaara kaise chalta hai

Answer dekho

Answer (C) Satellite ko Earth ka ek chakkar lagaane mein kitna time lagta hai

Kuch metre bada satellite, hazaaron kilometre radius wale orbit par, practically ek point hai. Baaki teen mein question yeh hai ki body ke alag parts kaise chalte hain, isliye uska size aur shape matter karte hain.

2.2 Instantaneous velocity aur speed

Average velocity details chhupa leti hai. Jo car ek ghante mein $60 k m$ cover karti hai, ho sakta hai woh pehle traffic mein reng rahi ho aur phir highway par tez daudi ho; average yeh nahi batata ki 10:15 par uska speedometer kya dikha raha tha. Kisi instant par velocity pakadne ke liye interval ko chhota karo. Time $t$ se shuru hone wale chhote interval $Δ t$ par average velocity nikaalo, phir $Δ t$ ko chhota aur chhota karte jao aur dekho ki average kis value par tik jaati hai.

Woh limiting value time $t$ par instantaneous velocity (तात्क्षणिक वेग) hai; jab physicists bina kisi qualifier ke 'velocity' kehte hain, to unka matlab yahi hota hai. Calculus notation mein yeh derivative $d x / d t$ hai:

Instantaneous velocity

v = \lim_{Δ t \to 0} \frac{Δ x}{Δ t} = \frac{d x}{d t}

(2.1)

jahan

$Δ x$: chhote interval $Δ t$ mein displacement
$d x / d t$: instant $t$ par position ke badalne ki rate

Is limit ko nikaalne ke teen tareeke hain. Graph se: jaise-jaise $x$ – $t$ graph par chord ka doosra point pehle point ki taraf khiskta hai, chord pehle point par tangent ban jaati hai (Fig. 2.2). Isliye $x$ – $t$ graph ke kisi point par tangent ka slope us pal ki velocity deta hai.

4/4Limit mein chord $P$ par tangent ban jaati hai: $v = 6.0 m / s$ .

Fig. 2.2Jaise-jaise

Δ t

chhota hota hai,

P

se nikli chord

P

par tangent ban jaati hai; uska slope

t = 2 s

par velocity hai.NCERT Fig. 2.1 · p. 14

Ek position-time graph, position 0 se 16 metre tak aur time 0 se 3.5 second tak, jismein upar jaata curve x = 1.5 t squared hai, aur point P 2 second aur 6 metre par hai. Step 1 mein P se 3 second wale point tak chord banti hai, slope 7.5 metre per second. Step 2 mein doosra point 2.5 second par aata hai, slope 6.75. Step 3 mein woh 2.1 second par aata hai, slope 6.15, aur chord curve ko lagbhag chhoo rahi hai. Step 4 mein P par tangent dikhta hai jiska slope 6.0 metre per second hai, aur pehle wali chords uske peeche faded hain.

Numbers se: chhote aur chhote $Δ t$ ke liye $Δ x / Δ t$ calculate karo. Traffic signal se chal padi ek motorbike ke liye $x = 1.5 t^{2}$ hai (metres mein, $t$ seconds mein); Table 2.1 $t = 2.0 s$ par uski velocity nikaalti hai. 2 s se aage ki taraf aur 2 s se peeche ki taraf liye gaye intervals ek hi value, $6.0 m / s$ , ki taraf badhte hain, ek upar se aur ek neeche se.

Table 2.1

t = 2.0 s

par chhote hote intervals mein motorbike (

x = 1.5 t^{2}

) ki average velocityNCERT Table 2.1 · p. 14

$Δ t$ (s)	aage ki taraf $Δ x / Δ t$ (m/s)	peeche ki taraf $Δ x / Δ t$ (m/s)
1.0	7.5	4.5
0.5	6.75	5.25
0.1	6.15	5.85
0.01	6.015	5.985
0.001	6.0015	5.9985

Calculus se: jab $x (t)$ ek formula ho, to use differentiate karo. $x = 1.5 t^{2}$ ke liye $v = d x / d t = 3.0 t$ , jo $t = 2.0 s$ par $6.0 m / s$ deta hai, theek wahi number jiski taraf table badh rahi thi. Calculus aksar sabse jaldi kaam karta hai; table dikhati hai ki derivative ka matlab kya hai. Uniform motion mein har chord pehle se hi seedhi $x$ – $t$ line par hoti hai, isliye kisi bhi instant par velocity kisi bhi interval ki average velocity ke barabar hoti hai.

Example 2.1NCERT Example 2.1 · p. 15

Wire par bead ki velocity

Ek bead seedhe wire par slide karta hai, jiski position $x = 6.0 t - 0.50 t^{3}$ hai, jahan $x$ metres mein aur $t$ seconds mein hai. Nikaalo (a) $t = 1.0 s$ aur $t = 2.0 s$ par uski velocity; (b) $t = 0$ aur $t = 2.0 s$ ke beech uski average velocity.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Do autorickshaws ek crossing par ek doosre ke paas se guzarte hain, ek north ki taraf aur ek south ki taraf, dono $8 m / s$ par. North ko positive lene par unki velocities $+ 8$ aur $- 8 m / s$ hain, phir bhi dono speedometers ek hi reading dikhate hain. Speedometer speed (चाल) dikhata hai, yaani velocity ka magnitude $∣ v ∣$ , jise direction se koi matlab nahi.

Quick check

3 mein se 1

x

–

t

graph se kisi khaas pal ki velocity kaise padhte hain?

Wahan tangent ka slope loWahan graph ki height padhoWahan tak graph ke neeche ka area nikaaloOrigin se chord ka slope lo

Answer dekho

Answer (A) Wahan tangent ka slope lo

Jab

Δ t \to 0

, chord tangent ban jaati hai, isliye uska slope

v

deta hai. Origin se chord

t = 0

se ab tak ki average velocity deti hai, us instant ki velocity nahi.

2.3 Acceleration

Velocity bhi aksar badalti hai: bus stop se chal padti hai, upar phenka gaya patthar dheema hota jaata hai. Kya hum badlaav ko per second naapein ya per metre? Khud Galileo ne pehle galat jawab chuna tha (box dekho). Rest se giraye gaye patthar ko dekho: 1 s ke baad woh $9.8 m / s$ par chalta hai, 2 s ke baad $19.6 m / s$ par, har second mein utni hi badhat. Per metre picture uljhi hui hai: girne ke pehle $4.9 m$ mein $9.8 m / s$ judta hai, lekin agle $14.7 m$ mein sirf $9.8 m / s$ aur. Jo rate constant rahe, woh kahin zyada useful hai, isliye acceleration ko per unit time define kiya jaata hai.

Is rate ko acceleration (त्वरण) measure karta hai. Agar velocity time $t_{1}$ par $v_{1}$ se badal kar time $t_{2}$ par $v_{2}$ ho jaati hai, to is interval mein average acceleration hai

Average acceleration

\overset{ˉ}{a} = \frac{v_{2} - v_{1}}{t_{2} - t_{1}} = \frac{Δ v}{Δ t}

(2.2)

Iski unit $(m / s) / s = m s^{- 2}$ hai. Velocity–time ( $v$ – $t$ ) graph par $\overset{ˉ}{a}$ ek chord ka slope hai, theek waise hi jaise $x$ – $t$ graph par $\overset{ˉ}{v}$ chord ka slope tha. Chord ko chhota karke tangent bana do aur tumhe instantaneous acceleration milta hai:

Instantaneous acceleration

a = \lim_{Δ t \to 0} \frac{Δ v}{Δ t} = \frac{d v}{d t}

(2.3)

Line par 'direction' ki sirf do values hoti hain, $+$ aur $-$ , isliye $a$ bas ek signed number hai. (Chapter 3 mein, plane mein motion par, velocity apna size badle bina ghoom sakti hai, aur yeh mudna bhi acceleration gina jaata hai.)

Example

Rest mein, phir bhi accelerate ho raha

Example 2.1 ke bead ke liye nikaalo (a) $t = 0$ se $t = 2.0 s$ tak uska average acceleration; (b) $t = 2.0 s$ par uska acceleration, jis instant par woh rest mein hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

$x$ – $t$ graph apne curvature se acceleration ka sign bhi bata deta hai. Velocity $x$ – $t$ graph ka slope hai. Positive acceleration mein yeh slope badhta rehta hai, isliye graph upar ki taraf mudta hai. Negative acceleration mein slope ghatta rehta hai aur graph neeche ki taraf mudta hai. Zero acceleration mein slope fixed rehta hai aur graph seedha hota hai (Fig. 2.3).

1/3Slope badh raha hai: graph upar ki taraf mudta hai, $a > 0$ .

Fig. 2.3

x

–

t

graph ka curvature acceleration ka sign dikhata hai.NCERT Fig. 2.2 · p. 16

Ek position-time axis jis par har step mein ek curve aur teen chhote tangent segments bane hain. Step 1: curve upar ki taraf mudta hai aur left se right tangents zyada steep hote jaate hain, label slope badhta rehta, yaani positive acceleration. Step 2: curve neeche ki taraf mudta hai aur tangents flat hote jaate hain, label slope ghatta rehta, yaani negative acceleration. Step 3: ek seedhi line jiske teeno tangents usi par same slope ke saath padte hain, yaani zero acceleration.

Ab se acceleration fixed rakha jaayega, jaise hawa mein udte patthar ka ya steady thrust se chal padi train ka. Constant acceleration mein $v$ – $t$ graph ki har chord aur har tangent ka slope same, $a$ , hota hai. Agar clock shuru hote waqt ( $t = 0$ ) velocity $v_{0}$ hai, to time $t$ ke baad woh $a t$ badh chuki hoti hai:

Constant acceleration mein velocity

v = v_{0} + a t

(2.4)

Isliye constant acceleration mein $v$ – $t$ graph seedha hota hai, slope $a$ aur intercept $v_{0}$ ke saath. Fig. 2.4 woh chaar tarah ki lines dikhata hai jo tumhe milengi. Har ek se do questions poochho: line time axis ki kis side par hai? (yeh motion ki direction hai) aur line axis se door ja rahi hai ya uski taraf aa rahi hai? (speed badh rahi hai ya ghat rahi hai).

4/4Poore samay $a < 0$ : pehle aage, jahan line axis cross karti hai wahan ek pal ke liye rest mein, phir peeche.

Fig. 2.4Axis ke upar matlab

+ x

direction mein motion; axis se door jaati line matlab speed badh rahi hai.NCERT Fig. 2.3 · p. 16

Ek velocity-time graph jismein plus values time axis ke upar aur minus values neeche hain, aur uske upar ek road par car hai jis par blue velocity arrow aur vermillion acceleration arrow hai. Step 1: plus 1 se plus 4 tak upar jaati line, car right ki taraf chal rahi hai aur dono arrows right ki taraf hain, speed badh rahi hai. Step 2: plus 4 se plus 1 tak neeche aati line, car right ki taraf chal rahi hai, acceleration arrow left ki taraf hai, speed ghat rahi hai. Step 3: minus 1 se minus 4 tak neeche jaati line, car ka mooh left ki taraf, dono arrows left ki taraf, speed badh rahi hai. Step 4: plus 3 se zero se hote hue minus 3 tak line; acceleration left ki taraf hai, velocity pehle right ki taraf aur baad mein left ki taraf, isliye jahan line axis cross karti hai wahan car rukti hai aur wapas mudti hai.

Time ke ek chhote hisse $Δ t$ mein velocity mushkil se badalti hai, isliye object lagbhag $v Δ t$ chalta hai. Yeh product $v$ – $t$ graph ke neeche height $v$ aur width $Δ t$ wali ek patli strip ka area hai. Ek interval mein aisi strips ko jodo to unke areas milkar poora displacement dete hain. Unhe aur patla karo to yeh seedhiyon jaisa shape curve se chipak jaata hai; limit mein sum ek integral ban jaata hai (Fig. 2.5). To: kisi interval mein $v$ – $t$ graph aur time axis ke beech ka area us interval ke displacement ke barabar hota hai. Constant velocity mein slices ki zaroorat nahi: $5 s$ tak $4 m / s$ par chalti car $20 m$ jaati hai, jo ek rectangle ka area hai.

3/3Axis ke upar area $+ 4.5 m$ hai, neeche $- 2.0 m$ : displacement $2.5 m$ , path length $6.5 m$ .

Fig. 2.5

v

–

t

graph ke neeche ka area displacement hai; axis ke neeche ka area negative gina jaata hai.NCERT Fig. 2.4 · p. 17

Metre per second mein velocity, seconds mein time ke against. Step 1: 5 second tak 4 metre per second par ek horizontal line, jiske neeche ka rectangle shaded hai aur us par area = 20 m likha hai. Step 2: ek upar jaata curve jiske neeche das patli rectangular strips hain, label thin strips. Step 3: ek seedhi line jo 3 metre per second se girti hai, 3 second par zero se guzarti hai aur 5 second par minus 2 tak jaati hai; axis ke upar wale triangle par plus 4.5 metre aur neeche wale dashed triangle par minus 2.0 metre likha hai, jisse displacement 2.5 metre aur path length 6.5 metre milti hai.

Time axis ke neeche ka area negative velocity se aata hai, isliye woh negative displacement gina jaata hai. Agar path length chahiye, to saare areas ke size jodo.

Quick check

4 mein se 1

Upar ko positive lene par, ek skydiver jisne abhi-abhi apna parachute khola hai, ab bhi neeche ja rahi hai lekin dheemi ho rahi hai.

v

aur

a

ke signs hain

v < 0

aur

a < 0

v > 0

aur

a < 0

v < 0

aur

a > 0

v > 0

aur

a > 0

Answer dekho

Answer (C) $v < 0$ aur $a > 0$

Neeche jaane ka matlab

v < 0

. Dheema hone ka matlab hai ki

a

ka sign

v

ka ulta hai, isliye

a > 0

: ek upar ki taraf acceleration.

a < 0

(option A) hone par uski speed badh rahi hoti.

2.4 Uniformly accelerated motion ke liye kinematic equations

Equation (2.4) uniformly accelerated motion ke liye 'kitni tezi se?' ka jawab deti hai. 'Kitni door?' seedhi $v$ – $t$ line ke neeche ke area se milta hai. Clock tab shuru karo ( $t = 0$ ) jab object origin par ho, taaki time $t$ par uski position $x$ hi uska displacement bhi ho. Shuru mein uski velocity $v_{0}$ hai aur time $t$ par $v$ .

Derivationv–t line ke neeche ke area se equations

Fig. 2.6 mein line time 0 par $v_{0}$ se badh kar time $t$ par $v$ tak jaati hai. Iske neeche ke area ko height $v_{0}$ wale ek rectangle mein aur uske upar height $v - v_{0}$ wale ek triangle mein baanto. Area, aur isliye displacement, hai

2/3Triangle: base $t$ aur height $v - v_{0} = a t$ , isliye area $\frac{1}{2} a t^{2}$ .

Fig. 2.6Rectangle plus triangle se

x = v_{0} t + \frac{1}{2} a t^{2}

milta hai; dashed line se

x = \overset{ˉ}{v} t

.NCERT Fig. 2.5 · p. 17

Velocity, time ke against; ek seedhi line time zero par v-nought se badh kar time t par v tak jaati hai. Step 1 height v-nought aur width t wala rectangle shade karta hai, label v-nought t. Step 2 upar triangle jodta hai, height v minus v-nought, label one half a t squared. Step 3 v-nought aur v ke average par ek dashed horizontal line banata hai; right mein uske upar kata chhota triangle left mein uske neeche ka gap theek bhar deta hai, isliye area average velocity guna t ke barabar hai.

x = v_{0} t + \frac{1}{2} (v - v_{0}) t

(2.5)

(2.4) se, $v - v_{0} = a t$ . Substitute karne par,

x = v_{0} t + \frac{1}{2} a t^{2}

(2.6)

Equation (2.5) ko $x = \frac{1}{2} (v_{0} + v) t$ ke roop mein bhi likh sakte hain. Object utni hi zameen cover karta hai jitni woh apni starting aur final velocities ke mean par steadily chal kar karta. Figure ke aakhri step mein, us height par dashed line apne upar ek triangle kaat-ti hai jo theek uske neeche ka gap bhar deta hai.

x = \overset{ˉ}{v} t

(2.7a)

sirf constant acceleration ke liye

\overset{ˉ}{v} = \frac{v_{0} + v}{2}

(2.7b)

Ek time-free equation un do results ko multiply karke milti hai jo hamare paas pehle se hain: (2.4) se $v - v_{0} = a t$ milta hai, aur (2.7a) ke saath (2.7b) se $v + v_{0} = 2 x / t$ . Product mein $t$ cancel ho jaate hain:

(v - v_{0}) (v + v_{0}) = a t \cdot \frac{2 x}{t}

v^{2} = v_{0}^{2} + 2 a x

(2.8)

Yeh milkar $v_{0}$ , $v$ , $a$ , $t$ aur $x$ ko jodti hain; koi bhi teen pata hon to baaki do nikal aate hain:

Kinematic equations

\begin{aligned} v & = v_{0} + a t \\ x & = v_{0} t + \frac{1}{2} a t^{2} \\ v^{2} & = v_{0}^{2} + 2 a x \end{aligned}

(2.9a)

Agar object origin ki jagah $x_{0}$ se shuru hota hai, to uska displacement $x - x_{0}$ hai. Pehli equation waisi hi rehti hai aur baaki do ban jaati hain

x = x_{0} + v_{0} t + \frac{1}{2} a t^{2}

(2.9b)

v^{2} = v_{0}^{2} + 2 a (x - x_{0})

(2.9c)

InteractiveMotion lab: x–t, v–t aur a–t ek saath

Ek interactive kinematics lab. Chaar presets (cruise, speed up, thrown up, backward start) ya teen sliders initial position minus 20 se 20 metre tak, initial velocity minus 20 se 20 metre per second tak aur constant acceleration minus 10 se 10 metre per second squared tak set karte hain. Play dabane par 4 second ka motion chalta hai: ek dot horizontal x-axis par chalta hai aur position-time, velocity-time aur acceleration-time graphs current time tak bante jaate hain. Velocity graph ke neeche ka area shaded hai, time axis ke upar green aur neeche dashes ke saath vermillion. Ek time slider kisi bhi instant par rok deta hai. Ek readout x, v aur a deta hai, yeh batata hai ki object ki speed badh rahi hai ya ghat rahi hai, ab tak ka displacement aur path length, aur average velocity aur average speed.

Fig. 2.7

x_{0}

v_{0}

aur

a

set karo, Play dabao, aur teeno graphs compare karo. Thrown up preset se shuru karo.

Lab mein object ko positive $v_{0}$ aur negative $a$ do aur use rukte aur wapas aate dekho. Turning point par $v$ – $t$ line time axis ko cross karti hai, usi instant par $x$ – $t$ curve apne peak par hota hai, aur uske baad path length displacement se aage nikal jaati hai.

Example 2.2NCERT Example 2.2 · p. 18

Calculus se equations

Constant acceleration $a$ wale ek object ke liye, jiski $t = 0$ par velocity $v_{0}$ aur position $x_{0}$ hai, kinematic equations ko integration se derive karo. Phir time ke saath badhte acceleration $a = 2 t$ (SI units), $v_{0} = 0$ aur $x_{0} = 0$ ke saath, dikhao ki yeh method zyada general kyun hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 2.3NCERT Example 2.3 · p. 18

Cliff se upar phenka gaya patthar

River se $20 m$ upar ek cliff ke kinaare se ek student ek patthar ko $15 m / s$ par seedha upar phenkta hai. Patthar upar jaata hai, cliff ke kinaare ke paas se neeche girta hai aur river mein ja girta hai. $g = 10 m / s^{2}$ lo aur air resistance ko ignore karo. Nikaalo (a) river se patthar ki sabse zyada height, (b) woh paani se kab takraata hai aur (c) us waqt woh kitni tezi se chal raha hai.

2/4 $t = 1.5 s$ : $y = 31.25 m$ , $v = 0$ , lekin $a$ ab bhi $- 10 m / s^{2}$ hai.

Fig. 2.8Paani se upar ki taraf

y

lo:

y_{0} = 20 m

v_{0} = + 15 m / s

a = - 10 m / s^{2}

.NCERT Fig. 2.6 · p. 19

Left mein river ke upar 20 metre ooncha cliff hai, aur right mein metres mein height scale. Cliff ke kinaare se ek dashed vertical path upar jaata hai aur wapas paani tak neeche aata hai. Acceleration arrow, minus 10 metre per second squared, poore samay neeche ki taraf hai. Step 1: t = 0 par patthar plus 15 metre per second velocity se kinaara chhodta hai, arrow upar. Step 2: 1.5 second par woh sabse upar hai, river se 31.25 metre upar, aur v zero hai. Step 3: 3.0 second par woh minus 15 metre per second se neeche jaate hue cliff ke kinaare se guzarta hai. Step 4: 4.0 second par woh minus 25 metre per second se paani tak pahunchta hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 2.4NCERT Example 2.4 · p. 19

Rest se free fall

Earth ki surface ke paas rest se chhodi gayi body ke motion ko describe karo, air resistance ko ignore karke. Chhodne ke $0.5$ , $1$ , $2$ aur $3 s$ baad uski velocity aur position nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Ek balcony se ek coin aur ek cricket ball saath mein giraao: agar hawa se koi fark na padta, to dono saath mein zameen par pahunchte. Earth ke paas sirf gravity ke under girne wali har body ki speed same rate se badhti hai, $g \approx 9.8 m / s^{2}$ neeche ki taraf. Aise motion ko free fall (मुक्त पतन) kehte hain. Height ke saath $g$ bahut dheere kam hota hai (har 10 km par lagbhag 0.3 %), isliye roz ke girne ke liye yeh constant hai aur free fall constant acceleration wala motion hai.

Table 2.2Rest se free fall, upar positive,

g = 9.8 m / s^{2}

$t$ (s)	$v$ (m/s)	$y$ (m)
0.5	−4.9	−1.23
1.0	−9.8	−4.9
2.0	−19.6	−19.6
3.0	−29.4	−44.1

Fig. 2.9 free fall ke teen graphs dikhata hai: ek flat $a$ – $t$ line, slope $- g$ wali ek seedhi $v$ – $t$ line, aur ek $y$ – $t$ curve jo neeche ki taraf aur zyada steep hota jaata hai.

2/3Velocity: ek seedhi line, $v = - 9.8 t$ .

Fig. 2.9Free fall, upar positive: constant

a = - g

, linear

v

, parabolic

y

.NCERT Fig. 2.7 · p. 20

0 se 3 second tak time ke against graphs, jo upar bane time axis ke neeche bane hain kyunki saari values negative hain. Step 1: acceleration minus 9.8 metre per second squared par ek flat line hai. Step 2: velocity ek seedhi line hai jo 0 se gir kar 3 second par minus 29.4 metre per second tak jaati hai. Step 3: position ek curve hai jo flat shuru hota hai aur tezi se neeche mudta hai, 3 second par minus 44.1 metre tak pahunchta hai.

Example 2.5NCERT Example 2.5 · p. 20

Galileo ka odd numbers wala rule

Galileo ne paaya ki rest se shuru karke, girti body time ke lagataar barabar hisson mein jo distances cover karti hai, woh odd numbers $1, 3, 5, 7, \dots$ ki tarah chalti hain. Is rule ko prove karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

4/4Phir 7 units (total 16 $= 4^{2}$ ).

Fig. 2.10Barabar times, badhte gaps:

1 : 3 : 5 : 7

, running totals

1, 4, 9, 16

ke saath.NCERT Table 2.2 · p. 20

Ek vertical scale jis par ball chhodne ke 0, tau, 2 tau, 3 tau aur 4 tau baad ke instants par dikhai gayi hai. Ball ki positions start se 0, 1, 4, 9 aur 16 units neeche hain. Step 1 pehli do positions aur pehla gap, 1 unit, dikhata hai. Step 2 teesri position aur 3 units ka gap jodta hai. Step 3 5 units ka gap jodta hai. Step 4 7 units ka gap aur summary jodta hai: barabar intervals mein distances 1 : 3 : 5 : 7 hain, jo chaar intervals ke baad total 16 units hoti hain.

Try karke dekho

Metronome ke saath Galileo ki slope

Tumhe chahiye

2 m ki curtain rail ya grooved plank
ek marble ya chhoti steel ball
ek sira uthaane ke liye ek kitaab
phone par ek metronome app
chalk ya sticky notes, ek measuring tape

Pehle guess karo

Agar pehli beat mein ball

4 c m

roll karti hai, to teesri beat mein use kitna roll karna chahiye?

Galileo ne slope se free fall ko dheema kiya tha. Tum ek ball aur ek steady beat ke saath unka test dohra sakte ho.

Ek seedhi 2 m curtain rail ya grooved plank ka ek sira kitaab par lagbhag 5 cm uthao, taaki upar se chhodi gayi ball neeche pahunchne mein lagbhag 5 s le.
Metronome app ko lagbhag 60 beats per minute par chalao. Ball ko theek ek beat par rest se chhodo.
Ek partner agli chaar beats mein se har beat par ball ki position ko chalk ya sticky note se mark kare. Har mark pakka karne ke liye kuch baar dohrao.
Har mark ko start se naapo, phir paas-paas ke marks ke beech ka gap naapo, aur har gap ko pehle gap se divide karo.

Table 2.3

Beat	Start se (cm)	Gap (cm)	Gap ÷ pehla
1
2
3
4

Tumhari entries apne aap save hoti hain.

Example 2.6NCERT Example 2.6 · p. 20

Stopping distance

Brakes lagne ke instant se vehicle ke rukne ke instant tak naapo: road ka woh hissa uski stopping distance hai. (a) Ise initial speed $v_{0}$ aur size $∣ a ∣$ wale steady deceleration ke terms mein nikaalo. (b) Ek bhari hui bus $6.0 m / s^{2}$ se brake karti hai. $15 m / s$ ( $54 k m / h$ ) par aur usse dugni speed par uski stopping distance nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Asli road par brakes turant kaam nahi karte. Driver ke reaction time $t_{r}$ ke dauraan vehicle apni poori speed par chalta rehta hai, jisse distance mein $v_{0} t_{r}$ jud jaata hai. Fig. 2.11 ise $t_{r} = 0.75 s$ ke liye dikhata hai: speed dugni karne par reaction distance dugni aur braking distance chaar guna ho jaati hai.

2/2 $t_{r} = 0.75 s$ ke liye $v_{0} t_{r}$ jodo: totals $30 m$ aur $97.5 m$ .

Fig. 2.11Braking distance

v_{0}^{2}

ki tarah badhti hai; reaction distance

v_{0}

ki tarah.

Road ki do lanes, har ek mein left par ek car. Upar wali lane: 15 metre per second; neeche wali lane: 30 metre per second. Step 1 sirf braking zones dikhata hai: upar wali lane mein 18.75 metre aur neeche wali lane mein 75 metre, chaar guna lamba. Step 2 har braking zone ke aage ek dashed reaction zone daalta hai, 0.75 second ke reaction time ke liye 11.25 metre aur 22.5 metre, jisse braking zones aur aage khisak jaate hain, aur totals 30 metre aur 97.5 metre dikhata hai.

Example 2.7NCERT Example 2.7 · p. 21

Girte ruler se reaction time

Ek dost ruler ko vertical pakadta hai, uska zero mark tumhare khule thumb aur finger ke level par, phir bina bataye chhod deta hai. Tum use $17.6 c m$ girne ke baad pakadte ho. Apna reaction time estimate karo ( $g = 9.8 m / s^{2}$ ).

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3/3 $d = 17.6 c m$ ke baad pakda: $t_{r} = \sqrt{2 d / g} \approx 0.19 s$ .

Fig. 2.12Pakadne se pehle ruler jitna girta hai, usse tumhara reaction time milta hai.NCERT Fig. 2.8 · p. 21

Centimetres mein marked ek vertical ruler, jiska zero mark neeche hai, dono taraf gol bars ki tarah bani do fingers ke level par. Step 1: dost chhodne wala hai; zero mark tumhari fingers par hai. Step 2: ruler girna shuru ho gaya hai, neeche khiska hua dikhaya gaya hai, aur arrow par freely girta likha hai. Step 3: fingers ruler par band hoti hain; ruler 17.6 centimetre gir chuka hai, isliye fingers par reading d = 17.6 centimetre hai, jisse lagbhag 0.19 second ka reaction time milta hai.

Try karke dekho

Apna reaction time naapo

Tumhe chahiye

ek 30 cm ruler
ek dost
ek calculator

Pehle guess karo

Shuru karne se pehle: agar tumhara reaction time dugna hota, to ruler girta

dugna doorchaar guna doorutni hi distanceaadha door

Apni forearm ko table par rakho aur haath kinaare se bahar, thumb aur index finger ke beech lagbhag 2 cm ka gap.
Tumhara dost ruler ko upar se aise pakade ki uska 0 cm mark tumhare thumb ke upar wale hisse ke level par ho.
Bina countdown ke tumhara dost ruler chhod deta hai. Ruler ko jitni jaldi ho sake pakdo aur apne thumb ke upar wale hisse par jo mark hai use padho.
Paanch baar dohrao, har baar ka reaction time nikaalo, phir roles badal lo.

Table 2.4

Trial	$d$ (cm)	$t_{r}$ (s)
1
2
3
4
5

Tumhari entries apne aap save hoti hain.

InteractiveReaction-time ruler

Ek ruler-drop game. Ek vertical 30 centimetre ruler latka hai, jiska zero mark do fingers ke beech hai. Get ready dabane ke baad, ruler ek se teen second ke beech kisi random pal par chhoda jaata hai aur g = 9.8 metre per second squared se girta hai. Catch dabane, space bar dabane ya stage par tap karne se woh ruk jaata hai; widget giri hui distance centimetres mein padhta hai aur use t = square root of 2d over g se reaction time mein badalta hai. Tumhare aakhri paanch attempts aur unka average ek table mein dikhte hain. Bahut jaldi dabana false start gina jaata hai, aur 30 centimetre se aage gir gaya ruler miss hai.

Fig. 2.13Get ready dabao, ruler ke girne ka intezaar karo, phir use pakdo. Woh jitna gira, usse

t_{r} = \sqrt{2 d / g}

ke through tumhara reaction time milta hai.

2.5 Relative velocity

Station par train mein baithe hue, tumhe kabhi laga hoga ki tumhari train chal padi, aur phir dikha ki asal mein bagal wale track ki train nikal rahi thi. Tumhe kya dikhta hai, yeh is par depend karta hai ki tum kaise chal rahe ho. Ek chalte object ko doosre chalte object se describe karne ke liye relative velocity (सापेक्ष वेग) chahiye.

Maan lo objects A aur B ek hi line par constant velocities $v_{A}$ aur $v_{B}$ se chal rahe hain, taaki $x_{A} (t) = x_{A} (0) + v_{A} t$ aur $x_{B} (t) = x_{B} (0) + v_{B} t$ . A ke relative B ki position ko gap $x_{B A} = x_{B} - x_{A}$ ke roop mein define karo, jo A se B tak naapa jaata hai. Dono equations ko subtract karne par milta hai

x_{B A} (t) = x_{B A} (0) + v_{B A} t

(2.10)

jahan

A ke respect mein B ki relative velocity

v_{B A} = v_{B} - v_{A}

(2.11)

A ke relative B ki velocity hai: woh velocity jo A, B ke liye naapega agar A khud ko rest mein maane. Isi tarah $v_{A B} = v_{A} - v_{B} = - v_{B A}$ : har ek ko doosra same speed se ulti direction mein chalta dikhta hai.

Fig. 2.14 ise $x$ – $t$ graph par dikhata hai. Agar $v_{A} = v_{B}$ , to lines parallel hoti hain, $v_{B A} = 0$ aur gap kabhi nahi badalta: same speed par saath-saath chalti do cars ek doosre ko khadi hui lagti hain. Agar velocities alag hain lekin sign same hai, to ek line zyada steep hoti hai aur gap $∣ v_{B} - v_{A} ∣$ ki rate se badalta hai; jahan lines cross karti hain, wahan ek doosre ko overtake karta hai. Agar velocities ke signs ulte hain, to relative speed dono speeds ka sum hoti hai, aur objects tezi se ek doosre ke paas aate hain.

2/3 $v_{B A} = + 10 m / s$ : B $200 m$ ka gap $20 s$ mein khatam karta hai.

Fig. 2.14Parallel lines: constant gap. Cross karti lines: crossing par vehicles barabar hote hain.

0 se 40 second tak time ke against 0 se 1000 metre tak position. Step 1: A 200 metre se aur B 0 se shuru hota hai, dono 20 metre per second par; lines parallel hain aur ek double arrow dikhata hai ki gap 200 metre hi rehta hai. Step 2: ab B 30 metre per second par chalta hai; uski zyada steep line 20 second aur 600 metre par A ki line se milti hai, jahan B, A ko pakad leta hai. Step 3: A 0 se 20 metre per second par chalta hai aur B 700 metre se minus 15 metre per second par; upar jaati aur neeche aati lines 20 second aur 400 metre par cross karti hain, jahan dono milte hain.

Example

Highway par overtake karna aur milna

Ek bus seedhe highway par $72 k m / h$ se chal rahi hai. (a) Bus se $200 m$ peeche ek car usi direction mein $108 k m / h$ se chal rahi hai. Woh bus ke barabar kab aur kahan pahunchti hai? (b) Baad mein, ulti taraf se $54 k m / h$ par aata ek truck bus se $700 m$ aage hai. Woh ek doosre ke paas se kab aur kahan guzarte hain?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Do trains parallel tracks par same direction mein

90 k m / h

aur

60 k m / h

se chal rahi hain. Dheemi train ke passenger ko tez train kis speed se aage nikalti dikhti hai?

km/h

Answer dekho

Answer 30 km/h

v_{B A} = 90 - 60 = 30 k m / h

, aage ki taraf.

Recap

Summary aur key points

Line ke along motion ko ek signed coordinate $x$ se describe kiya jaata hai, jo chune gaye origin se chuni gayi positive direction mein naapa jaata hai (recap).
Displacement $Δ x = x_{2} - x_{1}$ positive, negative ya zero ho sakta hai; path length cover kiya gaya har metre ginti hai, isliye average speed $\geq ∣ \overset{ˉ}{v} ∣$ (Fig. 2.1).
Instantaneous velocity $v = d x / d t$ , $Δ x / Δ t$ ki limit hai aur $x$ – $t$ graph ke tangent ka slope hai; speed $∣ v ∣$ hai (Section 2.2).
Average acceleration $Δ v / Δ t$ hai; instantaneous acceleration $a = d v / d t$ , $v$ – $t$ graph ka slope hai (Section 2.3).
Uniform motion: seedhi $x$ – $t$ line, horizontal $v$ – $t$ line. Constant acceleration: parabolic $x$ – $t$ curve, seedhi jhuki hui $v$ – $t$ line (Fig. 2.3, Fig. 2.4).
Jab $v$ aur $a$ ka sign same hota hai to object ki speed badhti hai aur alag hone par ghatti hai; $v = 0$ ka matlab $a = 0$ nahi hai (bead example).
Do instants ke beech $v$ – $t$ graph ke neeche ka area displacement hai; time axis ke neeche ka area negative gina jaata hai (Fig. 2.5).
Constant acceleration ke liye: $v = v_{0} + a t$ , $x - x_{0} = v_{0} t + \frac{1}{2} a t^{2}$ , $v^{2} = v_{0}^{2} + 2 a (x - x_{0})$ , $x - x_{0} = \frac{1}{2} (v_{0} + v) t$ ((2.9a)).
Free fall, upar positive: $a = - g$ ; rest se, lagataar barabar intervals mein distances $1 : 3 : 5 : \dots$ ke ratio mein hoti hain (Example 2.5).
Braking distance $v_{0}^{2} / 2 ∣ a ∣$ speed ke square ki tarah badhti hai; girta ruler reaction time ko $\sqrt{2 d / g}$ ke roop mein deta hai (Example 2.6, Example 2.7).
A ke relative B ki velocity $v_{B A} = v_{B} - v_{A}$ hai, yaani woh rate jisse gap $x_{B} - x_{A}$ badalta hai (Section 2.5).

Table 2.5Graph par har quantity kahan milti haiNCERT Summary table · p. 22

Quantity	Signed?	Graph par	SI unit
Displacement $Δ x$	haan	$v$ – $t$ ke neeche signed area	m
Path length	nahi	total area, sab $+$ gine gaye	m
Average velocity $\overset{ˉ}{v}$	haan	$x$ – $t$ par chord ka slope	$m s^{- 1}$
Velocity $v$	haan	$x$ – $t$ par tangent ka slope	$m s^{- 1}$
Speed $∣ v ∣$	nahi	$x$ – $t$ ki steepness	$m s^{- 1}$
Average acceleration $\overset{ˉ}{a}$	haan	$v$ – $t$ par chord ka slope	$m s^{- 2}$
Acceleration $a$	haan	$v$ – $t$ par tangent ka slope	$m s^{- 2}$

Practice · 27 questions

Practice questions answers ke saath (NCERT “Exercises”)

Questions 1–18 NCERT Exercises 2.1–2.18 ki skills ko nayi situations mein practise karaate hain; 19–27 entrance-style practice hain (JEE Main aur NEET pattern). Jab tak question kuch aur na kahe, $g = 9.8 m / s^{2}$ lo.

1
Quick answerEasyNCERT Exercise 2.1
Inmein se kis question mein chalti body ko ek point maan lena safe hai? Jo bhi sahi hon, sab chuno.
Ek passenger train ko Chennai se Bengaluru ka $350 k m$ ka safar poora karne mein kitna time lagta haiCeiling-fan blade ki tip kitni tezi se chalti haiEk skydiver ko $1 k m$ girne mein kitna time lagta haiGhoomta carrom striker board ke kinaare se kaise takra kar lautta hai
Saare sahi options chuno
Answer dekho
Answer
(A) Ek passenger train ko Chennai se Bengaluru ka $350 k m$ ka safar poora karne mein kitna time lagta hai
(C) Ek skydiver ko $1 k m$ girne mein kitna time lagta hai
Train (kuch sau metre lambi, sainkadon kilometre chalti hui) aur skydiver (2 m se kam lamba, 1000 m girta hua) cover ki gayi distances ke muqaable bahut chhote hain. Fan blade aur ghoomte striker mein question yeh hai ki alag parts kaise chalte hain, isliye size aur shape matter karte hain.
2
Short answerMediumNCERT Exercise 2.2
Kabir aur Riya ek hi market M se ek seedhi road par cycle chala kar ghar jaate hain. Unke $x$ – $t$ graphs diye gaye hain (Kabir solid, Riya dashed); rider ke ghar pahunchte hi har graph flat ho jaata hai.
Fig. 2.15
Minutes mein time ke against market se metres mein position. Kabir ki solid line time 0 par 0 se shuru hoti hai aur steadily badh kar 20 minute par 3000 metre tak jaati hai, phir flat rehti hai. Riya ki dashed line 4 minute par 0 se shuru hoti hai, zyada steep badh kar 14 minute par 2500 metre tak jaati hai, phir flat rehti hai. Lines 10 minute aur 1500 metre par cross karti hain, aur Kabir ki line lagbhag 16.7 minute par Riya ki flat line ko cross karti hai.
1. Market se zyada door kaun rehta hai?
  RiyaKabirDono barabar door rehte hain
  Answer dekho
  Answer (B) Kabir
  Kabir ka graph $3000 m$ par flat hota hai, Riya ka $2500 m$ par.
2. Market se pehle kaun nikalta hai, aur kitni der pehle?
  Riya, 4 min pehleDono saath nikalte hainKabir, 4 min pehle
  Answer dekho
  Answer (C) Kabir, 4 min pehle
  Kabir ki line $t = 0$ par upar jaana shuru karti hai, Riya ki $t = 4 m i n$ par.
3. Kaun tez chalata hai? Dono ki speeds nikaalo.
  Solution dikhao
  Slopes: Kabir $3000 / 20 = 150 m / m i n$ ( $2.5 m / s$ ); Riya $2500 / 10 = 250 m / m i n$ (lagbhag $4.2 m / s$ ). Riya tez hai.
4. Graphs do baar cross karte hain. Road par ek rider doosre ko kitni baar overtake karta hai?
  do baarek baarkabhi nahi
  Solution dikhao
  Ek baar. $t = 10 m i n$ par dono market se $1500 m$ par hain aur tez chalne wali Riya, Kabir se aage nikal jaati hai. Doosra crossing, lagbhag $16.7 m i n$ par, woh jagah hai jahan Kabir ki line Riya ki flat line se milti hai: Riya pehle hi ghar pahunch chuki hai aur Kabir bas uske ghar ke saamne se guzar raha hai.
DiagramEasyNCERT Exercise 2.3
Ek doodhwala subah 6:00 baje ghar se nikalta hai aur ek seedhi road par $12 k m / h$ se cycle chala kar $3.0 k m$ door dairy tak jaata hai. Woh wahan 30 min rukta hai, phir $18 k m / h$ se ghar lautta hai. Sahi scales chuno aur uski trip ka $x$ – $t$ graph banao.
Solution dikhao
AnswerCorners (6:00, 0), (6:15, $3.0 k m$ ), (6:45, $3.0 k m$ ) aur (6:55, 0) par.
Jaate hue: $3.0 / 12 = 0.25 h = 15 m i n$ , isliye woh 6:15 par dairy pahunchta hai aur 6:45 par nikalta hai. Wapsi: $3.0 / 18 = \frac{1}{6} h = 10 m i n$ , 6:55 par ghar.
Table 2.6
Time $x$ (km)
6:00 0
6:15 3.0
6:45 3.0
6:55 0
Scale: 5 min ke liye 1 cm aur 0.5 km ke liye 1 cm. Points ko seedhi lines se jodo: slope $12 k m / h$ ke saath upar jaati line, rukne ke dauraan flat, phir slope $- 18 k m / h$ ke saath zyada steep girti line.
NumericalMediumNCERT Exercise 2.4
Ek seedhe track par ek toy robot ek pattern dohraata hai: 6 steps aage, phir 4 steps peeche. Har step $0.5 m$ lamba hai aur $1 s$ leta hai. Track start se $6.0 m$ aage ek gap par khatam hota hai. $x$ – $t$ graph ka sketch banao aur nikaalo ki robot gap tak kab pahunchta hai.
s
Solution dikhao
Har cycle $10 s$ ka hai: $3.0 m$ aage aur $2.0 m$ peeche, yaani net $1.0 m$ . $x$ – $t$ graph ek zigzag hai: 6 s tak slope $+ 0.5 m / s$ wali upar jaati lines aur 4 s tak slope $- 0.5 m / s$ wali neeche aati lines, har peak pichhle se $1.0 m$ ooncha.
$n$ poore cycles ke baad robot $n$ metre par hai, aur agle forward run mein woh $(n + 3)$ metre tak pahunchta hai. $6.0 m$ wala gap pehli baar tab aata hai jab $n + 3 = 6$ : yaani 3 cycles (30 s) ke baad plus 6 forward steps.
$t = 30 + 6 = 36 s$
Graph par peaks ( $6$ , $16$ , $26$ , $36 s$ par $3$ , $4$ , $5$ , $6 m$ ) pehli baar $36 s$ par $6.0 m$ ko chhoote hain. $6.0 m$ ko net $0.1 m / s$ se divide karne par galat jawab $60 s$ aata.
5
NumericalEasyNCERT Exercise 2.5
Ek aircraft $288 k m / h$ par zameen chhoota hai aur rukne se pehle runway par $1.6 k m$ chalta hai. Uniform deceleration maan kar nikaalo:
1. deceleration ka size
  m/s²
  Solution dikhao
  $v_{0} = 288 / 3.6 = 80 m / s$ . $0 = v_{0}^{2} - 2 ∣ a ∣ x$ se:
  $∣ a ∣ = \frac{v_{0}^{2}}{2 x} = \frac{6400}{3200}$
  $∣ a ∣ = 2.0 m / s^{2}$
2. rukne mein lagne wala time
  s
  Solution dikhao
  $t = v_{0} / ∣ a ∣ = 80 / 2.0 = 40 s$ . Check: $\frac{1}{2} (80 + 0) (40) = 1600 m$ .
6
Short answerMediumNCERT Exercise 2.6
Ek patthar $24.5 m / s$ par vertically upar phenka jaata hai aur usi height par wapas pakda jaata hai. Air resistance ko ignore karo; $g = 9.8 m / s^{2}$ lo.
1. Patthar sabse upar ek pal ke liye rukta hai. Kaun sa pair use wahan describe karta hai?
  $v = 0$ , $a = 0$ $v = 24.5 m / s$ upar, $a = 0$ $v = 0$ , $a = 9.8 m / s^{2}$ upar $v = 0$ , $a = 9.8 m / s^{2}$ neeche
  Answer dekho
  Answer (D) $v = 0$ , $a = 9.8 m / s^{2}$ neeche
  Patthar ek pal ke liye rukta hai, lekin gravity ab bhi lag rahi hai: $a = g$ , neeche ki taraf, poore samay.
2. Origin ko flight ke sabse upar wale point par rakho aur clock wahin shuru karo ( $t = 0$ ), upar ko positive lekar. Top se pehle ( $t < 0$ ) aur uske baad ( $t > 0$ ) $x$ , $v$ aur $a$ ke kya signs hain?
  Solution dikhao
  Table 2.7
  $x$ $v$ $a$
  Top se pehle − + −
  Top ke baad − − −
  Flight ka har doosra point top se neeche hai, isliye dono hisson mein $x < 0$ . Upar jaate hue patthar positive direction mein chalta hai ( $v > 0$ ); neeche aate hue $v < 0$ . Gravity poore samay neeche ki taraf hai: $a = - 9.8 m / s^{2}$ . (Neeche ko positive lene par har sign ulat jaata.)
3. Woh kitna upar jaata hai, aur phenkne ke kitni der baad pakda jaata hai?
  Solution dikhao
  $h = \frac{v_{0}^{2}}{2 g} = \frac{{24.5}^{2}}{19.6}$
  $h \approx 30.6 m$
  Top tak ka time: $v_{0} / g = 2.5 s$ . Neeche aane mein bhi utna hi time lagta hai, isliye woh phenkne ke $5.0 s$ baad pakda jaata hai.
7
Short answerMediumNCERT Exercise 2.7
Seedhi line par chal rahe particle ke baare mein har statement true hai ya false? Koi reason ya example do.
1. Jo particle ek pal ke liye rest mein hai, us instant par uska acceleration zero hi hona chahiye.
  SahiGalat
  Answer dekho
  Answer Galat
  False. Flight ke top par ball ka $v = 0$ hai lekin $a = g$ .
2. Agar kisi instant par particle ki speed zero hai, to us instant par uski velocity bhi zero hai.
  SahiGalat
  Answer dekho
  Answer Sahi
  True. Speed $∣ v ∣$ hai, aur $∣ v ∣ = 0$ sirf tab jab $v = 0$ .
3. Agar koi particle ek interval ke har second mein same path length cover karta hai, to us interval mein uski velocity constant rehti hai.
  SahiGalat
  Answer dekho
  Answer Galat
  False. Do boards ke beech smooth ice par slide karta puck har second barabar distance cover karta hai, lekin har elastic bounce uski velocity ulat deta hai, jiske liye ek chhote pal ke liye bada acceleration chahiye.
4. Jis particle ki velocity aur acceleration dono negative hain, woh zaroor dheema ho raha hoga.
  SahiGalat
  Answer dekho
  Answer Galat
  False. Same signs ka matlab speed badhna: upar positive lene par, girte patthar ka $v < 0$ aur $a = - 9.8 m / s^{2}$ hai, aur woh tez hota jaata hai. Dheema hone ke liye ulte signs chahiye.
8
DiagramHardNCERT Exercise 2.8
Ek ball $20 m$ se ek sakht floor par giraayi jaati hai. Har bounce par woh floor ko us speed ke 80 % se chhodti hai jis speed se woh aayi thi. $g = 10 m / s^{2}$ lo aur har bounce ko instantaneous maano. $t = 0$ se $t = 8 s$ tak uska speed–time graph sketch karo.
1. Ball pehli baar floor se kab takraati hai, aur us waqt woh kitni tez hai?
  Solution dikhao
  $20 = \frac{1}{2} (10) t^{2}$ se, $t = 2.0 s$ ; phir $v = g t = 20 m / s$ .
2. Woh doosri baar floor se kab takraati hai?
  s
  Solution dikhao
  Woh $0.8 \times 20 = 16 m / s$ par rebound karti hai, $1.6 s$ upar jaati hai aur $1.6 s$ neeche aati hai: doosra impact $2.0 + 3.2 = 5.2 s$ par, phir se $16 m / s$ par.
3. $t = 8 s$ tak speed–time graph describe karo.
  Solution dikhao
  Rebound speeds: $16$ , $12.8$ , $10.24 m / s$ . Flights $3.2 s$ aur phir $2.56 s$ ki hain, isliye impacts $2.0$ , $5.2$ aur $7.76 s$ par aate hain.
  Table 2.8
  t (s) speed (m/s) event
  0 0 chhodi gayi
  2.0 20 → 16 bounce 1
  3.6 0 top, 12.8 m
  5.2 16 → 12.8 bounce 2
  6.48 0 top, 8.19 m
  7.76 12.8 → 10.24 bounce 3
  8.0 7.84 upar ja rahi
  In points ke beech speed har second $10 m / s$ badalti hai, har baar upar jaate hue zero tak girti hai aur neeche aate hue badhti hai. Graph chhote hote 'V' shapes ki ek line hai, jismein har bounce par achanak giraavat hai.
9
Long answerMediumNCERT Exercise 2.9
Ek lift ground floor se shuru hoti hai, 8th floor ( $32 m$ upar) tak jaati hai aur phir 3rd floor ( $12 m$ upar) tak neeche aati hai, kul $40 s$ leti hai. Iski madad se in dono ka fark samjhao:
1. trip ke liye displacement ka size aur path length;
  Solution dikhao
  Upar positive lene par displacement $+ 12 m$ hai, lekin lift $32 + 20 = 52 m$ chalti hai. General roop se path length chhote displacements ke sizes ka sum hai, $\sum ∣ Δ x_{i} ∣$ , jabki displacement unka signed sum hai, $\sum Δ x_{i}$ . Sum ka size kabhi sizes ke sum se zyada nahi hota, isliye path length $\geq ∣ Δ x ∣$ . Dono tabhi barabar hote hain jab har $Δ x_{i}$ ka sign same ho: yaani motion kabhi ulta na ho.
2. poori trip ke liye average velocity ka size aur average speed.
  Solution dikhao
  Dono ko same $Δ t = 40 s$ se divide karo: $∣ \overset{ˉ}{v} ∣ = 12 / 40 = 0.30 m / s$ , jabki average speed $52 / 40 = 1.3 m / s$ hai. Kyunki path length $\geq ∣ Δ x ∣$ , isliye average speed $\geq ∣ \overset{ˉ}{v} ∣$ , aur barabar sirf us motion ke liye jo wapas na mude.
10
NumericalMediumNCERT Exercise 2.10
Meera ghar se ek seedhi road par $1.8 k m$ door park gate tak $9.0 k m / h$ se jog karti hai. Gate band milne par woh turant mudti hai aur $6.0 k m / h$ se paidal ghar lautti hai. In intervals ke liye uski average velocity ka magnitude aur average speed nikaalo:
1. 0 se 12 min
  Solution dikhao
  Woh $1.8 / 9.0 = 0.20 h = 12 m i n$ par gate pahunchti hai, ab tak koi reversal nahi: $∣ \overset{ˉ}{v} ∣$ aur average speed dono $9.0 k m / h$ hain.
2. 0 se 30 min
  Solution dikhao
  Wapas paidal aane mein $1.8 / 6.0 = 0.30 h = 18 m i n$ lagte hain, isliye woh 30 min par ghar par hai. Displacement zero hai, isliye $∣ \overset{ˉ}{v} ∣ = 0$ ; average speed $3.6 k m / 0.50 h = 7.2 k m / h$ hai. 3.6 km ghoomne ke baad uski average speed zero kehna bekaar baat hogi, isiliye average speed path length use karti hai.
3. 0 se 20 min (average speed enter karo)
  km/h
  Solution dikhao
  Gate ke baad woh 8 min chalti hai, $6.0 \times 8 / 60 = 0.80 k m$ cover karti hai, isliye woh ghar se $1.0 k m$ door hai. $20 m i n = \frac{1}{3} h$ mein: $∣ \overset{ˉ}{v} ∣ = 3.0 k m / h$ ; uska path $2.6 k m$ hai, isliye average speed $7.8 k m / h$ hai.
Short answerMediumNCERT Exercise 2.11
Finite interval mein average speed average velocity ke size se zyada ho sakti hai. Samjhao ki kisi bhi ek instant par speed aur velocity ka size hamesha barabar kyun hote hain.
Solution dikhao
Average speed $∣ \overset{ˉ}{v} ∣$ se sirf isliye zyada hoti hai kyunki object interval ke dauraan mud sakta hai, jisse path length badhti hai par displacement nahi. Jab $Δ t \to 0$ , to mudne ka time hi nahi hota, isliye interval mein path length $∣ Δ x ∣$ ke barabar ho jaati hai (turning point par dono $Δ t$ se tez shrink hote hain). $Δ t$ se divide karke limit lene par instantaneous speed $∣ d x / d t ∣ = ∣ v ∣$ ke barabar hoti hai.
12
Quick answerMediumNCERT Exercise 2.12
Inmein se kaun se graphs aisi cheez dikhate hain jo seedhi line par chalne wala koi bhi particle nahi kar sakta? Jo bhi sahi hon, sab chuno.
Fig. 2.16
Chaar chhote graphs. (a) Time ke against position: right ki taraf khulne wale tedhe U jaisa curve, isliye kuch times par do positions hain. (b) Time ke against speed: ek wave jo time axis ke neeche chali jaati hai. (c) Time ke against path length: badhti hai aur phir flat rehti hai. (d) Time ke against position: ek jhuki hui line jiske beech mein vertical jump hai, phir ek aur jhuki hui line.
(a)(b)(c)(d)
Saare sahi options chuno
Answer dekho
Answer
(A) (a)
(B) (b)
(D) (d)
(a) ek hi instant par do positions deta hai. (b) negative speed dikhata hai, lekin speed ek magnitude hai. (d) zero time mein ek position se doosri position par jump karta hai, yaani infinite velocity. (c) possible hai: path length badhti hai, phir jab particle rest mein hai to fixed rehti hai.
Short answerMediumNCERT Exercise 2.13
Figure ek particle ka $x$ – $t$ graph dikhata hai: $t < 0$ ke liye ek seedhi line aur $t \geq 0$ ke liye curve $x = 4 t - t^{2}$ (SI units). Ek student kehta hai ki particle $t = 0$ se pehle seedhe path par aur uske baad curved path par chalta hai. Kya yeh sahi hai? Koi asli situation batao jise yeh graph describe kar sakta hai.
Fig. 2.17
Minus 2 se 4 second tak time ke against metres mein position. Negative times ke liye graph ek seedhi line hai jo 4 metre per second ke slope se upar jaati hai aur t = 0 par origin tak pahunchti hai. Uske baad woh ek curve par chalta hai jo dheere aur dheere badhta hua 2 second par 4 metre ke peak tak jaata hai, phir 4 second par wapas 0 tak girta hai.
Solution dikhao
Nahi. Particle poore samay $x$ -axis ke along chalta hai; graph sirf yeh dikhata hai ki $x$ , $t$ ke saath kaise badalta hai. $t = 0$ se pehle slope constant hai, $v = 4 m / s$ . Uske baad $v = 4 - 2 t$ aur $a = - 2 m / s^{2}$ : particle dheema hota hai, $t = 2 s$ par $x = 4 m$ par rukta hai aur wapas aata hai, $t = 4 s$ par $x = 0$ se guzarta hai. $4 m / s$ par chalti ek remote-controlled car, jise $t = 0$ par $2 m / s^{2}$ ke steady backward push par switch kiya jaaye, aisa hi karegi: woh brake karti hai, rukti hai aur phir reverse hoti hai.
NumericalMediumNCERT Exercise 2.14
Ek film shoot ke dauraan, $36 k m / h$ par chalti camera jeep usi road par $108 k m / h$ se aage daudti stunt car par aage ki taraf ek paintball fire karti hai. Paintball jeep ke relative $80 m / s$ par marker se nikalti hai. Air resistance ko ignore karke, woh car se kis speed se takraati hai? (Car se dekhi gayi paintball ki speed do, kyunki wahi tay karti hai ki woh kitni zor se lagti hai.)
m/s
Solution dikhao
Aage positive hai. Jeep $36 k m / h = 10 m / s$ ; car $108 k m / h = 30 m / s$ . Ground ke relative paintball: $10 + 80 = 90 m / s$ . Car ke relative: $90 - 30 = 60 m / s$ .
15
Short answerMediumNCERT Exercise 2.15
Aisa asli motion batao jo har graph (a)–(c) bana sake.
Fig. 2.18
Paas-paas teen chhote graphs. (a) Time ke against velocity: zero, phir achanak ek badi value tak vertical jump, phir seedhi line mein wapas zero tak giraavat, jiske baad woh zero hi rehti hai. (b) Time ke against position: ek curve jo pehle tezi se upar jaata hai, phir dheere aur dheere, aur aakhir mein flat ho jaata hai. (c) Time ke against acceleration: ek constant negative value, sirf ek chhote, oonche positive spike ko chhod kar.
1. Graph (a), time ke against velocity.
  Solution dikhao
  Rest mein padi hockey puck ko stick se maara jaata hai (velocity mein lagbhag turant jump), phir woh khurdari zameen par slide karti hai, steady rate se dheemi hoti hai jab tak ruk na jaaye, aur phir rest mein rehti hai.
2. Graph (b), time ke against position.
  Solution dikhao
  Brake laga kar rukti car, ya carpet par lurhak kar rukti ball: position pehle tezi se badhti hai, phir dheere, aur aakhir mein fixed ho jaati hai jab slope (velocity) gir kar zero ho jaata hai.
3. Graph (c), time ke against acceleration.
  Solution dikhao
  Freely girti ball (upar positive, $a = - g$ ) jo sakht floor se ek baar bounce karti hai: chhote se contact ke dauraan floor use zor se upar dhakelta hai, jisse bada positive spike banta hai, aur uske baad phir se $a = - g$ .
16
Short answerMediumNCERT Exercise 2.16
Figure ek line par oscillate karte particle ka $x$ – $t$ graph dikhata hai (simple harmonic motion, jo tum Class 11 mein aage padhoge): $x = A \cos (π t / 2)$ , jahan $t$ seconds mein hai. Har time par $x$ , $v$ aur $a$ ke signs batao.
Fig. 2.19
Minus 4 se 4 second tak time ke against position: amplitude A wali ek smooth cosine wave. t = 0 par yeh apne maximum A par hai, plus aur minus 1 second par zero cross karti hai, plus aur minus 2 second par minus A tak pahunchti hai, plus aur minus 3 second par phir zero cross karti hai aur plus aur minus 4 second par wapas A par aati hai.
1. $t = 0.5 s$
  $x > 0, v < 0, a < 0$ $x > 0, v > 0, a < 0$ $x < 0, v < 0, a > 0$ $x < 0, v > 0, a > 0$
  Answer dekho
  Answer (A) $x > 0, v < 0, a < 0$
  Axis ke upar ( $x > 0$ ) aur neeche ja raha ( $v < 0$ ); $a = - (π / 2)^{2} x < 0$ .
2. $t = 1.5 s$
  $x > 0, v < 0, a < 0$ $x > 0, v > 0, a < 0$ $x < 0, v < 0, a > 0$ $x < 0, v > 0, a > 0$
  Answer dekho
  Answer (C) $x < 0, v < 0, a > 0$
  Axis ke neeche aur ab bhi neeche ja raha: $x < 0$ , $v < 0$ , isliye $a > 0$ .
3. $t = - 0.5 s$
  $x > 0, v < 0, a < 0$ $x > 0, v > 0, a < 0$ $x < 0, v < 0, a > 0$ $x < 0, v > 0, a > 0$
  Answer dekho
  Answer (B) $x > 0, v > 0, a < 0$
  Axis ke upar aur $t = 0$ wale peak ki taraf upar ja raha: $x > 0$ , $v > 0$ , $a < 0$ .
17
Short answerMediumNCERT Exercise 2.17
Figure ek particle ka $x$ – $t$ graph dikhata hai, jismein teen barabar time intervals shaded hain. Graph se jawab do.
Fig. 2.20
Ek smooth position-time curve. Shaded interval 1 mein curve moderate slope se upar jaata hai. Baad mein woh ek peak tak pahunchta hai aur phir, shaded interval 2 mein, bahut tezi se time axis ke paar neeche girta hai. Axis ke neeche ek trough ke baad woh phir halke se upar jaata hai, aur shaded interval 3 mein sirf dheere-dheere chadhta hai.
1. Kis interval mein average speed sabse zyada hai?
  interval 1interval 2interval 3
  Answer dekho
  Answer (B) interval 2
  Interval 2 ka slope sabse steep hai, isliye same time mein sabse bada |Δx|.
2. Kis interval mein average speed sabse kam hai?
  interval 1interval 2interval 3
  Answer dekho
  Answer (C) interval 3
  Interval 3 ka slope sabse halka hai.
3. Intervals 1, 2 aur 3 mein average velocity ke signs kya hain?
  $+, -, +$ $+, +, -$ $-, +, +$ $+, -, -$
  Answer dekho
  Answer (A) $+, -, +$
  $x$ interval 1 mein badhta hai, 2 mein ghatta hai aur 3 mein badhta hai. Curve kisi bhi interval ke andar wapas nahi mudta, isliye har average speed $∣ \overset{ˉ}{v} ∣$ ke barabar hai.
18
Short answerHardNCERT Exercise 2.18
Ek particle kabhi apni direction nahi badalta; us direction ko positive kaho. Neeche uski speed ka time ke against graph hai, jismein barabar width ki teen shaded strips aur chaar marked instants P, Q, R aur S hain.
Fig. 2.21
Ek smooth speed-time curve jo time axis ke upar hi rehta hai. Yeh P par ek halke minimum tak girta hai, shaded interval 1 mein tezi se badh kar Q par ek gol maximum tak jaata hai, shaded interval 2 mein Q ke theek baad halka girta hai, phir shaded interval 3 mein R se hote hue zyada tezi se girta hai, ek trough mein flat hota hai, aur S par ek doosre, neeche wale gol maximum tak badhta hai.
1. Sign ko ignore karke, kis shaded interval mein average acceleration sabse zyada hai?
  interval 1interval 2interval 3
  Answer dekho
  Answer (A) interval 1
  Interval 1 mein speed sabse zyada badalti hai, ek steep rise ke saath.
2. Kis interval mein average speed sabse zyada hai?
  interval 1interval 2interval 3
  Answer dekho
  Answer (B) interval 2
  Interval 2 curve ke sabse ooche hisse ke paas hai.
3. Har interval mein $v$ aur $a$ ke signs batao.
  Solution dikhao
  Teeno mein $v > 0$ , kyunki motion hamesha positive direction mein hai. Interval 1 mein $a > 0$ (speed badh rahi hai) aur intervals 2 aur 3 mein $a < 0$ (speed ghat rahi hai, 2 mein dheere aur 3 mein tezi se).
4. P, Q, R aur S par acceleration kya hai?
  Solution dikhao
  P (minimum), Q aur S (maxima) par tangent horizontal hai, isliye $a = 0$ . R par curve tezi se gir raha hai, isliye wahan $a$ negative hai, zero nahi.
Quick answerMedium
Ek particle $x$ -axis ke along $x = 3 t^{3} - 9 t + 2$ (SI units) ke hisaab se chalta hai. Pehle $2 s$ mein woh jo distance travel karta hai, woh hai
$6 m$ $12 m$ $18 m$ $24 m$
Answer dekho
Answer (C) $18 m$
$t = 1 s$ par $v = 9 t^{2} - 9 = 0$ , jahan particle wapas mudta hai. $x (0) = 2$ , $x (1) = - 4$ , $x (2) = 8 m$ , isliye distance $6 + 12 = 18 m$ hai. Displacement sirf $6 m$ hai.
Quick answerEasy
Assertion (A): Ek body ki ek hi instant par velocity zero aur acceleration non-zero ho sakta hai. Reason (R): Acceleration is par depend karta hai ki velocity kitni tezi se badal rahi hai, velocity ki value par nahi.
A aur R dono true hain, aur R, A ko sahi explain karta haiA aur R dono true hain, lekin R, A ko explain nahi kartaA true hai lekin R false haiA false hai lekin R true hai
Answer dekho
Answer (A) A aur R dono true hain, aur R, A ko sahi explain karta hai
Vertical throw ke top par $v = 0$ hota hai jabki $d v / d t = - g$ . R theek yahi reason deta hai.
Quick answerMedium
Assertion (A): Vertically upar phenki gayi ball ka apni flight ke sabse ooche point par acceleration $g$ , neeche ki taraf, hota hai. Reason (R): Sabse ooche point par ball ek pal ke liye rest mein hoti hai, isliye wahan us par koi force nahi lagta.
A aur R dono true hain, aur R, A ko sahi explain karta haiA aur R dono true hain, lekin R, A ko explain nahi kartaA true hai lekin R false haiA false hai lekin R true hai
Answer dekho
Answer (C) A true hai lekin R false hai
A true hai. R false hai: gravity har instant par lagti hai, top par bhi. Ek pal ke liye rest mein hone se force khatam nahi hota.
NumericalMedium
Tower ke top se giraya gaya patthar apne girne ke aakhri second mein $25 m$ cover karta hai. Tower ki height nikaalo ( $g = 10 m / s^{2}$ ).
m
Answer dekho
Answer 45 m
$5 (2 n - 1) = 25$ se $n = 3$ milta hai, isliye girna $3 s$ chalta hai aur $h = \frac{1}{2} (10) (3)^{2} = 45 m$ .
NumericalEasy
Ek car rest se $10 s$ mein uniformly accelerate hokar $20 m / s$ tak pahunchti hai, $20 s$ tak usi speed par chalti hai, phir uniformly brake karke $5 s$ mein ruk jaati hai. Uske $v$ – $t$ graph ke neeche ke area se total distance nikaalo.
m
Answer dekho
Answer 550 m
$\frac{1}{2} (10) (20) + 20 (20) + \frac{1}{2} (5) (20) = 100 + 400 + 50 = 550 m$ .
Quick answerMedium
Do patthar same height se giraaye jaate hain, doosra pehle ke $1.0 s$ baad. Jab tak dono gir rahe hain, unke beech ki distance
constant rehti haisteady rate se badhti hai $t^{2}$ ki tarah, aur tezi se badhti haighatti hai
Answer dekho
Answer (B) steady rate se badhti hai
Dono $g$ se accelerate karte hain, isliye relative acceleration zero hai aur relative velocity $g \times 1.0 s = 9.8 m / s$ hi rehti hai. Isliye gap time ke saath linearly badhta hai.
25
Quick answerEasy
Har graph ko us motion se match karo jo woh dikhata hai.
Left mein koi item tap karo, phir right mein uska match.
Answer dekho
Answer
$x$ – $t$ : horizontal line : rest mein
$v$ – $t$ : origin se guzarti upar jaati line : rest se, constant positive acceleration
$a$ – $t$ : $- g$ par horizontal line : free fall, upar positive
$x$ – $t$ : seedhi line, negative slope : constant velocity, negative direction mein
Pehle vertical axis wali quantity padho: flat x–t line ka matlab x nahi badal raha; origin se guzarti seedhi v–t line ka matlab v zero se steadily badh raha hai.
NumericalHard
Ek particle ka acceleration $a = 6 t$ (SI units) hai aur $t = 0$ par uski velocity $2 m / s$ hai. $t = 2 s$ par uski velocity nikaalo.
m/s
Answer dekho
Answer 14 m/s
$v = 2 + \int_{0}^{t} 6 t^{'} d t^{'} = 2 + 3 t^{2}$ , isliye $v (2) = 14 m / s$ . $a (2) = 12$ ke saath $v = v_{0} + a t$ use karne par galat jawab $26 m / s$ aata.
Quick answerEasy
$36 k m / h$ par chalti car ko $8.0 m$ mein roka ja sakta hai. Same braking ke saath, $72 k m / h$ par uski stopping distance hai
$64 m$ $32 m$ $24 m$ $16 m$
Answer dekho
Answer (B) $32 m$
$d_{s} \propto v_{0}^{2}$ : speed dugni karne par $8.0 m$ 4 se multiply ho jaata hai.

Time	$x$ (km)
6:00	0
6:15	3.0
6:45	3.0
6:55	0

	$x$	$v$	$a$
Top se pehle	−	+	−
Top ke baad	−	−	−

t (s)	speed (m/s)	event
0	0	chhodi gayi
2.0	20 → 16	bounce 1
3.6	0	top, 12.8 m
5.2	16 → 12.8	bounce 2
6.48	0	top, 8.19 m
7.76	12.8 → 10.24	bounce 3
8.0	7.84	upar ja rahi

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 2.1	Practice Q1
NCERT Exercise 2.2	Practice Q2
NCERT Exercise 2.3	Practice Q3
NCERT Exercise 2.4	Practice Q4
NCERT Exercise 2.5	Practice Q5
NCERT Exercise 2.6	Practice Q6
NCERT Exercise 2.7	Practice Q7
NCERT Exercise 2.8	Practice Q8
NCERT Exercise 2.9	Practice Q9
NCERT Exercise 2.10	Practice Q10
NCERT Exercise 2.11	Practice Q11
NCERT Exercise 2.12	Practice Q12
NCERT Exercise 2.13	Practice Q13
NCERT Exercise 2.14	Practice Q14
NCERT Exercise 2.15	Practice Q15
NCERT Exercise 2.16	Practice Q16
NCERT Exercise 2.17	Practice Q17
NCERT Exercise 2.18	Practice Q18

Quick check: MCQs

Position, displacement aur average velocity: ek recap3 questions
2.2 Instantaneous velocity aur speed3 questions
2.3 Acceleration4 questions
2.4 Uniformly accelerated motion ke liye kinematic equations4 questions
2.5 Relative velocity3 questions

Glossary · 15

Key terms (English + Hindi)

accelerationत्वरणtvaran: Time ke saath velocity ke badalne ki rate: average $Δ v / Δ t$ , instantaneous $d v / d t$ ; SI unit $m s^{- 2}$ .Chapter mein pehli baar yahan aaya →
average speedऔसत चालausat chaal: Total path length divided by lagne wala time; kabhi average velocity ke size se kam nahi.Chapter mein pehli baar yahan aaya →
average velocityऔसत वेगausat veg: Displacement divided by lagne wala time, $\overset{ˉ}{v} = Δ x / Δ t$ ; x–t graph par chord ka slope.Chapter mein pehli baar yahan aaya →

displacementविस्थापनvisthapan: Position mein badlaav, $Δ x = x_{2} - x_{1}$ ; iska sign chune gaye axis ke along direction batata hai.Chapter mein pehli baar yahan aaya →

free fallमुक्त पतनmukt patan: Sirf gravity ke under motion, air resistance ko neglect karke; Earth ke paas acceleration $g \approx 9.8 m / s^{2}$ neeche ki taraf hota hai.Chapter mein pehli baar yahan aaya →

instantaneous velocityतात्क्षणिक वेगtatkshanik veg: Woh value jiski taraf average velocity jaati hai jab time interval zero tak chhota hota hai, $v = d x / d t$ ; graph mein, x–t graph par tangent ka slope.Chapter mein pehli baar yahan aaya →

kinematicsशुद्धगतिकीshuddhgatiki: Motion (position, velocity, acceleration) ka description, yeh pooche bina ki use kaun cause karta hai.Chapter mein pehli baar yahan aaya →

path lengthपथ-लंबाईpath-lambai: Route ke along asal mein travel ki gayi total distance; kabhi negative nahi aur kabhi displacement ke size se kam nahi.Chapter mein pehli baar yahan aaya →
point objectबिंदु वस्तुbindu vastu: Aisi body jiska size, study kiye gaye time mein uski chali hui distance ke muqaable negligible ho, isliye use ek single point maana ja sakta hai.Chapter mein pehli baar yahan aaya →

reaction time: Kisi signal ko notice karne aur us par respond karna shuru karne ke beech ki deri.Chapter mein pehli baar yahan aaya →
rectilinear motionसरल रेखीय गतिsaral rekhiya gati: Ek hi seedhi line ke along motion, isliye ek akela coordinate $x$ position bata deta hai.Chapter mein pehli baar yahan aaya →
relative velocityसापेक्ष वेगsapeksh veg: Ek object ki velocity jo doosre object se naapi jaaye; line ke along, $v_{B A} = v_{B} - v_{A}$ .Chapter mein pehli baar yahan aaya →

speedचालchaal: Instantaneous velocity ka magnitude, $∣ v ∣$ .Chapter mein pehli baar yahan aaya →
stopping distance: Brakes lagne ke pal se rukne tak vehicle jo distance cover karta hai; steady braking ke liye $v_{0}^{2} / 2 ∣ a ∣$ .Chapter mein pehli baar yahan aaya →

uniformly accelerated motionएकसमान त्वरित गतिeksaman tvarit gati: Aisa motion jiska acceleration size aur direction mein same rehta hai, isliye uska v–t graph seedha hota hai.Chapter mein pehli baar yahan aaya →

2.1 Introduction

Position, displacement aur average velocity: ek recap

2.2 Instantaneous velocity aur speed

2.3 Acceleration

2.4 Uniformly accelerated motion ke liye kinematic equations

2.5 Relative velocity

Apna NCERT exercise question dhoondo

Quick check: MCQs

Tum chapter ke end tak pahunch gaye