Class 11 Physics · Chapter 2
Short answer: "
Class 11 Physics Chapter 2 — Motion in a Straight Line mechanics ka pehla real chapter hai (rationalised 2026-27 syllabus me, jahan purana \"Physical World\" chapter drop ho chuka hai aur Units and Measurements Chapter 1 hai). Yahan se distance, displacement, average/instantaneous speed aur velocity, acceleration, aur teen kinematic equations (v=u+at, s=ut+½at², v²=u²+2as) shuru hoti hain — jo aage Motion in a Plane, Laws of Motion aur Gravitation ke har numerical ki base banti hain. Neeche full step-by-step NCERT-style solutions, formulas table, common mistakes aur exam-important questions diye gaye hain.
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Chapter 1 (Units and Measurements) ne tumhe measurement, significant figures aur errors sikhaye. Ab Chapter 2 — Motion in a Straight Line — se asli physics shuru hoti hai: koi object move kaise karta hai, uski position time ke saath kaise change hoti hai, aur is change ko describe karne ke liye kaunse tools chahiye.
\nYe chapter kinematics ka foundation hai — sirf 1-D (ek straight line) motion tak limited, lekin jo concepts yahan banenge — distance vs displacement, average vs instantaneous velocity, acceleration, position-time aur velocity-time graphs, aur teen kinematic equations — wahi Motion in a Plane (jahan projectile motion aata hai), Laws of Motion, aur Gravitation ke numericals me baar-baar repeat hote hain.
\nAgar tum \"class 11 physics ncert solutions\" ka poora set dhoondh rahe ho: rationalised 2026-27 session ka syllabus 14 chapters ka hai — Units and Measurements, Motion in a Straight Line, Motion in a Plane, Laws of Motion, System of Particles and Rotational Motion (jiske important formulas alag topic hain), Gravitation (numericals ke saath), Mechanical Properties of Solids aur Fluids, Thermodynamics, Kinetic Theory of Gases (derivation-heavy), Oscillations, aur Waves (jisme Doppler Effect ka formula aata hai). Purana \"Physical World\" chapter ab standalone examinable chapter nahi raha — is wajah se Units and Measurements ab Chapter 1 hai aur ye chapter Chapter 2 ban gaya hai.
\nIs page pe niche in-text worked examples, full exercise solutions (step-by-step, marking-scheme style), key formulas ki table, common mistakes, aur important questions milenge — jo class 11 physics chapter 2 motion in a straight line ncert solutions specifically dhoondh rahe students ke liye hai.
\n"Chapter 2 Summary — 5 Minute Revision
"Position ek fixed reference point (origin) se measure hoti hai. Displacement ek vector hai — sirf initial se final position tak ka shortest straight-line change; distance/path length ek scalar hai — total path jo tay kiya gaya. Dono barabar tabhi hote hain jab motion ek hi direction me, bina wapas mude, ho.
\nAverage velocity = total displacement ÷ total time; average speed = total distance ÷ total time — non-uniform ya round-trip motion me ye do alag results dete hain. Instantaneous velocity Δt → 0 hone par average velocity ki limiting value hai, yaani dx/dt.
\nAcceleration velocity ke change ki rate hai (a = dv/dt). Jab acceleration uniform (constant) ho, to teen kinematic equations — v = u + at, s = ut + ½at², v² = u² + 2as — poora motion solve kar deti hain.
\nGraphs: position-time graph ka slope velocity deta hai; velocity-time graph ka slope acceleration deta hai aur us graph ke under ka area displacement deta hai. Relative velocity ek object ki velocity dusre chalte hue object ke reference frame me hoti hai (vAB = vA − vB) — same direction me chal rahe objects ke liye ye subtract hoti hai, opposite direction ke liye add.
\n"In-Text Questions — Solutions
Example A. Ek particle ki position x(t) = 3t² − 2t + 5 (SI units) se di gayi hai. t = 2 s pe iski instantaneous velocity nikaalo.
Concept: Instantaneous velocity = dx/dt (differentiation)
v(t) = dx/dt = 6t − 2
At t = 2 s:
v(2) = 6(2) − 2 = 10 m/s
Example B. Ek aadmi 4 m East chalta hai, phir 3 m West chalta hai. Uski total distance aur displacement nikaalo.
Distance (path length) = 4 + 3 = 7 m (dono legs add hoti hain, direction ignore)
Displacement (East ko positive maan ke) = +4 + (−3) = +1 m East
Notice: distance (7 m) > |displacement| (1 m) kyunki motion me direction change hui.
Example C. Ek aadmi ghar se 2.5 km door market tak 5 km/h ki speed se jaata hai, aur wapas 7.5 km/h ki speed se aata hai. Poore trip ki average speed aur average velocity nikaalo.
Time to market: t₁ = 2.5/5 = 0.5 h
Time back: t₂ = 2.5/7.5 = 1/3 h
Total time = 0.5 + 1/3 = 5/6 h
Total distance = 2.5 + 2.5 = 5 km
Average speed = 5 ÷ (5/6) = 6 km/h
Total displacement = 0 (wapas ghar aa gaya)
Average velocity = 0 ÷ (5/6) = 0 km/h
Ye classic example hai jo dikhata hai ki average speed aur average velocity kitne different ho sakte hain jab motion round-trip ho.
Example D. Ek car 30 m/s se chal rahi thi, brake lagne par 5 s me ruk jaati hai. Iski acceleration nikaalo (deceleration).
Given: u = 30 m/s, v = 0, t = 5 s
a = (v − u)/t = (0 − 30)/5 = −6 m/s²
Negative sign batata hai ki ye deceleration hai (motion ki direction ke opposite acceleration) — magnitude 6 m/s².
Example E. Ek bike rest se 4 m/s² se accelerate hoti hai. 10 s me uski final velocity aur is duration me tay ki gayi distance nikaalo.
Given: u = 0, a = 4 m/s², t = 10 s
Final velocity: v = u + at = 0 + 4×10 = 40 m/s
Distance: s = ut + ½at² = 0 + ½×4×100 = 200 m
Cross-check via v² = u² + 2as: 1600 = 0 + 2×4×s ⟹ s = 200 m ✓
Example F. Do gaadiyan same point se same direction me chalti hain — car P 10 m/s constant velocity se, car Q rest se 2 m/s² acceleration se, dono same time pe start karte hain. Car Q, car P ko kab pakdegi (overtake karne se pehle same position pe aayegi)?
Positions (t seconds baad):
xP = 10t; xQ = 0 + ½ × 2 × t² = t²
Milne ki condition: xP = xQ
10t = t² ⟹ t² − 10t = 0 ⟹ t(t − 10) = 0
t = 0 (start point) ya t = 10 s — ye second solution hai jab Q, P ko pakadti hai.
Poore Class 11 Physics ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q15)
Q1. Distance aur displacement me difference batao. Kya kisi motion me dono equal ho sakte hain?
Distance (path length) scalar hai — object ne total kitna path tay kiya, wo batata hai, kabhi negative nahi hota aur decrease nahi ho sakta. Displacement vector hai — sirf initial se final position tak ka shortest straight-line change, jisme direction (sign) bhi include hota hai.
Dono equal tab hote hain jab object ek hi direction me, bina peeche mude, straight line me move kare — us case me path bhi seedha hota hai aur end-to-end change bhi wahi hota hai. Jaise hi object direction badalta hai (peeche aata hai), distance > |displacement| ho jaata hai.
Q2. Kya displacement zero ho sakta hai jabki distance travelled zero na ho? Example do.
Haan, bilkul. Agar koi object apne starting point pe wapas aa jaaye, to net displacement = 0 hoga chahe usne kitna bhi path cover kiya ho.
Example: Ek runner 400 m ke circular track ka ek poora chakkar laga ke wahi start point pe wapas aata hai. Distance travelled = 400 m (non-zero), lekin displacement = 0 m, kyunki final position = initial position.
Q3. Kya average speed kabhi average velocity ke magnitude ke barabar ho sakti hai? Kab?
Haan, sirf ek special case me: jab motion pura time interval me ek hi direction me ho (path seedha ho aur wapas na aaye). Tab total distance = |total displacement|, isliye:
Average speed = total distance / t = |total displacement| / t = |Average velocity|
Agar object beech me direction badalta hai (jaise aage jaake wapas aana), to average speed hamesha |average velocity| se bada ya barabar hoga, kabhi chhota nahi.
Q4. x-t graph diya gaya hai jisme ek particle pehle positive x-direction me jaata hai, phir ruk ke wapas origin ki taraf aata hai. Is motion ke liye velocity ka sign kaise decide karoge har phase me?
x-t graph me velocity = slope hoti hai (v = dx/dt), instantaneous rate of change of position.
- Jab graph upar ki taraf badh raha ho (x increase ho raha ho time ke saath) → slope positive → velocity positive (aage ki taraf motion).
- Jab graph horizontal (flat) ho → slope zero → velocity zero (object ruka hua hai, momentary rest).
- Jab graph neeche ki taraf jaa raha ho (x decrease ho raha ho) → slope negative → velocity negative (wapas origin ki taraf motion).
Sign hamesha chosen positive direction ke reference me hota hai — isliye graph padhte waqt pehle apna positive-direction convention fix karna zaroori hai.
Q5. Ek sharabi (drunkard) seedhi gali me chal raha hai — har second me 5 kadam aage aur turant 3 kadam peeche leta hai. Har kadam 1 m ka hai aur 1 s lagta hai. Uske ghar se 13 m door ek gaddha (pit) hai. Wo gaddhe me kab girega?
Pattern samjho: Har "cycle" me 8 kadam (8 s) lagte hain, net displacement = 5 − 3 = 2 m aage.
Cycle-wise net position: cycle 1 baad = 2 m, cycle 2 baad = 4 m, cycle 3 baad = 6 m, cycle 4 baad = 8 m.
5th cycle me wo pehle 5 kadam aage leta hai — is point tak position = 8 + 5 = 13 m ho jaati hai. Yaani peeche ke 3 kadam lene se pehle hi wo pit tak pahunch jaata hai.
Time = 4 poore cycles (4 × 8 s = 32 s) + 5 kadam aage (5 s) = 37 s.
Total time = 37 s
Q6. Do cars A aur B ek hi road pe same direction me 72 km/h aur 45 km/h ki uniform speed se chal rahi hain. Car A se dekhne par car B ki relative velocity kitni hai?
Given: vA = 72 km/h = 20 m/s, vB = 45 km/h = 12.5 m/s (dono same direction, positive maan lo)
vBA = vB − vA = 12.5 − 20 = −7.5 m/s
Negative sign batata hai ki car A ke reference frame me, car B peeche ki taraf 7.5 m/s (= 27 km/h) ki speed se ja rahi dikhegi — matlab A, B ko overtake kar raha hai.
Q7. Do trains ek dusre ki taraf opposite directions se 90 km/h aur 54 km/h se aa rahi hain, jab unke beech distance 720 m hai. Kitni der baad woh mil jaayengi?
Given: u1 = 90 km/h = 25 m/s, u2 = 54 km/h = 15 m/s (opposite directions)
Relative velocity (approach) = u1 + u2 = 25 + 15 = 40 m/s
t = distance / relative velocity = 720 / 40 = 18 s
Q8. Ek stone kisi tower ki chhat se rest se drop kiya jaata hai jo 44.1 m unchi hai. (g = 9.8 m/s², air resistance ignore karo) Stone ground pe kitni der me pahunchega aur usse final velocity kya hogi?
Given: u = 0, h = 44.1 m, a = g = 9.8 m/s² (downward positive)
Time: using s = ut + ½at²
44.1 = 0 + ½ × 9.8 × t² ⟹ t² = 9 ⟹ t = 3 s
Final velocity: using v = u + at
v = 0 + 9.8 × 3 = 29.4 m/s
Cross-check: v² = u² + 2as = 0 + 2×9.8×44.1 = 864.36 ⟹ v = 29.4 m/s ✓
Q9. Ek ball ko ground se seedha upar 29.4 m/s ki speed se throw kiya jaata hai. (g = 9.8 m/s²) Maximum height, time to reach it, aur total time of flight nikaalo.
Given: u = 29.4 m/s (upward positive), a = −g = −9.8 m/s², top pe v = 0
Time to max height: v = u + at
0 = 29.4 − 9.8t ⟹ t = 3 s
Max height: v² = u² + 2as
0 = (29.4)² − 2 × 9.8 × h ⟹ h = 864.36 / 19.6 = 44.1 m
Total time of flight (up + down, symmetric motion) = 2 × 3 s = 6 s
Q10. Ek car rest se 2 m/s² ki uniform acceleration se chalti hai. 5th second me wo kitni distance tay karegi?
Distance in nth second ka formula:
sn = u + (a/2)(2n − 1)
Given: u = 0, a = 2 m/s², n = 5
s5 = 0 + (2/2)(2×5 − 1) = 1 × 9 = 9 m
Car 5th second me 9 m cover karegi.
Q11. Ek bus rest se 1 m/s² se uniformly accelerate hoke straight road pe 10 s tak chalti hai, uske baad woh uniform velocity se chalti rehti hai. Bus ki distance-covered-per-second (nth second) ka plot n = 1 se 10 tak accelerated phase me kis type ka curve dega — straight line ya parabola?
Accelerated phase (0 se 10 s) me: sn = u + (a/2)(2n − 1) = (a/2)(2n − 1) [kyunki u = 0]
Ye n ka linear function hai (sn ∝ n, ek straight line jiska form sn = an − a/2 hai), parabola nahi.
Note: total distance x vs t ka graph parabola hota hai (kyunki x = ½at²), lekin per-second-distance vs n ka graph straight line hota hai — dono ko confuse mat karo.
Uniform-velocity phase (10 s ke baad) me sn constant reh jaayega — horizontal line.
Q12. Ek particle ki position x = 2t² − 3t + 4 (SI units) se di gayi hai. Iski velocity aur acceleration nikaalo, aur t = 2 s pe unki value batao.
Velocity: v = dx/dt = 4t − 3
Acceleration: a = dv/dt = 4 m/s² (constant)
At t = 2 s:
v = 4(2) − 3 = 5 m/s; a = 4 m/s²
Q13. v-t graph me ek particle ki velocity 0 s se 6 s tak 0 se 12 m/s tak uniformly badhti hai. Is interval me particle ne kitna displacement tay kiya?
v-t graph ke under ka area = displacement hota hai. Yahan graph ek right-angled triangle hai (base = time, height = velocity):
Displacement = ½ × base × height = ½ × 6 × 12 = 36 m
(Cross-check via kinematics: a = (12−0)/6 = 2 m/s²; s = ut + ½at² = 0 + ½×2×36 = 36 m ✓)
Q14. Ek ladka open-top lift me khada hai jo rest me hai. Wo ball ko upar 29.4 m/s ki max speed se throw kar sakta hai. Ball wapas uske haath me kitni der me aayega? Ab lift 4.9 m/s ki uniform speed se upar move karne lage aur ladka wahi max speed se ball throw kare — is baar ball kitni der me wapas aayega?
Case 1 — Lift rest me: Ball relative to ground: u = 29.4 m/s, a = −g = −9.8 m/s². Total time of flight (up+down):
t = 2u/g = 2×29.4/9.8 = 6 s
Case 2 — Lift uniform velocity se upar move kar raha: Yahan important cheez ye hai ki lift uniform velocity se move kar raha hai, koi acceleration nahi. Isliye lift ek non-accelerating (inertial) frame hai, aur is frame ke andar ball ka motion bilkul waisa hi dikhega jaisa rest wale lift me dikhta tha — kyunki relative velocity of ball w.r.t. boy sirf 29.4 m/s (up) hi hai, lift ki uniform velocity dono (ball aur boy) me common hai aur cancel ho jaati hai.
Isliye ball wapas usi 6 s me uske haath me aayegi — answer nahi badalta, sirf tab badalta jab lift accelerate kar rahi hoti.
Q15. Ek particle apne total time ka first half distance velocity u₁ se aur second half distance velocity u₂ se tay karta hai (same direction me). Poore trip ka average velocity nikaalo, aur dikhao ki ye (u₁+u₂)/2 ke barabar NAHI hota.
Maan lo total distance = 2s, isliye har half = s.
Time for first half: t₁ = s/u₁; time for second half: t₂ = s/u₂
Total time = t₁ + t₂ = s/u₁ + s/u₂ = s(u₁+u₂)/(u₁u₂)
Average velocity = Total distance / Total time = 2s ÷ [s(u₁+u₂)/(u₁u₂)] = 2u₁u₂ / (u₁+u₂)
Ye harmonic mean hai, simple average (u₁+u₂)/2 nahi. (u₁+u₂)/2 formula sirf tab valid hota hai jab uniform acceleration ho aur u₁, u₂ initial-final velocities ho — jab equal time ke liye alag speeds ho, tabhi simple average sahi hota hai, equal distance ke case me nahi.
Important Equations — Ek Nazar Me
"| Concept | Formula / Rule |
|---|---|
| Displacement | Δx = x₂ − x₁ (vector, can be +, −, or 0) |
| Distance / Path length | total path covered (scalar, always ≥ 0, never decreases) |
| Average velocity | vavg = Δx / Δt = (x₂ − x₁) / (t₂ − t₁) |
| Average speed | speedavg = total distance / total time |
| Instantaneous velocity | v = limΔt→0 (Δx/Δt) = dx/dt |
| Average acceleration | aavg = Δv / Δt |
| Instantaneous acceleration | a = dv/dt = d²x/dt² |
| 1st kinematic equation | v = u + at |
| 2nd kinematic equation | s = ut + ½at² |
| 3rd kinematic equation | v² = u² + 2as |
| Distance in nth second | sn = u + (a/2)(2n − 1) |
| Relative velocity (1-D) | vAB = vA − vB |
| Free fall (up positive, drop) | v = −gt, h = ½gt², v² = 2gh |
| Vertical throw upward | time of flight = 2u/g, max height = u²/2g |
| v-t graph meaning | slope = acceleration; area under curve = displacement |
↔ Table ko side me swipe karein
\n"Common Mistakes — Yahan Marks Kat te Hain
- Distance aur displacement ko same maan lena — displacement ek vector hai (sign/direction matter karta hai), distance sirf magnitude hai; non-straight motion me dono kabhi barabar nahi hote.
- Average velocity = (u+v)/2 formula ko HAR situation me laga dena — ye formula sirf uniform (constant) acceleration ke case me valid hai, kisi bhi general motion me nahi.
- Deceleration (braking/upar jaate hue gravity) ke liye acceleration ka negative sign lagana bhool jana — sign convention poori problem me consistent rakhna zaroori hai, warna kinematic equation ka answer galat aayega.
- v-t graph ka slope aur area confuse kar dena — slope hamesha acceleration deta hai, area under the curve hamesha displacement deti hai; dono ko ulta samajh lena bahut common mistake hai.
- Ye maan lena ki speed kabhi bhi average velocity ke magnitude ke barabar hoti hai — ye sirf tab sach hai jab motion pure interval me ek hi direction me ho; direction badalne par average speed hamesha |average velocity| se zyada ya barabar hoti hai, kam kabhi nahi.
- Free-fall/projectile problems me positive direction fix na karna aur beech mein sign switch kar dena (kabhi g positive, kabhi negative) — ek hi problem ke andar convention badalne se displacement aur velocity dono ke sign galat ho jaate hain.
Board-Style Important Questions
- 1 mark: Ek particle ka displacement kisi time interval me zero hai — kya iska matlab hai ki distance travelled bhi zero hoga? Ek example ke saath justify karo.
- 2 marks: Velocity-time graph ka use karke kinematic equation v = u + at derive karo.
- 3 marks: Ek car 20 m/s se chal rahi hai aur uniform braking se 4 s me rok di jaati hai. (a) Acceleration nikaalo (b) Rukne se pehle car ne kitni distance tay ki, nikaalo.
- 3 marks: Do parallel rail tracks pe train A aur train B opposite directions me respectively 54 km/h aur 72 km/h se chal rahi hain. Train A me baithe observer ke liye train B ki velocity nikaalo.
- 5 marks: Ek ball ground se seedha upar 29.4 m/s ki speed se throw kiya jaata hai. Kinematic equations use karke (a) maximum height (b) us tak pahunchne ka time aur (c) total time of flight nikaalo. Iske sath saath is motion ka v-t graph bhi sketch karo.
Aksar Poochhe Jaane Wale Sawaal
Class 11 Physics Chapter 2 Motion in a Straight Line ke NCERT solutions kahan milenge jo poore step-by-step working ke saath hon?
Isi page pe upar in-text examples aur exercise solutions dono full CBSE marking-scheme style step-by-step diye gaye hain — sirf final answer nahi, har step ka reasoning bhi. Ye motion in a straight line ke class 11 physics ncert solutions dhoondh rahe students ke liye directly usable format hai.
Class 11 Physics all chapters PDF 2026-27 session ke liye kahan se lena chahiye?
Sabse reliable source hamesha official ncert.nic.in hai — wahi rationalised 2026-27 session ka current, updated textbook PDF publish karta hai. Third-party PDFs kabhi outdated ya wrong-edition ho sakte hain, isliye chapter list aur content match karne se pehle official site cross-check karna best practice hai.
Motion in a Plane ka projectile motion topic is chapter se kaise connect hota hai?
Motion in a Straight Line sirf 1-D (ek direction) motion cover karta hai. Agla chapter — Motion in a Plane — inhi concepts (displacement, velocity, acceleration) ko 2-D vectors me extend karta hai, jahan projectile motion jaisa combined horizontal+vertical motion aata hai. Isliye ye chapter uska direct foundation hai — jo bhi motion in a plane projectile motion class 11 solutions me karoge, uske base formulas yahin se aate hain.
Class 11 Physics ka chapter-wise weightage CBSE exam ke liye kya hota hai?
CBSE koi fixed, saal-dar-saal lock hui per-chapter weightage officially publish nahi karta — sirf overall unit-wise blueprint sample paper ke saath aata hai jo har saal thoda vary kar sakta hai. Isliye is page pe koi invented number nahi diya gaya hai; exact latest weightage ke liye current session ka official CBSE curriculum document/sample paper hi check karo.
Class 11 Physics ke deleted syllabus 2026-27 me is chapter se kya-kya hata hai?
Rationalised 2026-27 syllabus me sabse bada change ye hai ki purana standalone "Physical World" chapter fully drop ho gaya hai, isliye Units and Measurements ab Chapter 1 hai aur Motion in a Straight Line Chapter 2 ban gaya hai. Is chapter ke andar, kuch relative-velocity ka extended/complex treatment trim kiya gaya hai — core concepts (displacement, velocity, acceleration, kinematic equations) poori tarah retained hain. (Note: ye REGISTRY-grade third-party confirmation hai, official ncert.nic.in PDF se final visual cross-check karna recommended hai numerically precise claims se pehle.)
Chapter 1 Units and Measurements ke important questions is chapter ke numericals me kaise kaam aate hain?
Units and Measurements class 11 important questions me jo significant figures aur dimensional consistency practice hoti hai, wo Motion in a Straight Line ke har numerical me directly use hoti hai — jaise units convert karna (km/h → m/s), ya final answer ko sahi significant figures me round karna. Ye do chapters isliye ek dusre ki practical foundation hain, sirf syllabus order me alag lagte hain.
Class 11 Physics — Saare Chapters
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