Class 9 Science · Chapter 4
Short answer:
Class 9 Science Chapter 4 — "Describing Motion Around Us" physics ka pehla chapter hai jo motion describe karna sikhata hai: distance vs displacement, speed vs velocity, acceleration, distance-time aur velocity-time graphs, aur teen equations of motion (v = u + at, s = ut + ½at², v² = u² + 2as). Yehi chapter Class 9 science motion chapter formulas ki neev hai jo Class 11 Kinematics tak use hoti hai — isliye numericals pe extra focus zaroori hai.
Chapter 4 "Describing Motion Around Us" poore Class 9 Physics ki neev hai — yahin se distance-displacement, speed-velocity, acceleration, aur equations of motion jaise concepts shuru hote hain jo aage Work-Energy aur Sound chapters se hote hue Class 11 Kinematics tak har jagah use honge. Ye chapter conceptual (definitions, difference-based questions) aur numerical (formula-based calculation) dono tarah ke questions ka mix hai, isliye CBSE marking scheme me har step dikhana zaroori hai — sirf final answer likhne se full marks nahi milte.
Ye chapter Class 9 science new syllabus 2026-27 chapter list ke hisaab se rationalised "Exploration" textbook ka Chapter 4 hai. Neeche di gayi solutions me har in-text aur exercise question ka step-by-step working diya gaya hai, taaki concept clear ho aur calculation bhi CBSE marking-scheme style me match kare.
Chapter 4 Summary — 5 Minute Revision
Is chapter me hum seekhte hain ki kisi bhi object ki motion ko kaise describe karte hain — sabse pehle ek reference point (origin) tay karna padta hai. Distance path ki total length hoti hai (scalar), jabki displacement shortest straight-line path hota hai direction ke saath (vector) — displacement kabhi bhi distance se zyada nahi ho sakta, aur zero bhi ho sakta hai agar object apni starting position par wapas aa jaye.
Speed (distance/time, scalar) aur velocity (displacement/time, vector) me fark uniform circular motion me sabse clear dikhta hai — speed constant rehte hue bhi velocity continuously change hoti hai kyunki direction badalti rehti hai, isliye ye accelerated motion hoti hai.
Acceleration velocity ke change ki rate hai (uniform ya non-uniform). Distance-time graph ka slope speed batata hai, aur velocity-time graph ke neeche ka area distance/displacement batata hai. In dono graphs se hi teeno equations of motion (v = u + at, s = ut + ½at², v² = u² + 2as) derive hoti hain, jo poore chapter ka numerical backbone hain.
In-Text Questions — Solutions
Ek object kisi distance tak move karta hai, lekin kya uska displacement zero ho sakta hai? Ek example ke saath samjhao.
Haan, possible hai. Displacement sirf initial aur final position ke beech ka shortest straight-line distance (direction sahit) hota hai, jabki distance actual path length hai.
Example: Ek athlete 200 m ke circular track ka ek pura chakkar laga kar wapas apne starting point par aa jata hai. Yahan distance covered = 200 m, lekin displacement = 0 m, kyunki initial aur final position same hai.
Ek kisan 15 m side wale square khet ki boundary ke around 50 s me ek chakkar poora karta hai. 4 minute ke baad uska displacement ka magnitude kitna hoga?
Perimeter of square = 4 × 15 = 60 m.
Time for 1 round = 50 s ⇒ 4 min = 240 s
Number of rounds = 240 / 50 = 4.8 rounds
4 poore rounds (200 s) ke baad farmer wapas starting corner A par hota hai (displacement = 0). Bacha hua time = 240 − 200 = 40 s, jisme kisan 40 s × (60/50) m/s = 48 m aur chalta hai.
48 m me 3 poori sides (45 m: A→B→C→D) cover ho jaati hain, aur 4th side (D→A) par 3 m aur chal jata hai. Coordinates: A(0,0), D(0,15), farmer ki final position = (0,12).
Displacement magnitude = 12 m (side AD ke along)
In statements ko check karo: (a) Displacement kabhi bhi distance se zyada nahi ho sakta. (b) Displacement kabhi zero nahi ho sakta jab object move kare.
(a) True — displacement (shortest straight path) hamesha distance (actual path length) se kam ya barabar hota hai, kabhi zyada nahi ho sakta.
(b) False — jaisa upar dikhaya gaya, object move kar sakta hai aur phir bhi displacement zero ho sakta hai agar wo apni starting position par wapas aa jaye (jaise circular track ka ek chakkar).
Speed aur velocity me fark batao.
| Speed | Velocity |
|---|---|
| Scalar quantity (sirf magnitude) | Vector quantity (magnitude + direction) |
| Speed = distance / time | Velocity = displacement / time |
| Kabhi negative nahi hoti | Direction ke hisaab se positive/negative ho sakti hai |
↔ Table ko side me swipe karein
Kis condition me average velocity ka magnitude average speed ke barabar hota hai?
Jab object ek hi straight line me, ek hi direction me (bina direction change kiye) move kare — tab total distance aur displacement ka magnitude equal ho jaate hain, isliye average speed = average velocity ka magnitude.
Car ka odometer kya measure karta hai?
Odometer car dwara tay ki gayi total distance (path length) measure karta hai, displacement nahi.
Uniform motion me object ka path kaisa dikhta hai?
Uniform motion (constant speed, straight line) me object equal time intervals me equal distance cover karta hai, isliye iska path ek straight line hota hai aur distance-time graph bhi straight line hota hai.
Ek ground station tak satellite se signal 4 minute me pahunchta hai. Agar signal ki speed light ki speed (3 × 108 m/s) ke barabar hai, to satellite ki ground station se distance kitni hai?
t = 4 min = 240 s
distance = speed × time = 3 × 108 × 240
distance = 7.2 × 1010 m
Uniform aur non-uniform acceleration me fark batao.
Uniform acceleration: velocity equal time intervals me equal amount se change hoti hai (jaise free-fall me gravity ke under motion).
Non-uniform acceleration: velocity ka change rate equal time intervals me alag-alag hota hai (jaise traffic me chalti car).
Ek bike 5 second me apni speed 90 km/h se ghata kar 60 km/h kar deti hai. Bike ka acceleration nikaalo.
u = 90 km/h = 25 m/s, v = 60 km/h = 16.67 m/s, t = 5 s
a = (v − u) / t = (16.67 − 25) / 5
a = −1.67 m s-2 (retardation)
Ek train station se start ho kar uniform acceleration se 8 minute me 45 km/h ki speed pakadti hai. Uska acceleration nikaalo.
u = 0, v = 45 km/h = 12.5 m/s, t = 8 min = 480 s
a = (v − u) / t = 12.5 / 480
a ≈ 0.026 m s-2
Uniform aur non-uniform motion ke liye distance-time graph ka nature kaisa hota hai?
Uniform motion ke liye distance-time graph ek straight line hota hai (constant slope). Non-uniform motion ke liye graph ek curve hota hai, kyunki speed lagatar change ho rahi hoti hai.
Agar distance-time graph time-axis ke parallel straight line hai, to object ki motion ke baare me kya bata sakte ho?
Object rest (stationary) me hai — time badh raha hai lekin position/distance change nahi ho raha, matlab speed = zero.
Agar speed-time graph time-axis ke parallel straight line ho, to kya matlab hai?
Object uniform (constant) speed se move kar raha hai — speed change nahi ho rahi, isliye acceleration = zero.
Velocity-time graph ke neeche jo area hoti hai, wo kya represent karti hai?
Vah area object dwara us time interval me tay ki gayi distance (ya displacement) ko represent karti hai.
Ek bus rest se start ho kar 0.2 m s-2 ke uniform acceleration se 1.5 minute chalti hai. (a) Final speed (b) distance covered nikaalo.
u = 0, a = 0.2 m s-2, t = 90 s
(a)
v = u + at = 0 + 0.2 × 90 = 18 m/s
(b)
s = ut + ½at² = 0 + ½ × 0.2 × 90²
s = 0.1 × 8100 = 810 m
Uniform circular motion kya hai? Ek example do.
Jab koi object circular path par constant speed se move kare, lekin direction continuously change hoti rahe, use uniform circular motion kehte hain. Ismein velocity constantly change hoti hai (kyunki direction badalti hai), isliye ye accelerated motion hai chahe speed constant ho.
Example: Ek stone ko dori se bandh kar circular path me ghumana.
Ek athlete 200 m circumference wale circular track ka ek chakkar 40 s me poora karta hai. Uski speed nikaalo.
distance = circumference = 200 m, time = 40 s
speed = distance / time = 200 / 40 = 5 m/s
Exercise Questions — Solutions (Q1–Q13)
Ek athlete 16 m radius wale circular track ka ek chakkar 40 s me poora karta hai. 2 minute 20 second ke baad uska (a) distance covered aur (b) displacement kya hoga?
circumference = 2πr = 2 × 3.14 × 16 = 100.48 m
total time = 2 min 20 s = 140 s
number of rounds = 140 / 40 = 3.5 rounds
(a) Distance:
distance = 3.5 × 100.48 = 351.68 m
(b) Displacement:
3.5 rounds ke baad athlete track ke diametrically opposite point par hota hai, isliye displacement = diameter.
displacement = 2r = 2 × 16 = 32 m
Ek athlete 200 m straight track ko 24 s me tay karta hai. Uski average speed nikaalo.
average speed = total distance / total time = 200 / 24
average speed ≈ 8.33 m/s
Ek motorboat nadi ko ek taraf 24 km straight-line distance 2 h me paar karta hai, aur usi route se wapas 3 h me aata hai. Poore trip ke liye (a) average speed aur (b) average velocity nikaalo.
total distance = 24 + 24 = 48 km
total time = 2 + 3 = 5 h
(a) Average speed:
average speed = 48 / 5 = 9.6 km/h
(b) Average velocity:
Boat wapas apni starting position par aa jaata hai, isliye total displacement = 0.
average velocity = 0 / 5 = 0 km/h
Ek object ki distance-time reading di gayi hai: t(s) = 0, 2, 4, 6, 8 aur x(m) = 0, 4, 8, 12, 16. Batao ki motion uniform hai ya non-uniform, iski speed nikaalo, aur batao graph ka shape kaisa hoga.
Har 2 s me distance 4 m se badh raha hai — ek constant rate se, isliye ye uniform motion hai.
speed = 4 m / 2 s = 2 m/s (constant)
Distance-time graph origin (0,0) se guzarne wali ek straight line hoga, jiska slope = 2 m/s.
Ek car ka odometer reading drive shuru hone se pehle 2000 km tha, aur 4 hours ki drive ke baad 2200 km ho gaya. Car ki average speed km/h aur m/s dono me nikaalo.
distance = 2200 − 2000 = 200 km, time = 4 h
speed = 200 / 4 = 50 km/h
m/s me convert karne ke liye:
50 km/h = 50 × (1000/3600) = 13.89 m/s
Ek motorcycle rest se start ho kar uniform acceleration se 20 s me 8 m/s ki speed pakadti hai, phir 8 m/s ki constant speed se 20 s aur chalti hai, aur uske baad uniform retardation se 5 s me ruk jaati hai. Poore safar me tay ki gayi total distance nikaalo.
Phase 1 (accelerating): u = 0, v = 8 m/s, t = 20 s
distance1 = ½(u+v)t = ½(0+8)(20) = 80 m
Phase 2 (constant speed): v = 8 m/s, t = 20 s
distance2 = 8 × 20 = 160 m
Phase 3 (retarding): u = 8 m/s, v = 0, t = 5 s
distance3 = ½(8+0)(5) = 20 m
Total distance:
total = 80 + 160 + 20 = 260 m
Ek bus 100 km/h ki speed se chal rahi hai. Brakes lagne par uniform retardation 0.5 m s-2 lagta hai. Bus ko rukne me kitna time aur kitni distance lagegi?
u = 100 km/h = 27.78 m/s, v = 0, a = −0.5 m s-2
Time:
t = (v − u)/a = (0 − 27.78)/(−0.5) = 55.56 s
Distance (v² = u² + 2as):
0 = (27.78)² − 2(0.5)s
s = 771.6 / 1 = 771.6 m
Ek trolley, inclined plane par 2 cm/s initial velocity ke saath, 3 cm s-2 ke uniform acceleration se neeche jaa rahi hai. Start ke 4 s baad iski velocity kya hogi?
u = 2 cm/s, a = 3 cm s-2, t = 4 s
v = u + at = 2 + (3 × 4)
v = 14 cm/s
Ek racing car 5 m s-2 ke uniform acceleration se start (rest se) hoti hai. 8 s me ye kitni distance cover karegi?
u = 0, a = 5 m s-2, t = 8 s
s = ut + ½at² = 0 + ½(5)(64)
s = 160 m
Ek ball 8 m/s ki velocity se vertically upar throw ki jaati hai. Gravity ke kaaran acceleration 9.8 m s-2 neeche ki taraf lagta hai. Ball kitni height tak jayegi aur waha pahunchne me kitna time lagega?
Time to max height (v = 0 at highest point):
v = u + at ⇒ 0 = 8 − 9.8t
t = 8/9.8 ≈ 0.82 s
Height:
v² = u² − 2gh ⇒ 0 = 64 − 2(9.8)h
h = 64/19.6 ≈ 3.27 m
Kya uniform circular motion ek accelerated motion hoti hai? Reason ke saath explain karo.
Haan, uniform circular motion accelerated motion hoti hai, kyunki velocity ek vector quantity hai (magnitude + direction). Circular motion me speed (magnitude) constant rehta hai, lekin direction continuously badalti rehti hai — isliye velocity continuously change hoti hai, jiska matlab hai object ka acceleration (centripetal acceleration) non-zero hai.
True/False batao, reason ke saath: "Ek object jo non-zero acceleration se move kar raha hai, uski velocity kisi bhi instant par zero nahi ho sakti."
False. Object ki velocity kisi instant par zero ho sakti hai chahe acceleration non-zero ho — jaise ball ko upar throw karne par highest point par uski velocity zero hoti hai, lekin gravity ke kaaran acceleration (−9.8 m s-2) us instant par bhi laga rehta hai.
Ek train rest se start hoti hai aur uniform acceleration se 20 s me 40 m/s ki speed pakadti hai, phir 60 s tak 40 m/s ki constant speed se chalti hai, uske baad 10 s me uniform retardation se ruk jaati hai. Velocity-time graph ka use karke total distance nikaalo.
Phase 1 (triangle area):
area1 = ½ × 20 × 40 = 400 m
Phase 2 (rectangle area):
area2 = 40 × 60 = 2400 m
Phase 3 (triangle area):
area3 = ½ × 10 × 40 = 200 m
Total distance = area under velocity-time graph:
total = 400 + 2400 + 200 = 3000 m = 3 km
Important Equations — Ek Nazar Me
| Quantity | Formula | SI Unit |
|---|---|---|
| Speed | v = distance / time | m/s |
| Average speed | total distance / total time | m/s |
| Velocity | v = displacement / time | m/s |
| Average velocity (uniform acceleration) | (u + v) / 2 | m/s |
| Acceleration | a = (v − u) / t | m s-2 |
| 1st equation of motion | v = u + at | — |
| 2nd equation of motion | s = ut + ½at² | — |
| 3rd equation of motion | v² = u² + 2as | — |
| Uniform circular motion speed | v = 2πr / t | m/s |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Distance aur displacement ko same maan lena — displacement direction-sensitive hota hai aur distance se kam ya barabar hota hai, kabhi zyada nahi.
- Speed aur velocity ko interchangeably use karna — speed scalar hai, velocity vector; circular motion me speed constant hoke bhi velocity change ho sakti hai (isliye acceleration exist karti hai).
- Numerical solve karne se pehle units convert karna bhool jana (jaise km/h ko m/s me convert na karna) — isse final numerical answer galat aata hai.
- Equations of motion (v = u + at, etc.) ko har situation me apply kar dena — ye equations sirf uniform (constant) acceleration ke case me hi valid hoti hain, non-uniform acceleration me directly apply nahi hoti.
- Retardation/deceleration ko positive number treat kar dena, negative acceleration ka sign convention miss kar dena — isse final answer ka sign galat aata hai.
- Velocity-time graph ke area ko hamesha simple distance samajh lena, jabki agar velocity negative ho jaaye (direction reverse), to area ka sign bhi consider karke net displacement nikalna padta hai.
Board-Style Important Questions
- Distance aur displacement me antar spasht karo, ek udaharan ke saath.
- Ek car 0 se 20 m/s ki speed 10 second me pakadti hai (uniform acceleration se). Acceleration aur is dauran tay ki gayi distance nikaalo.
- Uniform circular motion ko accelerated motion kyun kaha jaata hai? Karan sahit samjhao.
- Ek train rest se start hoti hai aur 15 s me uniform acceleration se 30 m/s ki speed pakadti hai, phir 45 s tak constant speed se chalti hai, uske baad 10 s me uniform retardation se ruk jaati hai. Velocity-time graph describe karo aur total distance nikaalo.
- Speed-time graph, jo time-axis ke parallel straight line ho, kis type ki motion darshata hai? Reason do.
Aksar Poochhe Jaane Wale Sawaal
Class 9 Science Chapter 4 "Describing Motion Around Us" me exam ke liye sabse important topics kaunse hain?
Equations of motion ke numericals (v = u + at, s = ut + ½at², v² = u² + 2as), distance vs displacement ka conceptual difference, aur distance-time / velocity-time graphs — ye teen areas sabse zyada scoring aur frequently-asked hote hain. Class 9 science motion chapter formulas ka table ache se yaad rakhna chahiye.
Kya "class 9 science chapter 3 ncert solutions" bhi isi step-by-step CBSE marking-scheme style me milte hain?
Haan, poore Class 9 Science syllabus ke solutions isi tarah step-by-step CBSE marking-scheme style me diye jaate hain, taaki concept aur calculation dono clear ho aur full marks milne me aasani ho.
"Class 9 science exploration book pdf download" kaha se karein — koi official source hai?
Best practice yahi hai ki official NCERT website (ncert.nic.in) se hi free PDF download karo, kyunki wahi authentic aur updated textbook content hota hai. Third-party sites se pirated PDF download avoid karna chahiye.
Class 9 science new syllabus 2026-27 chapter list me total kitne chapters hain?
Rationalised "Exploration" textbook me total 13 chapters hain (Chapter 1 introductory hai, isliye kai third-party sites galti se "12 chapters" bolti hain). Motion wala chapter is list me Chapter 4 hai.
Motion chapter ke baad kaunsa chapter aata hai — kya "work energy and simple machines class 9 numericals" bhi isi tarah important hain?
Haan, Work, Energy, and Simple Machines chapter bhi numerical-heavy hai, khaaskar pulley, lever, aur inclined plane ke numericals — Motion chapter ke concepts (velocity, distance) yahan bhi indirectly use hote hain.
Kya class 9 science all chapters notes pdf ek jagah milenge?
Chapter-wise solutions (jaise ye Chapter 4 spec) ko compile karke ek consolidated notes PDF banaya ja sakta hai — har chapter ke formulas table, in-text/exercise solutions, aur common mistakes ek jagah rakhne se revision fast hoti hai.
Class 9 Science — Saare Chapters
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