Describing Motion Around Us (Class 9): Notes और MCQ Practice
Class 9 Science • Class 9
आसान भाषा में:
जब कोई object समय के साथ किसी fixed reference point के respect में अपनी position बदलता है, तो वह motion में होता है। Straight line motion सबसे simple motion है। इसमें हम यह मापते हैं कि कोई चीज कितनी दूर गई (distance), अपनी शुरुआत से असल में कितनी हटी (displacement), कितनी तेजी से चली (speed/velocity), और उसकी रफ्तार कितनी तेजी से बदली (acceleration)।
Key Points & Formulas:
- Reference Point (Origin 'O'): किसी object की position बताने के लिए चुना गया fixed point।
- Distance vs Displacement:
- Distance: कुल तय किया गया path length (दिशा की जरूरत नहीं)।
- Displacement: Initial और final position के बीच का net change (magnitude + direction दोनों जरूरी)। SI unit: metre (m).
- Displacement का magnitude हमेशा distance से कम या उसके बराबर होता है (Displacement ≤ Distance)।
- Average Speed: Total distance travelled ÷ Time interval (SI unit: m s⁻¹ या m/s).
- Average Velocity (vav): Displacement ÷ Time interval = s ÷ t (SI unit: m s⁻¹).
- Average Acceleration (a): Rate of change of velocity = (v − u) ÷ (t₂ − t₁). SI unit: m s⁻².
- Speeding up: Acceleration velocity की दिशा में होता है।
- Slowing down: Acceleration velocity की उल्टी दिशा में होता है (negative sign).
- Earth की gravity में नीचे गिरती वस्तु का acceleration constant होता है: g = 9.8 m s⁻².
Memory Tricks:
- "Displacement = Direct Short-cut": Distance पूरा चक्कर है, पर Displacement सिर्फ initial से final point तक का 'direct' फासला है।
- "Brake माने Minus": Accelerator दबाओगे तो a और v 'साथ-साथ' (same direction, positive); Brake दबाओगे तो a 'उलटा' (opposite direction, negative)!
Common Board-Exam Mistakes:
- Distance और Displacement को एक समझना: अगर कोई जाकर वापस उसी point पर आ जाए, तो Displacement = 0 m होता है, लेकिन Distance zero नहीं होता!
- Units convert न करना: Speed अगर km h⁻¹ में दी हो और time seconds में, तो पहले km h⁻¹ को m s⁻¹ में बदलें (1 km h⁻¹ = 1000 m ÷ 3600 s)!
Describing Motion Around Us
1. Chapter in Short
This chapter introduces the fundamental concepts needed to describe motion in our everyday world. Motion is the change in the position of an object over time with respect to a fixed reference point. The chapter focuses on the simplest type of motion—linear motion (motion in a straight line). It explains the clear differences between distance and displacement, speed and velocity, and introduces the concept of acceleration (how fast velocity changes). It also shows how motion can be represented using numbers, formulas, and graphs.
2. Key Concepts Explained Simply
A. Reference Point and Position
- To say whether an object is moving or where it is located, we always need a fixed starting point called the reference point (or origin 'O').
- The position of an object is described by both its distance from the reference point and its direction.
- Motion vs. Rest: If an object's position changes with time relative to the reference point, it is in motion. If its position does not change, it is at rest.
- Everyday Example: An athlete running on a straight track starting from a line marked '0 m'. The starting line is the reference point. Points to the right are taken as positive (+), and points to the left as negative (−).
B. Instant of Time vs. Time Interval
- An instant of time is a single clock reading at a specific moment (e.g., at 10 seconds).
- A time interval is the duration between two clock readings (e.g., from 4 seconds to 10 seconds, the interval is 10 − 4 = 6 seconds).
C. Distance Travelled vs. Displacement
- Distance travelled: The actual total path length covered by an object during its motion. It needs only a numerical value (magnitude) and unit, not direction.
- Displacement: The net change in position of an object between two instants of time (the shortest straight-line distance from the initial position to the final position, along with direction).
- Everyday Example: If an athlete runs 100 m forward from point O to point A, and then runs 60 m back to point B (which is 40 m from O):
- Total distance travelled = 100 m + 60 m = 160 m
- Displacement = 40 m in the positive direction (from O to B).
- Important Rule: Total distance and the magnitude of displacement are equal only if an object moves in a straight line in one direction without turning back.
D. Average Speed and Average Velocity
- Average Speed: Tells how fast an object moves regardless of direction. It is the total distance covered divided by the time taken.
- Average Velocity: Tells how fast the position of an object changes and in which direction. It is the displacement divided by the time interval.
- Everyday Example: If a swimmer swims 25 m to the other end of a pool and returns 25 m back to the starting point in 50 s:
- Total distance = 50 m → Average speed = 50 m ÷ 50 s = 1 m/s
- Displacement = 0 m → Average velocity = 0 m ÷ 50 s = 0 m/s
E. Uniform and Non-Uniform Motion
- Uniform motion in a straight line: An object covers equal distances in equal intervals of time (moves at a constant speed).
- Non-uniform motion in a straight line: An object covers unequal distances in equal intervals of time (moves with increasing or decreasing speed).
F. Average Acceleration
- Acceleration measures how quickly the velocity of an object changes.
- If an object speeds up, the acceleration is in the direction of velocity.
- If an object slows down (like when brakes are applied), the acceleration is opposite to the direction of velocity (represented with a minus sign).
- Key Fact: An object moving at a constant high velocity has zero acceleration because its velocity is not changing at all.
3. Important Definitions, Formulas, Facts, and Dates
Definitions
- Linear Motion: Motion of an object in a straight line.
- Reference Point: A fixed point used to describe the location and motion of an object.
- Displacement: The net change in the position of an object between two given instants of time.
- Average Speed: The total distance travelled divided by the time interval.
- Average Velocity: The displacement (change in position) divided by the time interval.
- Average Acceleration: The change in velocity divided by the time interval.
- Rate of Change: The ratio of change in one physical quantity to the corresponding change in time.
Important Formulas
- Average Speed:
Average speed = Total distance travelled ÷ Time interval
- Average Velocity:
vav = Displacement ÷ Time interval = s ÷ t
- Average Acceleration:
a = (Final velocity − Initial velocity) ÷ Time interval = (v − u) ÷ (t2 − t1)
- Unit Conversion:
1 km/h = 1000 m ÷ 3600 s = 5 ÷ 18 m/s
Units
- Distance and Displacement: Metre (m)
- Speed and Velocity: Metre per second (m s⁻¹ or m/s); also measured in km h⁻¹
- Acceleration: Metre per second squared (m s⁻² or m/s²)
Historical Facts
- The concept of speed as distance divided by time is found in ancient Indian treatises, including the Aryabhatiya (5th century CE) and the mathematical text Ganitakaumudi (14th century CE).
- Acceleration of an object falling freely under Earth's gravity near the surface is constant and denoted by g = 9.8 m s⁻².
4. Solved Examples (Board-Exam Style)
Example 1 (Ancient Indian Problem - Ganitakaumudi)
Question: Two postmen start walking towards each other from a distance of 210 yojanas. One travels 9 yojanas per day and the other covers 5 yojanas per day. In how many days will they meet each other?
Answer:
- Combined distance covered by both postmen in one day = 9 yojanas + 5 yojanas = 14 yojanas.
- Total distance to be covered = 210 yojanas.
- Time taken = Total distance ÷ Combined speed per day
- Time = 210 ÷ 14 = 15 days.
- Therefore, they will meet each other after 15 days.
Example 2 (Average Speed vs. Average Velocity)
Question: A family drives 200 km north in 3 hours. Afterwards, they drive 200 km south back to their starting position in 2 hours. Calculate their average speed and average velocity for the entire trip.
Answer:
- Total distance travelled = 200 km + 200 km = 400 km
- Total time taken = 3 h + 2 h = 5 h
- Average speed = Total distance ÷ Total time = 400 km ÷ 5 h = 80 km/h
- Since they returned to the starting position, net displacement = 0 km
- Average velocity = Displacement ÷ Total time = 0 km ÷ 5 h = 0 km/h
Example 3 (Acceleration and Deceleration)
Question: A bus is travelling along a straight highway at 36 km/h. The driver accelerates for 10 s, increasing the velocity to 54 km/h. Later, the driver applies brakes and the bus stops in 5 s. Find the average acceleration in both situations.
Answer:
- Part (i): When accelerating
- Initial velocity (u) = 36 km/h = 36 × (1000 ÷ 3600) = 10 m/s
- Final velocity (v) = 54 km/h = 54 × (1000 ÷ 3600) = 15 m/s
- Time (t) = 10 s
- a = (v − u) ÷ t = (15 − 10) ÷ 10 = 5 ÷ 10 = 0.5 m s⁻² (in the direction of velocity)
- Part (ii): When brakes are applied
- Initial velocity (u) = 15 m/s
- Final velocity (v) = 0 m/s (bus comes to a stop)
- Time (t) = 5 s
- a = (v − u) ÷ t = (0 − 15) ÷ 5 = −15 ÷ 5 = −3 m s⁻²
- The negative sign shows that acceleration acts opposite to the direction of motion.
Example 4 (Free Fall Acceleration)
Question: An object is dropped from rest from a height. Its velocity after 1 s is 9.8 m/s, after 2 s is 19.6 m/s, and after 3 s is 29.4 m/s. Calculate the acceleration between 1 s and 2 s, and between 2 s and 3 s. Is it constant?
Answer:
- Acceleration between 1 s and 2 s:
a = (19.6 − 9.8) ÷ (2 − 1) = 9.8 ÷ 1 = 9.8 m s⁻²
- Acceleration between 2 s and 3 s:
a = (29.4 − 19.6) ÷ (3 − 2) = 9.8 ÷ 1 = 9.8 m s⁻²
- Yes, the acceleration is constant and equals 9.8 m s⁻², directed vertically downward (acceleration due to gravity, g).
5. Common Mistakes and Quick Revision Points
- Mistake: Confusing distance with displacement.
- Correction: Distance is always positive and cannot decrease as an object moves. Displacement can be positive, negative, or zero.
- Mistake: Thinking high speed means high acceleration.
- Correction: Acceleration depends only on the rate of change of velocity. A train moving at a constant speed of 120 km/h on a straight track has zero acceleration.
- Mistake: Forgetting to convert units to SI units.
- Correction: If speed is given in km/h and time in seconds, convert km/h to m/s by multiplying by (1000 ÷ 3600) before using acceleration formulas.
- Displacement Magnitude vs. Distance: Magnitude of displacement is always less than or equal to the total distance travelled (it is never greater).
- When are they equal? Distance = magnitude of displacement, and average speed = magnitude of average velocity only when an object moves along a straight line in a single direction without reversing.
Practice MCQ (15)
औसत त्वरण का SI मात्रक क्या है?
- (A) m s⁻¹
- (B) m s⁻²
- (C) km h⁻¹
- (D) m s
उत्तर: (B)
पाठ्यपुस्तक 'समय के एक क्षण' और 'समय अंतराल' में किस प्रकार भेद करती है?
- (A) An instant of time is a duration between two readings, while a time interval is a single clock reading.
- (B) An instant of time is a single reading of a clock at a given point of time, whereas a time interval is the duration between two instants.
- (C) Both terms mean exactly the same thing and can be used interchangeably.
- (D) An instant of time requires specifying direction, while a time interval does not.
उत्तर: (B)
एक सरल रेखा में गति के लिए, किस स्थिति में तय की गई कुल दूरी और विस्थापन का परिमाण बराबर होते हैं?
- (A) Only when the object comes to rest at the origin
- (B) When the object moves without turning back, that is, in one direction
- (C) When the average acceleration is strictly zero
- (D) Only when the object moves with increasing speed
उत्तर: (B)
14वीं शताब्दी ईस्वी के किस प्राचीन भारतीय गणितीय ग्रंथ में दो डाकियों के एक-दूसरे की ओर चलने पर उनके मिलने का समय ज्ञात करने की समस्या शामिल है?
- (A) Aryabhatiya
- (B) Ganitakaumudi
- (C) Siddhanta Shiromani
- (D) Brahmasphutasiddhanta
उत्तर: (B)
किसी वाहन के स्पीडोमीटर का पाठ्यांक किसी दिए गए क्षण पर लगभग किस भौतिक राशि को दर्शाता है?
- (A) Average acceleration over the entire trip
- (B) Total displacement from the origin
- (C) Magnitude of the velocity at that instant
- (D) Average speed divided by the total time taken
उत्तर: (C)
एक एथलीट मूल बिंदु O (0 m) से दौड़ना शुरू करता है, 100 m पर बिंदु A तक पहुँचता है, और फिर उसी रेखा पर वापस दौड़ते हुए 40 m पर बिंदु B तक आता है। तय की गई कुल दूरी और प्रारंभिक बिंदु से विस्थापन क्या है?
- (A) Distance = 100 m, Displacement = 40 m
- (B) Distance = 160 m, Displacement = 40 m in the positive direction
- (C) Distance = 140 m, Displacement = 60 m in the negative direction
- (D) Distance = 160 m, Displacement = 160 m in the positive direction
उत्तर: (B)
गणितकौमुदी के उदाहरण 4.1 में, दो डाकिया एक-दूसरे से 210 योजन की दूरी से क्रमशः 9 योजन प्रति दिन और 5 योजन प्रति दिन की चाल से एक-दूसरे की ओर चलना शुरू करते हैं। उनके मिलने से पहले पहला डाकिया कितने योजन की दूरी तय करता है?
- (A) 75 yojanas
- (B) 105 yojanas
- (C) 135 yojanas
- (D) 150 yojanas
उत्तर: (C)
एक पारिवारिक यात्रा के दौरान, एक कार 3 घंटे में 200 km उत्तर की ओर चलती है, और फिर 2 घंटे में 200 km दक्षिण की ओर चलती है। पूरी यात्रा के लिए औसत चाल और औसत वेग क्या हैं?
- (A) Average speed = 80 km h⁻¹, Average velocity = 0 km h⁻¹
- (B) Average speed = 0 km h⁻¹, Average velocity = 80 km h⁻¹
- (C) Average speed = 100 km h⁻¹, Average velocity = 40 km h⁻¹ north
- (D) Average speed = 80 km h⁻¹, Average velocity = 40 km h⁻¹ south
उत्तर: (A)
NCERT अध्याय के क्रियाकलाप 4.1 के आधार पर, विस्थापन और तय की गई कुल दूरी के संबंध में निम्नलिखित में से कौन सा कथन सदैव सत्य है?
- (A) Displacement is never zero.
- (B) The magnitude of displacement can be greater than the total distance travelled.
- (C) The magnitude of displacement is less than or equal to the total distance travelled.
- (D) The magnitude of displacement is strictly less than the total distance travelled in all cases.
उत्तर: (C)
एक सीधे राजमार्ग पर 36 km h⁻¹ के वेग से चल रही एक बस 54 km h⁻¹ का वेग प्राप्त करने के लिए 10 s तक त्वरित होती है। इस समयांतराल के दौरान इसका औसत त्वरण क्या है?
- (A) 1.8 m s⁻²
- (B) 0.5 m s⁻²
- (C) 5.0 m s⁻²
- (D) 0.05 m s⁻²
उत्तर: (B)
क्रियाकलाप 4.1 में, एक गेंद को मूल बिंदु O से लंबवत ऊपर की ओर अधिकतम ऊँचाई B (140 cm पर) तक फेंका जाता है, और यह वापस O से 80 cm ऊपर बिंदु C पर गिरती है। O से C तक गेंद द्वारा तय की गई कुल दूरी और विस्थापन क्या हैं?
- (A) Distance = 80 cm, Displacement = 80 cm upward
- (B) Distance = 200 cm, Displacement = 80 cm upward
- (C) Distance = 200 cm, Displacement = 60 cm downward
- (D) Distance = 140 cm, Displacement = 80 cm downward
उत्तर: (B)
एक बस 54 km h⁻¹ की चाल से चलती है। चालक एक बाधा देखता है और ब्रेक लगाता है, जिससे बस 5 s में पूरी तरह रुक जाती है। बस का औसत त्वरण क्या है, और इसका चिह्न क्या दर्शाता है?
- (A) 3 m s⁻², acting in the direction of velocity
- (B) −3 m s⁻², acting opposite to the direction of velocity
- (C) −10.8 m s⁻², acting in the direction of velocity
- (D) −0.6 m s⁻², acting opposite to the direction of velocity
उत्तर: (B)
एक वस्तु को विरामावस्था से लंबवत नीचे गिराया जाता है। इसका वेग t = 2 s पर 19.6 m s⁻¹ और t = 4 s पर 39.2 m s⁻¹ है। इस 2-सेकंड के समयांतराल में औसत त्वरण का परिमाण क्या है?
- (A) 4.9 m s⁻²
- (B) 9.8 m s⁻²
- (C) 19.6 m s⁻²
- (D) 29.4 m s⁻²
उत्तर: (B)
एक कार एक सीधी सड़क पर 120 km h⁻¹ के अत्यधिक उच्च नियत वेग से चलती है। इसका औसत त्वरण क्या है, और क्यों?
- (A) It is high because velocity is 120 km h⁻¹.
- (B) It is zero because acceleration depends on the rate of change of velocity, not on how fast an object moves.
- (C) It is positive and equal to 12 m s⁻² due to high momentum.
- (D) It cannot be determined without knowing the total distance travelled.
उत्तर: (B)
सारंग 25 m लंबे एक स्विमिंग पूल में एक छोर से दूसरे छोर तक तैरता है और 50 s में प्रारंभिक बिंदु पर वापस आ जाता है। उसकी औसत चाल अशून्य (non-zero) क्यों है जबकि उसका औसत वेग शून्य है?
- (A) Because total distance travelled is 0 m, while displacement is 50 m.
- (B) Because his speed changed continuously while his direction remained identical.
- (C) Because total distance travelled is 50 m (yielding 1 m s⁻¹), but net change in position is 0 m (yielding 0 m s⁻¹).
- (D) Because velocity requires specifying time in hours rather than seconds.
उत्तर: (C)