📘 Class 9 Science – Chapter 7: Motion – Solutions (NCERT Based)

🔹 In-Text Questions & Solutions

📍 Page 74

Q1. An object has moved through a distance. Can it have zero displacement?

✅ Answer: Yes. If an object moves and returns to its original position, its displacement is zero.
📝 Example: Walking 5 m forward and then 5 m backward. Distance = 10 m, Displacement = 0 m.

Q2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the displacement at the end of 2 minutes 20 seconds?

✅ Solution:
Side = 10 m → Perimeter = 40 m
Total time = 140 s → Rounds = 140 / 40 = 3.5
➡ 3 rounds → displacement = 0
➡ Half round = diagonal = √(10² + 10²) = √200 ≈ 14.14 m
✅ Answer: Displacement = 14.14 m

Q3. Which of the following is true for displacement?
(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.

✅ Answer: ❌ Both are incorrect.
✔️ Displacement can be zero and is always ≤ distance.

📍 Page 76

Q1. Distinguish between speed and velocity.

Feature	Speed	Velocity
Definition	Distance per unit time	Displacement per unit time
Quantity	Scalar	Vector
Can be zero?	No	Yes

Q2. When is magnitude of average velocity equal to average speed?

✅ Answer: When the object moves in a straight line without changing direction.

Q3. What does an odometer measure?

✅ Answer: It measures the distance travelled by a vehicle.

Q4. What does the path of an object look like when it is in uniform motion?

✅ Answer: A straight line.

Q5. A signal from a spaceship reaches ground in 5 min. What is the distance if speed = 3 × 10⁸ m/s?

✅ Solution:
Time = 5 min = 300 s
Distance = 3 × 10⁸ × 300 = 9 × 10¹⁰ m

📍 Page 77

Q1. When is a body in:
(i) Uniform acceleration?
(ii) Non-uniform acceleration?

✅ Answer:
(i) When velocity changes equally in equal time intervals (e.g., free fall)
(ii) When velocity changes unequally (e.g., car in traffic)

Q2. A bus decreases speed from 80 km/h to 60 km/h in 5 s. Find acceleration.

✅ Solution:
Convert to m/s: 80 km/h = 22.22, 60 km/h = 16.67
a = (v – u) / t = (16.67 – 22.22)/5 = -1.11 m/s²

Q3. A train starts from rest, attains 40 km/h in 10 minutes. Find acceleration.

✅ Solution:
v = 40 km/h = 11.11 m/s, t = 600 s
a = (11.11 – 0)/600 = 0.0185 m/s²

🔹 Chapter-End Exercise Solutions

Q1. Circular track of 200 m diameter. In 2 min 20 sec (140 s), how much distance and displacement?

✅ Solution:
Radius r = 100 m → Circumference = 2πr ≈ 628.57 m
Rounds = 140 / 40 = 3.5
➡ Distance = 3.5 × 628.57 = 2200 m
➡ Displacement = diameter = 200 m

Q2. Joseph jogs A→B (300 m in 150 s), then 100 m back to C in 60 s.

(a) From A to B:
Speed = 300/150 = 2 m/s, Displacement = 300 m → Velocity = 2 m/s

(b) From A to C:
Total Distance = 400 m, Time = 210 s → Speed = 400/210 = 1.90 m/s
Displacement = 200 m → Velocity = 200/210 = 0.95 m/s

Q3. Abdul goes at 20 km/h, returns at 30 km/h. Find average speed.

✅ Solution:
Let one-way distance = D
Average Speed = 2D / (D/20 + D/30) = 24 km/h

Q4. Motorboat at 3 m/s² for 8 s. Find distance covered.

✅ Solution:
u = 0, a = 3, t = 8
s = ut + ½ at² = 0 + 0.5×3×64 = 96 m

Q5. Car speed-time graph (no image). Distance = area under graph.
(b) Uniform motion = straight horizontal line on graph

Q6. From Fig 7.10 (not shown):
(a) Fastest = Object with steepest graph (e.g., A)
(b) If all intersect, then yes
(c)(d) Use graph intersection points to find distances

Q7. Ball falls from 20 m, a = 10 m/s². Find velocity & time.

✅ Solution:
v² = u² + 2as = 0 + 2×10×20 = 400 → v = 20 m/s
v = u + at → 20 = 0 + 10t → t = 2 s

Q8. Speed-time graph (not shown):
(a) Distance = Area under graph (use triangle or trapezium area)
(b) Uniform motion = flat horizontal portion

Q9. State whether situations are possible:

(a) Constant acceleration, zero velocity → Yes (e.g., top of upward throw)
(b) Acceleration with uniform speed → Yes (e.g., circular motion)
(c) Acceleration ⊥ to direction → Yes (e.g., satellite in orbit)

Q10. Satellite radius = 42250 km, time = 24 hrs. Find speed.

✅ Solution:
r = 4.225×10⁷ m, T = 86400 s
v = 2πr / T = 2×3.14×4.225×10⁷ / 86400 ≈ 3070 m/s

📝 Prepared by: Vinayak Study Junction Beawar

Class 9 Science Chapter 7 – Motion (Solutions in Hindi)

कक्षा 9 विज्ञान – अध्याय 7: गति (Motion)

NCERT प्रश्नों के उत्तर और संख्यात्मक प्रश्न हल सहित

अध्याय के बीच में पूछे गए प्रश्न

Q1. क्या कोई वस्तु दूरी तय करके भी विस्थापन शून्य रख सकती है?

हाँ, अगर वस्तु प्रारंभिक बिंदु से चलकर फिर उसी बिंदु पर लौट आती है, तो विस्थापन शून्य होगा। उदाहरण के लिए, कोई व्यक्ति 100 मीटर जाकर फिर लौटकर वापस आ जाता है, तब उसकी कुल दूरी 200 मीटर होगी पर विस्थापन 0 होगा।

Q2. एक किसान एक वर्ग के आकार के खेत की परिधि पर चलता है जिसकी प्रत्येक भुजा 10 मीटर है और वह 2 मिनट 20 सेकंड में खेत के चारों ओर घूमता है। प्रारंभिक बिंदु से उसका विस्थापन कितना होगा?

पूरा समय = 2 मिनट 20 सेकंड = 140 सेकंड
प्रत्येक चक्कर में विस्थापन = 0 (क्योंकि वह वहीं लौट आता है)
यदि उसने 1 चक्कर = 40 सेकंड में लिया हो, तो 140 सेकंड में = 3.5 चक्कर
इसका मतलब है कि वह चौथे चक्कर के आधे रास्ते (विपरीत कोने) पर है।
इसलिए विस्थापन = विकर्ण = √(10² + 10²) = √200 = 14.14 मीटर

Q3. विस्थापन के बारे में कौन सा कथन सही है?

(a) यह शून्य नहीं हो सकता – गलत
(b) इसकी परिमाण दूरी से अधिक होती है – गलत
(c) विस्थापन शून्य हो सकता है जबकि दूरी शून्य नहीं होती है – सही

संख्यात्मक प्रश्न हल (Examples)

उदाहरण 7.1: कोई वस्तु 4 सेकंड में 16 मीटर और फिर 2 सेकंड में 16 मीटर चलती है। औसत गति ज्ञात कीजिए।

कुल दूरी = 16 + 16 = 32 मीटर
कुल समय = 4 + 2 = 6 सेकंड
औसत गति = कुल दूरी / कुल समय = 32 / 6 = 5.33 m/s

उदाहरण 7.3: उषा 90 मीटर लंबे तालाब में एक मिनट में 180 मीटर तैरती है। औसत गति और औसत वेग ज्ञात कीजिए।

दूरी = 180 मीटर, विस्थापन = 0 (वह वहीं वापस आती है)
औसत गति = 180/60 = 3 m/s, औसत वेग = 0/60 = 0 m/s

उदाहरण 7.4: राहुल अपनी साइकिल 0 से 6 m/s तक 30 सेकंड में गति करता है, फिर ब्रेक लगाकर उसे 4 m/s पर 5 सेकंड में लाता है। दोनों अवस्थाओं में त्वरण ज्ञात करें।

पहला भाग:
= 0, v = 6 m/s, t = 30 s
a = (v – u)/t = (6 – 0)/30 = 0.2 m/s²
दूसरा भाग:
u = 6, v = 4, t = 5
a = (4 – 6)/5 = -0.4 m/s²

अध्याय के अंत में अभ्यास प्रश्नों के उत्तर

Q1. व्यास 200 मीटर वाले वृत्ताकार ट्रैक पर एक धावक 2 मिनट 20 सेकंड में कितनी दूरी और विस्थापन तय करेगा?

परिधि = π × D = 3.14 × 200 = 628 मीटर
समय = 2 मिनट 20 सेकंड = 140 सेकंड
चक्कर = 140 / 40 = 3.5 चक्कर
दूरी = 3.5 × 628 = 2198 मीटर
विस्थापन = व्यास = 200 मीटर (क्योंकि आधा चक्कर)

Q4. एक मोटरबोट स्थिर स्थिति से 3 m/s² के त्वरण से 8 सेकंड में कितना चलेगी?

u = 0, a = 3, t = 8
s = ut + (1/2)at² = 0 + (1/2)(3)(64) = 96 मीटर

Q10. 42250 किमी त्रिज्या वाले परिक्रमा पथ पर कृत्रिम उपग्रह 24 घंटे में एक परिक्रमा पूरी करता है। उसकी गति ज्ञात कीजिए।

s = 2πr = 2 × 3.14 × 42250 = 265,220 किमी
v = s/t = 265220 / 24 = 11050.83 km/h

सभी प्रश्नों के उत्तर और सूत्रों को सावधानीपूर्वक जाँचा गया है ताकि विद्यार्थी इन्हें आसानी से समझ सकें।

Class 9 Science Chapter 7 – Motion (Solutions)

Class 9 Science – Chapter 7: Motion

NCERT In-Text and End-Exercise Questions with Solutions

In-Text Questions

Q1. An object has moved through a distance. Can it have zero displacement?

Yes. If the object moves and returns to its original position, the displacement is zero though distance is covered. Example: Walking 100 m forward and then 100 m back — total distance is 200 m, displacement is 0.

Q2. A farmer moves along the boundary of a square field of side 10 m in 2 min 20 s. What will be the magnitude of displacement at the end of 2 min 20 s?

Total time = 140 seconds. Time for one round (perimeter = 40 m): Assume one round = 40 s → 3.5 rounds in 140 s. After 3.5 rounds, the farmer will be at the opposite corner → displacement = diagonal of square = √(10² + 10²) = 14.14 m.

Q3. Which of the following is true for displacement?

(a) It cannot be zero – False
(b) Its magnitude is greater than the distance travelled – False
(c) It is zero or less than or equal to the distance – True

Numerical Questions (Examples)

Example 7.1: An object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed?

Total distance = 32 m, Total time = 6 s → Average speed = 32/6 = 5.33 m/s

Example 7.3: Usha swims in a 90 m pool. She covers 180 m in one minute from one end to the other and back. Find average speed and velocity.

Distance = 180 m, Displacement = 0 → Average speed = 180/60 = 3 m/s, Average velocity = 0/60 = 0 m/s

Example 7.4: Rahul paddles from rest to 6 m/s in 30 s, then brakes to 4 m/s in 5 s. Find acceleration in both cases.

First: a = (6 – 0)/30 = 0.2 m/s²
Second: a = (4 – 6)/5 = -0.4 m/s²

End-Exercise Questions

Q1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement after 2 min 20 s?

Time = 140 s, Time per round = 40 s → 3.5 rounds
Distance = 3.5 × 628 m = 2198 m
Displacement = diameter = 200 m (after half round)

Q4. A motorboat starting from rest on a lake accelerates at 3.0 m/s² for 8.0 s. How far does it travel?

s = ut + ½at² = 0 + ½ × 3 × 64 = 96 m

Q10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to complete one revolution.

Circumference = 2πr = 2 × 3.14 × 42250 = 265220 km
Speed = Distance/Time = 265220 / 24 = 11050.83 km/h

All answers are based on the NCERT Class 9 Science Chapter 7 textbook for accuracy and clarity.