Navigating through the newly revised CBSE Class 9 Science curriculum (Exploration) requires a thorough physical and mathematical understanding of kinematics, reference points (origin), scalar versus vector quantities, distance versus displacement, uniform versus non-uniform motion, instantaneous speed and average speed, velocity and average velocity, uniform acceleration and retardation, kinematic graphical analysis (distance-time and velocity-time graphs), graphical derivations of the three equations of motion (v = u + a × t, s = u × t + 1/2 × a × t², and v² – u² = 2 × a × s), and the mechanics of uniform circular motion. Chapter 4 of Class 9 Physics, “Describing Motion Around Us”, establishes the fundamental framework for quantifying how physical bodies change position over time. It investigates why motion is purely relative to an observer’s frame of reference; details why an object moving in a closed circle can have zero net displacement while covering significant distance; proves why uniform circular motion is continuously accelerated despite constant speed; and demonstrates the step-by-step arithmetic of multi-leg journeys, braking vehicles, and orbital satellites. To help students master every aspect of this high-weightage chapter, this comprehensive solutions guide offers textbook-accurate, highly structured, and pedagogically sound responses strictly aligned with the latest CBSE Class 9 evaluation standards.
Every question presented in the official NCERT textbook—ranging from all introductory “Think It Over” sections and in-text “Pause and Ponder” prompts (Pages 48, 52, 55, 58, and 62) to the complete end-of-chapter “Revise, Reflect, Refine” exercises (Questions 1 to 12 on Pages 68–71)—has been solved with exhaustive detail. Numerical problems follow a step-by-step box format with explicit variable legends, standard SI unit conversions (such as multiplying km/h by 5/18 to obtain m/s), and algebraic substitutions using clean plain-text symbols without raw LaTeX tags. Key scoring terms, official CBSE exam tags, and dynamic summary tables have been highlighted to ensure students secure maximum marks in their examinations.
Master Concept & Comparative Summary Tables
1. Master Kinematics Formula Sheet
| Physical Quantity / Law | Standard Mathematical Formula | Standard SI Unit & Symbol | Key Operational Notes |
|---|---|---|---|
| Speed (v) | Speed = Total Distance / Time Taken = s / t | Metres per second (m/s) | Scalar quantity; always positive or zero; never negative. |
| Average Speed (v_{avg}) | Average Speed = Total Distance Travelled / Total Time Taken | Metres per second (m/s) | For equal distance legs: v_avg = (2 × v₁ × v₂) / (v₁ + v₂). |
| Velocity (v) | Velocity = Displacement / Time Taken = s / t | Metres per second (m/s) | Vector quantity; possesses magnitude and direction; can be zero or negative. |
| Average Velocity (v_{avg}) | Average Velocity = (Initial Velocity + Final Velocity) / 2 = (u + v) / 2 | Metres per second (m/s) | Valid strictly when acceleration is constant (uniform acceleration). |
| Acceleration (a) | Acceleration = (Final Velocity - Initial Velocity) / Time = (v - u) / t | Metres per second squared (m/s²) | Positive for speeding up; negative (retardation/deceleration) for slowing down. |
| First Equation of Motion | v = u + (a × t) | Velocity in m/s | Relates initial velocity (u), final velocity (v), acceleration (a), and time (t). |
| Second Equation of Motion | s = (u × t) + (1/2 × a × t²) | Distance/Displacement in m | Gives position/distance covered (s) in time (t) under constant acceleration. |
| Third Equation of Motion | v² - u² = 2 × a × s | Velocity² in m²/s² | Time-independent equation relating velocities, acceleration, and distance. |
| Circular Speed (v_{circ}) | Speed = (2 × π × r) / t | Metres per second (m/s) | Distance is the circumference (2\pi r) of the circular orbit of radius r. |
2. Comparison: Distance vs. Displacement
| Characteristic Parameter | Distance Travelled | Displacement (Shortest Path) |
|---|---|---|
| Physical Definition | The actual length of the total path traversed by a moving body irrespective of direction. | The shortest straight-line distance measured from the initial position to the final position. |
| Quantity Nature | Scalar Quantity (Magnitude only). | Vector Quantity (Requires both Magnitude and Direction). |
| Sign & Value Possibility | Always Positive (> 0) for a moving body; never zero or negative. | Can be Positive, Negative, or Zero. |
| Closed Path Value | For a body returning to its starting point, distance is non-zero (2\pi r or perimeter). | For a body returning to its starting point, displacement is strictly Zero (0). |
| Magnitude Relationship | `Distance ≥ | Displacement |
NCERT In-Text Questions: “Think It Over”
Page No. 48: Think It Over (Questions 1 & 2)
Question 1 Can an object be in motion relative to one observer while simultaneously at rest relative to another observer? Explain with an everyday example. [Exam Favorite]
Answer: Yes, motion is strictly relative and depends on the chosen frame of reference (observer’s position).
- Illustrative Real-World Example:
- Consider passengers sitting inside a moving train carriage.
- Relative to a co-passenger sitting next to them: A passenger is at rest, because their relative position inside the carriage does not change with time.
- Relative to an observer standing stationary on the railway platform: The exact same passenger is in rapid motion, because the distance and position of the passenger relative to the platform changes continuously as the train travels past.
- Conclusion: An absolute state of rest or motion does not exist in the universe; motion can only be described relative to a specified reference point.
Question 2 An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the magnitude of displacement at the end of 2 minutes 20 seconds? [Exam Favorite]
Answer:
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NUMERICAL SOLUTION (CIRCULAR TRACK DISTANCE & DISPLACEMENT):
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GIVEN DATA:
• Diameter of circular track (d) = 200 m
• Radius of circular track (r) = d / 2 = 200 / 2 = 100 m
• Time to complete 1 round = 40 s
• Total Time of Run (t) = 2 minutes 20 seconds
= (2 × 60) + 20 = 140 seconds
STEP 1: Calculating Total Number of Rounds Completed
Number of Rounds = Total Time / Time for 1 Round
Number of Rounds = 140 / 40 = 3.5 rounds (3 complete rounds + 0.5 half round)
STEP 2: Calculating Total Distance Covered
Distance for 1 Round = Circumference of circle = 2 × π × r
= 2 × (22 / 7) × 100 m = 4400 / 7 ≈ 628.57 m
Total Distance = 3.5 × (2 × π × r)
Total Distance = 3.5 × 2 × (22 / 7) × 100
Total Distance = 7 × (22 / 7) × 100 = 22 × 100 = 2200 m (or 2.2 km)
STEP 3: Calculating Magnitude of Displacement
• After 3 complete rounds, the athlete returns to the initial starting point (Displacement = 0).
• In the remaining 0.5 round, the athlete moves to the diametrically opposite point of the track.
• Shortest straight-line distance between starting point and diametrically opposite point
= Diameter of the circle = 200 m.
FINAL ANSWER:
• Total Distance Covered = 2200 metres (2.2 km)
• Magnitude of Displacement = 200 metres
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NCERT In-Text Questions: “Pause and Ponder”
Page No. 52: Pause and Ponder (Questions 1 to 3)
Question 1 Under what physical condition is the magnitude of the average velocity of an object equal to its average speed? [Exam Favorite]
Answer: The magnitude of average velocity is strictly equal to average speed when an object moves along a straight line in a constant, single direction without turning back.
- Reasoning: In unidirectional straight-line motion, the total distance covered is exactly equal to the magnitude of net displacement (
Distance = |Displacement|). Dividing both by the identical time interval (t) gives:Average Speed = Total Distance / t = Magnitude of Displacement / t = Magnitude of Average Velocity.
Question 2 What does the odometer of an automobile measure? [Exam Favorite]
Answer: The odometer of an automobile measures the total cumulative Distance travelled by the vehicle in kilometres (km). (The speedometer measures instantaneous speed in km/h).
Question 3 What does the path of an object look like when it is in uniform motion? [Exam Favorite]
Answer: When an object is in uniform motion (covering equal distances in equal intervals of time along a fixed direction), its path is a Straight Line.
Page No. 55 & 58: Pause and Ponder (Questions 4 & 5)
Question 4 A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus. [Exam Favorite]
Answer:
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NUMERICAL SOLUTION (BUS DECELERATION / RETARDATION):
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GIVEN DATA:
• Initial Velocity (u) = 80 km/h = 80 × (5 / 18) = 400 / 18 = 22.22 m/s
• Final Velocity (v) = 60 km/h = 60 × (5 / 18) = 300 / 18 = 16.67 m/s
• Time Taken (t) = 5 s
CALCULATION:
Formula: Acceleration (a) = (Final Velocity - Initial Velocity) / Time
a = (v - u) / t
a = (16.67 - 22.22) / 5
a = -5.55 / 5 = -1.11 m/s²
FINAL ANSWER:
The acceleration of the bus is -1.11 m/s² (or a Retardation / Deceleration of 1.11 m/s²).
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Question 5 A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 minutes. Find its acceleration in standard SI units. [Exam Favorite]
Answer:
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NUMERICAL SOLUTION (TRAIN ACCELERATION):
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GIVEN DATA:
• Initial Velocity (u) = 0 m/s (Starting from rest at station)
• Final Velocity (v) = 40 km/h = 40 × (5 / 18) = 200 / 18 = 11.11 m/s
• Time Taken (t) = 10 minutes = 10 × 60 = 600 s
CALCULATION:
Formula: Acceleration (a) = (v - u) / t
a = (11.11 - 0) / 600
a = 11.11 / 600 = (200 / 18) / 600 = 200 / 10800 = 1 / 54 ≈ 0.0185 m/s²
FINAL ANSWER:
The acceleration of the train is 0.0185 m/s² (or 1/54 m/s²).
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Page No. 62: Pause and Ponder (Question 6)
Question 6 What is the nature of the distance-time graphs for: (i) uniform motion, and (ii) non-uniform motion of an object? [Exam Favorite]
Answer:
- For Uniform Motion: The distance-time graph is a Straight Line sloping upwards at a constant angle passing through the origin. The constant slope represents constant speed.
- For Non-Uniform Motion: The distance-time graph is a Curved Line (parabolic or irregular). A changing slope indicates that speed is varying over time (accelerated or decelerated motion).
NCERT Chapter-End Exercises: “Revise, Reflect, Refine” (Pages 68–71)
Question 1 An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and displacement at the end of 2 minutes 20 s? [Exam Favorite]
Answer:
================================================================================ EXERCISE NUMERICAL 1 SOLUTION: -------------------------------------------------------------------------------- • Total time = 2 min 20 s = 140 s • Time for 1 round = 40 s • Total rounds completed = 140 / 40 = 3.5 rounds 1. Total Distance: Distance = 3.5 × Circumference = 3.5 × (2 × 22/7 × 100) = 2200 m 2. Magnitude of Displacement: After 3.5 rounds, the athlete is at the diametrically opposite point. Displacement = Diameter = 200 m FINAL ANSWER: • Distance Covered = 2200 m (2.2 km) • Displacement = 200 m ================================================================================
Question 2 Joseph jogs from one end A to the other end B of a straight 300 m road in 2 min 30 s and then turns around and jogs 100 m back to point C in another 1 min. What are Joseph’s average speeds and velocities in jogging: (a) from A to B, and (b) from A to C? [Exam Favorite]
Answer:
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NUMERICAL SOLUTION (AVERAGE SPEED & AVERAGE VELOCITY):
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CASE (a): JOGGING FROM A TO B:
• Distance = 300 m, Displacement = 300 m (Straight line A to B)
• Time taken = 2 min 30 s = (2 × 60) + 30 = 150 s
1. Average Speed (A to B) = Total Distance / Time = 300 m / 150 s = 2.0 m/s
2. Average Velocity (A to B) = Displacement / Time = 300 m / 150 s = 2.0 m/s
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CASE (b): JOGGING FROM A TO C (VIA B):
• Total Distance = Path AB + Path BC = 300 m + 100 m = 400 m
• Net Displacement = Straight-line distance AC = AB - BC = 300 m - 100 m = 200 m
• Total Time = Time(AB) + Time(BC) = 150 s + 60 s = 210 s
1. Average Speed (A to C) = Total Distance / Total Time
= 400 m / 210 s ≈ 1.90 m/s
2. Average Velocity (A to C) = Net Displacement / Total Time
= 200 m / 210 s ≈ 0.95 m/s
FINAL ANSWER SUMMARY:
• From A to B : Average Speed = 2.0 m/s, Average Velocity = 2.0 m/s
• From A to C : Average Speed = 1.90 m/s, Average Velocity = 0.95 m/s
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Question 3 Abdul, while driving to school, computes the average speed for his trip to be 20 km/h. On his return trip along the same route, there is less traffic and the average speed is 30 km/h. What is the average speed for Abdul’s round trip? [Exam Favorite]
Answer:
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NUMERICAL SOLUTION (TWO-WAY ROUND TRIP AVERAGE SPEED):
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Let the one-way distance from home to school be d km.
• Total round-trip distance = d + d = 2d km
STEP 1: Calculating Time for Forward Trip (t₁)
Speed v₁ = 20 km/h ===> t₁ = Distance / v₁ = d / 20 hours
STEP 2: Calculating Time for Return Trip (t₂)
Speed v₂ = 30 km/h ===> t₂ = Distance / v₂ = d / 30 hours
STEP 3: Calculating Total Round-Trip Time (t_total)
t_total = t₁ + t₂ = (d / 20) + (d / 30) = (3d + 2d) / 60 = 5d / 60 = d / 12 hours
STEP 4: Calculating Average Speed
Average Speed = Total Distance / Total Time
Average Speed = 2d / (d / 12) = 2d × (12 / d) = 2 × 12 = 24 km/h
Harmonic Mean Formula Shortcut:
Average Speed = (2 × v₁ × v₂) / (v₁ + v₂) = (2 × 20 × 30) / (20 + 30)
= 1200 / 50 = 24 km/h
FINAL ANSWER:
The average speed for Abdul's entire round trip is 24 km/h.
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Question 4 A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s² for 8.0 s. How far does the boat travel during this time? [Exam Favorite]
Answer:
================================================================================ NUMERICAL SOLUTION (DISTANCE UNDER UNIFORM ACCELERATION): -------------------------------------------------------------------------------- GIVEN DATA: • Initial Velocity (u) = 0 m/s (Starting from rest) • Acceleration (a) = 3.0 m/s² • Time Duration (t) = 8.0 s CALCULATION: Using Second Equation of Motion: Distance (s) = (u × t) + (1/2 × a × t²) s = (0 × 8.0) + (1/2 × 3.0 × (8.0)²) s = 0 + (1/2 × 3.0 × 64) s = 1.5 × 64 = 96 metres FINAL ANSWER: The motorboat travels a distance of 96 metres during this time. ================================================================================
Question 5 A driver of a car travelling at 52 km/h applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km/h in another car applies brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied? [Exam Favorite]
Answer:
================================================================================ NUMERICAL & GRAPHICAL ANALYSIS (AREA UNDER V-T GRAPH): -------------------------------------------------------------------------------- CAR 1 (First Car): • Initial Speed (u₁) = 52 km/h = 52 × (5 / 18) = 260 / 18 = 14.44 m/s • Stopping Time (t₁) = 5 s, Final Speed (v₁) = 0 m/s • Distance Travelled (s₁) = Area under Triangle 1 on Speed-Time Graph s₁ = 1/2 × Base × Height = 1/2 × 5 s × 14.44 m/s = 36.1 metres CAR 2 (Second Car): • Initial Speed (u₂) = 3 km/h = 3 × (5 / 18) = 15 / 18 = 0.833 m/s • Stopping Time (t₂) = 10 s, Final Speed (v₂) = 0 m/s • Distance Travelled (s₂) = Area under Triangle 2 on Speed-Time Graph s₂ = 1/2 × Base × Height = 1/2 × 10 s × 0.833 m/s = 4.17 metres COMPARISON: Distance of Car 1 (36.1 m) > Distance of Car 2 (4.17 m) FINAL ANSWER: The FIRST CAR (travelling at 52 km/h) travelled MUCH FARTHER after braking (36.1 m vs 4.17 m). ================================================================================
Question 6 Fig. 4.11 shows the distance-time graph of three objects A, B, and C. Study the graph and answer the following questions: (a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C? [Exam Favorite]
Answer:
- (a) Which is travelling fastest?Object B is travelling the fastest.
- Reason: On a distance-time graph, the slope of the line represents speed (Speed = Distance / Time). Object B has the steepest slope (largest angle with the time axis), indicating the highest speed.
- (b) Are all three ever at the same point? No. All three lines never intersect at a single common point on the graph.
- (c) Distance travelled by C when B passes A: Locate the intersection point of line B and line A on the graph. Draw a vertical line to meet the curve of C, and trace horizontally to the distance axis to read approximately 6.5 km to 7.0 km.
- (d) Distance travelled by B when it passes C: Locate the intersection point of line B and line C. The distance reading on the vertical axis shows that B has travelled approximately 5.5 km to 6.0 km from its origin.
Question 7 A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m/s², with what velocity will it strike the ground? After what time will it strike the ground? [Exam Favorite]
Answer:
================================================================================ NUMERICAL SOLUTION (FREE FALL UNDER GRAVITY): -------------------------------------------------------------------------------- GIVEN DATA: • Initial Velocity (u) = 0 m/s (Gently dropped from rest) • Height / Distance (s) = 20 m • Acceleration (a = g) = +10 m/s² (Downwards) STEP 1: Finding Striking Velocity (v) Using Third Equation of Motion: v² - u² = 2 × a × s v² - (0)² = 2 × 10 × 20 v² = 400 v = Square root of (400) = 20 m/s STEP 2: Finding Time Taken to Strike (t) Using First Equation of Motion: v = u + (a × t) 20 = 0 + (10 × t) 10 × t = 20 t = 20 / 10 = 2 seconds FINAL ANSWER: • Final Velocity on striking the ground = 20 m/s • Time taken to strike the ground = 2 seconds ================================================================================
Question 8 State which of the following physical situations are possible and provide an illustrative everyday example for each: (a) An object moving with a constant acceleration but with zero velocity. (b) An object moving with an acceleration but with uniform speed. (c) An object moving in a certain direction with an acceleration in the perpendicular direction. [Exam Favorite]
Answer:
- (a) Constant Acceleration with Zero Velocity:POSSIBLE.
- Example: When a ball is thrown vertically upward, at the highest point of its trajectory, its instantaneous velocity is momentarily zero (v = 0), yet it experiences a constant downward gravitational acceleration (a = g = 9.8\text{ m/s}²).
- (b) Acceleration with Uniform Speed:POSSIBLE.
- Example: An object in Uniform Circular Motion (e.g., a satellite orbiting Earth or a stone whirled on a string). Its speed remains constant, but because its direction of motion changes continuously at every point, its velocity changes, producing a continuous inward centripetal acceleration.
- (c) Motion in One Direction with Perpendicular Acceleration:POSSIBLE.
- Example: An airplane flying horizontally at constant speed while gravity exerts a downward acceleration perpendicular to its horizontal flight path, or an object in circular motion where centripetal acceleration is perpendicular to the tangential velocity vector at all points.
Question 9 An artificial satellite is moving in a circular orbit of radius 42,250 km. Calculate its linear orbital speed if it takes 24 hours to complete one revolution around the Earth. [Exam Favorite]
Answer:
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NUMERICAL SOLUTION (SATELLITE ORBITAL SPEED):
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GIVEN DATA:
• Orbit Radius (r) = 42,250 km = 42,250,000 m
• Time Period (t) = 24 hours = 24 × 60 × 60 = 86,400 seconds
STEP 1: Distance Covered in One Complete Revolution
Distance (s) = Circumference of circular orbit = 2 × π × r
Distance = 2 × 3.1416 × 42,250 km ≈ 265,464 km
STEP 2: Calculating Orbital Speed in km/s and m/s
Formula: Orbital Speed (v) = Distance / Time = (2 × π × r) / t
v = 265,464 km / 86,400 s ≈ 3.0725 km/s
v = 3.0725 × 1000 m/s ≈ 3072.5 m/s (or ~3.1 km/s)
FINAL ANSWER:
The orbital speed of the artificial satellite is 3.07 km/s (or 3072.5 m/s).
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Question 10 Derive the three kinematic equations of motion graphically from a velocity-time graph for an object moving with uniform acceleration. [Exam Favorite]
Answer:
================================================================================ GRAPHICAL DERIVATION OF THE THREE EQUATIONS OF MOTION: -------------------------------------------------------------------------------- Consider a body moving with initial velocity u at time t = 0 (Point A). Under uniform acceleration a, its velocity increases to v at time t (Point B). • OA = DC = u (Initial Velocity) • BC = v (Final Velocity) • OC = AD = t (Time Interval) • Change in Velocity BD = BC - CD = v - u 1. DERIVATION OF FIRST EQUATION (v = u + a·t): Acceleration a = Slope of line AB = BD / AD a = (v - u) / t a × t = v - u v = u + (a × t) -------------------------------------------- (Equation 1) 2. DERIVATION OF SECOND EQUATION (s = u·t + 1/2·a·t²): Distance s = Area of Trapezium OABC = Area of Rectangle OADC + Area of Triangle ABD s = (OA × OC) + (1/2 × AD × BD) s = (u × t) + [1/2 × t × (v - u)] From Eq 1, substitute (v - u) = a × t: s = (u × t) + [1/2 × t × (a × t)] s = (u × t) + (1/2 × a × t²) ------------------------------- (Equation 2) 3. DERIVATION OF THIRD EQUATION (v² - u² = 2·a·s): Distance s = Area of Trapezium OABC = [ (Sum of Parallel Sides) / 2 ] × Height s = [ (OA + BC) / 2 ] × OC s = [ (u + v) / 2 ] × t From Eq 1, express t = (v - u) / a: s = [ (v + u) / 2 ] × [ (v - u) / a ] s = (v² - u²) / (2 × a) 2 × a × s = v² - u² v² - u² = 2 × a × s ---------------------------------------- (Equation 3) ================================================================================
Question 11 Why is uniform circular motion referred to as an accelerated motion? Explain. [Exam Favorite]
Answer:
- Explanation:
- Velocity is a vector quantity defined by both magnitude (speed) and direction.
- In uniform circular motion, an object moves along a circular path with constant speed.
- However, at every single point on the circle, the direction of motion changes continuously along the tangent to that point.
- Because direction changes constantly, the velocity vector changes continuously over time.
- Any change in velocity with respect to time represents acceleration.
- Hence, uniform circular motion is an accelerated motion, driven by an inward centripetal acceleration directed toward the center of the circular circle (a = v² / r).
Question 12 A train travelling at a speed of 90 km/h is brought to rest by applying brakes that produce a uniform acceleration of -0.5 m/s². Find how far the train will go before it is brought to rest. [Exam Favorite]
Answer:
================================================================================ NUMERICAL SOLUTION (STOPPING DISTANCE OF TRAIN): -------------------------------------------------------------------------------- GIVEN DATA: • Initial Velocity (u) = 90 km/h = 90 × (5 / 18) = 5 × 5 = 25 m/s • Final Velocity (v) = 0 m/s (Brought to rest) • Acceleration (a) = -0.5 m/s² (Retardation) CALCULATION: Using Third Equation of Motion: v² - u² = 2 × a × s (0)² - (25)² = 2 × (-0.5) × s -625 = -1.0 × s s = -625 / -1.0 = 625 metres FINAL ANSWER: The train will travel a distance of 625 metres before coming to a complete stop. ================================================================================
Frequently Asked Questions (FAQs) – Class 9 Science Chapter 4
Question 1: Define Speed and Velocity. What is the fundamental difference? [Exam Favorite] Answer:
- Speed: The rate of distance covered per unit time (
Speed = Distance / Time). It is a scalar quantity with magnitude only. - Velocity: The rate of displacement per unit time in a specified direction (
Velocity = Displacement / Time). It is a vector quantity with both magnitude and direction.
Question 2: Can an object have zero displacement but non-zero distance? Give an example. [Exam Favorite] Answer: Yes. When an object completes one full round of a circular track of radius r and returns to its starting point, its total distance is 2\pi r, while its net displacement is Zero (0).
Question 3: What is Uniform Acceleration? Give an everyday example. [Exam Favorite] Answer: Uniform acceleration occurs when an object’s velocity changes by equal amounts in equal intervals of time. Example: A body falling freely under gravity in a vacuum (a = g = 9.8\text{ m/s}²).
Question 4: What is Retardation or Deceleration? What is its SI unit? [Exam Favorite] Answer: Retardation is negative acceleration that occurs when the velocity of a moving body decreases over time (e.g., applying brakes to a speeding vehicle). Its SI unit is metres per second squared (m/s²).
Question 5: How is speed calculated from a Distance-Time graph? [Exam Favorite] Answer: Speed is equal to the Slope (Gradient) of the Distance-Time graph: Slope = Change in Distance / Change in Time = (s₂ - s₁) / (t₂ - t₁) = Speed.
Question 6: How is distance calculated from a Velocity-Time graph? [Exam Favorite] Answer: The total distance (or displacement) travelled by an object is equal to the total Area enclosed under the Velocity-Time graph and the time axis.
Question 7: What is the acceleration of an object moving with uniform velocity? [Exam Favorite] Answer: Zero (0\text{ m/s}²), because the change in velocity is zero (a = (v - u) / t = (v - v) / t = 0).
Question 8: Convert a speed of 72 km/h into m/s. [Exam Favorite] Answer: Speed = 72 × (5 / 18) = 4 × 5 = 20 m/s.
Question 9: Convert a speed of 15 m/s into km/h. [Exam Favorite] Answer: Speed = 15 × (18 / 5) = 3 × 18 = 54 km/h.
Question 10: State the SI units of: (a) Distance, (b) Speed, (c) Acceleration. [Exam Favorite] Answer:
- (a) Distance: Metre (m)
- (b) Speed / Velocity: Metres per second (m/s)
- (c) Acceleration: Metres per second squared (m/s²)
Question 11: Under what condition is displacement negative? [Exam Favorite] Answer: Displacement is negative when an object moves in a direction opposite to the originally chosen positive reference direction.
Question 12: Why is the motion of the tip of the second’s hand of a watch considered uniform circular motion? [Exam Favorite] Answer: Because the tip of the second’s hand covers equal circular angular distances in equal intervals of time at constant speed, while its direction of motion changes continuously.
Question 13: What does a horizontal line parallel to the time axis represent in: (a) Distance-Time graph, (b) Velocity-Time graph? [Exam Favorite] Answer:
- (a) In a Distance-Time graph: The object is Stationary (At Rest).
- (b) In a Velocity-Time graph: The object is moving with Constant / Uniform Velocity (zero acceleration).
Question 14: What is the average speed of a car that travels 60 km in the first hour and 40 km in the second hour? [Exam Favorite] Answer: Average Speed = Total Distance / Total Time = (60 + 40) km / (1 + 1) h = 100 / 2 = 50 km/h.
Question 15: What is the slope of a Velocity-Time graph called? [Exam Favorite] Answer: The slope of a Velocity-Time graph represents Acceleration (Slope = Change in Velocity / Time = Acceleration).
Mastering the NCERT Solutions for Class 9 Science Chapter 4 (Exploration), “Describing Motion Around Us”, equips students with the kinematic formulas, graphical derivations, vector concepts, and numerical problem-solving techniques required for top performance in CBSE physics evaluations. Review the numerical box solutions, the master summary tables, and the 15 high-yield FAQs above to secure full marks in your examinations.
Master Formula & Concept Summary Tables
1. Master Formula Sheet for Kinematics
| Standard Formula | SI Unit & Symbol | Key Mathematical Notes | |
|---|---|---|---|
| Speed (v) | Speed = Distance / Time = s / t | Metre per second (m/s or m s⁻¹) | Scalar quantity; always positive (v \ge 0). |
| Velocity (v) | Velocity = Displacement / Time = Δx / t | Metre per second (m/s or m s⁻¹) | Vector quantity; can be positive, negative, or zero. |
| Average Speed (v_{av}) | v_av = Total Distance / Total Time | Metre per second (m/s) | v_av = (s₁ + s₂ + ...) / (t₁ + t₂ + ...). |
| Average Velocity (v_{av}) | v_av = Total Displacement / Total Time = (u + v) / 2 | Metre per second (m/s) | Valid for uniform linear acceleration. |
| Acceleration (a) | a = (v - u) / t = Change in Velocity / Time | Metre per second squared (m/s² or m s⁻²) | Vector quantity; negative acceleration is Retardation / Deceleration. |
| First Equation of Motion | v = u + a · t | Speed/Velocity in m/s | Velocity-Time relation. |
| Second Equation of Motion | s = u · t + 1/2 · a · t² | Displacement/Distance in m | Position-Time relation. |
| Third Equation of Motion | v² - u² = 2 · a · s (or v² = u² + 2as) | Distance/Speed in m & m/s | Position-Velocity relation (Independent of time). |
| Circular Speed (v_{circ}) | v = 2 · π · r / T | Metre per second (m/s) | Distance per revolution = Circumference (2\pi r). |
| Stopping Distance (d_{stop}) | d_stop = d_reaction + d_braking = (u · t_r) + (u² / 2a) | Metre (m) | Reaction distance + Braking distance. |
2. Comparison: Distance vs. Displacement
| Parameter / Basis | Distance Travelled (s) | Displacement (\Delta x or \vec{s}) |
|---|---|---|
| Definition | The actual total length of the path traversed by a moving body irrespective of direction. | The shortest straight-line vector distance directed from the initial position to the final position. |
| Physical Nature | Scalar Quantity (has only magnitude). | Vector Quantity (has both magnitude and direction). |
| Sign / Value | Always positive (s > 0) for a moving body; never zero or negative. | Can be positive, negative, or zero depending on the reference frame. |
| Path Dependence | Depends entirely on the specific path taken between two points. | Independent of the path followed; depends only on initial and final positions. |
| Magnitude Relation | Distance \ge Magnitude of Displacement (`Distance = | Displacement |
3. Graphical Interpretation of Motion
| Type of Graph | Physical Feature Analyzed | Derived Physical Quantity | Graphical Behavior / Significance |
|---|---|---|---|
| Position-Time Graph (x-t) | Slope of the graph (\Delta x / \Delta t) | Velocity (v) | • Horizontal line (\text{Slope} = 0): Object is at Rest. • Constant straight line: Uniform Velocity. • Curved upward: Accelerated Motion. |
| Velocity-Time Graph (v-t) | Slope of the graph (\Delta v / \Delta t) | Acceleration (a) | • Horizontal line (\text{Slope} = 0): Zero Acceleration (Constant Velocity). • Positive straight slope: Uniform Acceleration. • Negative straight slope: Uniform Retardation. |
| Velocity-Time Graph (v-t) | Area under the graph line | Magnitude of Displacement (s) | Split into geometric shapes (triangles, rectangles, or trapeziums) to calculate total distance/displacement covered. |
NCERT In-Text Questions: “Think It Over” (Page No. 48)
Question 1 How much distance should we maintain from the truck ahead to avoid a collision if it suddenly applies the brakes? [Exam Favorite]
Answer: To prevent a rear-end collision, the trailing vehicle must maintain a safe following distance equal to or greater than its Total Stopping Distance:
- Total Stopping Distance = Reaction Distance + Braking Distance (d_{stop} = u \cdot t_{reaction} + \frac{u^2}{2a}).
- The distance must account for: (i) the driver’s physiological perception-reaction time (typically 0.5 to 1.5 seconds during which the vehicle travels distance u \cdot t_r before the brake pedal is depressed), and (ii) the mechanical braking distance required to bring the vehicle to a complete stop under friction.
- Under standard driving guidelines, maintaining a “2-Second Rule” (or at least one car length per 15 km/h of speed) ensures sufficient stopping distance.
Question 2 Does this distance depend upon the speed with which we are moving? [Exam Favorite]
Answer: Yes, the safe following distance depends strongly on the initial speed (u) of the vehicle.
- Kinematic Relationship: Braking distance is proportional to the square of initial velocity (d_{braking} \propto u^2, derived from v^2 – u^2 = 2as \implies s = u^2 / 2a).
- Practical Implication: If the vehicle’s speed is doubled (from 30\text{ km/h} to 60\text{ km/h}), the braking distance increases by four times (2^2 = 4). Furthermore, the reaction distance (u \cdot t_r) increases linearly with speed. Therefore, at higher speeds, a substantially larger distance must be maintained.
NCERT In-Text Questions: “Pause and Ponder”
Page No. 51: Pause and Ponder (Questions 1 to 3)
Question 1 In the example of an athlete running back and forth on a straight track, when will the displacement of the athlete be zero? What will be the total distance travelled in that case? [Exam Favorite]
Answer:
- Condition for Zero Displacement: The displacement of the athlete will be zero when the athlete completes the round trip and returns to the exact initial starting point. Displacement is the straight-line vector difference between final and initial coordinates (\Delta x = x_f – x_i = 0).
- Total Distance Travelled: If the length of the straight running track from starting point A to turnaround point B is L, the total distance travelled is the sum of both paths: \text{Total Distance} = \text{Path } AB + \text{Path } BA = L + L = \mathbf{2L}.
Question 2 Fuel used up in a vehicle depends on which of the following: distance travelled or displacement? Justify your answer. [Exam Favorite]
Answer: Fuel consumed by a vehicle depends directly on the Total Distance Travelled, not on displacement.
================================================================================ SCIENTIFIC JUSTIFICATION: -------------------------------------------------------------------------------- 1. Work Done Against Resistive Forces: The internal combustion engine (or electric motor) must continuously burn fuel to perform mechanical work against opposing road friction and aerodynamic air drag along every metre of the actual path driven. 2. Work Formula: Work Done = Friction Force × Total Path Length (Distance). 3. Example: If a delivery van drives 50 km delivering goods around a city and returns to its starting depot, its net displacement is ZERO. However, it burns a significant volume of petrol corresponding to the full 50 km distance covered. ================================================================================
Question 3 As shown in textbook figures, a ball is thrown vertically upwards from O. It moves up straight till B and then falls back to O. Can this be considered a motion in a straight line? [Exam Favorite]
Answer: Yes, this is strictly a motion in a straight line (one-dimensional rectilinear motion).
- Reason: The ball moves along a single, continuous vertical linear axis (Y-axis). Although the velocity vector reverses its direction at the peak (B) from upward (+v) to downward (-v), the spatial trajectory of the object remains confined entirely to a single one-dimensional straight path.
Page No. 53: Pause and Ponder (Question 4)
Question 4 An elevator moves from ground floor to 4th floor (12\text{ m} high), then moves down to 2nd floor (6\text{ m} high from ground). What is the total distance travelled and the net displacement? [Exam Favorite]
Answer:
================================================================================ CALCULATION (DISTANCE VS. DISPLACEMENT IN ELEVATOR): -------------------------------------------------------------------------------- GIVEN DATA: • Initial Position (Ground Floor) = 0 m • Peak Position (4th Floor) = +12 m • Final Position (2nd Floor) = +6 m STEP 1: Calculating Total Distance Travelled (s) Path 1 (Ground to 4th Floor) : 12 - 0 = 12 m Path 2 (4th to 2nd Floor) : 12 - 6 = 6 m Total Distance = 12 m + 6 m = 18 m STEP 2: Calculating Net Displacement (Δx) Displacement = Final Position - Initial Position Δx = (+6 m) - (0 m) = +6 m (in the upward direction) FINAL ANSWER: • Total Distance Travelled = 18 m • Net Displacement = 6 m (Upward) ================================================================================
Page No. 56: Pause and Ponder (Question 5)
Question 5 A girl is riding her scooter and finds that the speedometer reading is constant. Is it possible for the scooter to be accelerating, and if so, how? [Exam Favorite]
Answer: Yes, it is entirely possible for the scooter to be accelerating even if the speedometer reading remains constant.
- Physical Explanation: Speedometer measures only instantaneous speed (magnitude of velocity). Acceleration is defined as the rate of change of velocity (\vec{a} = \Delta \vec{v} / \Delta t). Velocity is a vector quantity that has both magnitude and direction.
- Mechanism: If the girl steers her scooter along a curved road, circular roundabout, or makes a turn while maintaining a steady speed of 40\text{ km/h}, her direction of motion changes continuously. This change in directional velocity produces a non-zero Centripetal Acceleration directed toward the center of curvature.
Page No. 60: Pause and Ponder (Question 6)
Question 6 A car travels 200\text{ km} North in 4\text{ hours}, and returns 200\text{ km} South in 4\text{ hours}. Calculate total distance, displacement, average speed, and average velocity. [Exam Favorite]
Answer:
================================================================================ NUMERICAL SOLUTION (ROUND TRIP KINEMATICS): -------------------------------------------------------------------------------- GIVEN DATA: • Forward Trip : Distance = 200 km North, Time t₁ = 4 h • Return Trip : Distance = 200 km South, Time t₂ = 4 h • Total Time (T) = t₁ + t₂ = 4 + 4 = 8 hours CALCULATIONS: 1. Total Distance Travelled: s = 200 km + 200 km = 400 km 2. Net Displacement: Δx = (+200 km North) + (-200 km South) = 0 km 3. Average Speed: v_av = Total Distance / Total Time = 400 km / 8 h = 50 km/h 4. Average Velocity: v_av = Net Displacement / Total Time = 0 km / 8 h = 0 km/h FINAL ANSWER: • Total Distance = 400 km • Net Displacement = 0 km • Average Speed = 50 km/h • Average Velocity = 0 km/h ================================================================================
NCERT Chapter-End Exercises: “Revise, Reflect, Refine” (Pages 68–72)
Question 1 My father went to a shop from home which is located at a distance of 250 m on a straight road. On reaching there, he discovered that he forgot to carry a cloth bag. He came home to take it, went to the shop again, bought provisions and came back home. How much was the total distance travelled by him? What was his displacement from home? [Exam Favorite]
Answer:
================================================================================ NUMERICAL SOLUTION (DISTANCE AND DISPLACEMENT): -------------------------------------------------------------------------------- GIVEN DATA: Distance between Home and Shop = 250 m (Straight road) STEP 1: Path Traversed by Father 1. Trip 1 (Home to Shop) : 250 m 2. Trip 2 (Shop to Home) : 250 m (Returned for cloth bag) 3. Trip 3 (Home to Shop again) : 250 m (Went for shopping) 4. Trip 4 (Shop to Home final) : 250 m (Returned home with provisions) Total Distance Travelled = 250 + 250 + 250 + 250 = 1000 m (or 1.0 km) STEP 2: Net Displacement from Home Displacement = Final Position - Initial Position Since father started from Home and finally returned to Home: Displacement = 0 m FINAL ANSWER: • Total Distance Travelled = 1000 m (1 km) • Net Displacement from Home = 0 m ================================================================================
Question 2 A bus is travelling at 36 km h⁻¹ when the driver sees an obstacle 30 m ahead. The driver takes 0.5 seconds to react before pressing the brake. Once the brake is applied, the velocity of the bus reduces with constant acceleration of 2.5 m s⁻². Will the bus be able to stop before reaching the obstacle? [Exam Favorite]
Answer:
================================================================================ NUMERICAL SOLUTION (REACTION & BRAKING DISTANCE): -------------------------------------------------------------------------------- GIVEN DATA: • Initial Speed (u) = 36 km/h = 36 × (5/18) = 10 m/s • Distance to obstacle (D) = 30 m • Reaction Time (t_r) = 0.5 s • Deceleration (a) = -2.5 m/s² (opposite to motion) • Final Velocity (v) = 0 m/s (Bus comes to rest) STEP 1: Distance Covered During Reaction Time (d₁) During reaction time, brakes are not yet applied; bus moves at uniform speed: d₁ = u × t_r = 10 m/s × 0.5 s = 5 m STEP 2: Braking Distance After Applying Brakes (d₂) Using Third Equation of Motion: v² - u² = 2 · a · d₂ (0)² - (10)² = 2 × (-2.5) × d₂ -100 = -5 × d₂ d₂ = -100 / -5 = 20 m STEP 3: Total Stopping Distance (d_total) d_total = d₁ + d₂ = 5 m + 20 m = 25 m STEP 4: Conclusion Since the total stopping distance (25 m) is LESS than the distance to the obstacle (30 m), the bus stops (30 - 25 = 5 m) before the obstacle. FINAL ANSWER: Yes, the bus will be able to stop safely 5 m before reaching the obstacle. ================================================================================
Question 3 Is it possible for a body to have zero velocity and non-zero acceleration? Give an example. [Exam Favorite]
Answer: Yes, it is physically possible.
- Real-World Example: When a ball is thrown vertically upwards into the air:
- At the highest point (peak of its trajectory), the vertical velocity of the ball momentarily becomes zero (v = 0) as it changes direction.
- However, at that exact instant, the ball is still subject to the Earth’s gravitational force, experiencing a constant non-zero acceleration due to gravity directed downward (a = g \approx 9.8\text{ m/s}^2).
Question 4 A car starts from rest and its velocity reaches 24 m s⁻¹ in 6 s. Find the average acceleration and the distance travelled in these 6 s. [Exam Favorite]
Answer:
================================================================================
NUMERICAL SOLUTION (UNIFORM ACCELERATION & DISTANCE):
--------------------------------------------------------------------------------
GIVEN DATA:
• Initial Velocity (u) = 0 m/s (Starts from rest)
• Final Velocity (v) = 24 m/s
• Time Taken (t) = 6 s
STEP 1: Calculating Average Acceleration (a)
Formula: a = (v - u) / t
a = (24 - 0) / 6 = 24 / 6 = 4 m/s²
STEP 2: Calculating Distance Travelled (s)
Formula (Second Equation of Motion): s = u · t + 1/2 · a · t²
s = (0 × 6) + 1/2 × (4) × (6)²
s = 0 + 2 × 36 = 72 m
FINAL ANSWER:
• Average Acceleration = 4 m/s²
• Distance Travelled = 72 m
================================================================================
Question 5 Under what condition is the magnitude of average velocity equal to the average speed? [Exam Favorite]
Answer: The magnitude of average velocity is equal to average speed only when an object moves along a perfectly straight line in a single, fixed direction without reversing or turning back.
Under this condition, the total distance travelled equals the magnitude of net displacement (\text{Distance} = \vert{}\text{Displacement}\vert{}), resulting in equal numerical values.
Question 6 In textbook Fig. 4.27, position-time graphs for two objects A and B are shown as two straight lines of different slopes that intersect at a point. Do objects A and B ever have equal velocity? Explain. [Exam Favorite]
Answer: No, objects A and B never have equal velocity.
- Scientific Reason: On a position-time (x-t) graph, the slope of the line represents the velocity of the object (\text{Velocity} = \text{Slope} = \Delta x / \Delta t).
- Since the two graphs are straight lines with distinct, constant, and different slopes, object A and object B maintain different constant velocities throughout their motion.
- The point where the two lines intersect merely indicates that both objects occupy the same physical position at that specific instant of time (one overtakes the other), but their velocities at that point remain unequal.
Question 7 A graph in textbook Fig. 4.28 shows the change in position with time for two objects, A and B, moving in a straight line from 0 to 10 seconds. Choose the correct option(s): (i) The average velocity of both over the 10 s time interval is equal since they have the same initial and final positions. (ii) The average speeds of both over the 10 s time interval are equal since both cover equal distances in equal time. (iii) The average speed of A over the 10 s time interval is lower than that of B since it covers a shorter distance than B in 10 seconds. (iv) The average speed of A over the 1st time interval is greater than that of B since B’s speed is lower than A’s in some segments.
Answer: Options (i) and (iii) are correct. Explanation:
- Option (i) is correct: Both objects start at the same initial position at t = 0\text{ s} and finish at the same final position at t = 10\text{ s}. Because net displacement (\Delta x) and total time (10\text{ s}) are identical, their average velocities are equal (v_{av} = \Delta x / \Delta t).
- Option (iii) is correct: Object A moves monotonically in a straight line, while Object B moves back and forth, traversing a greater total path length (distance). Thus, the average speed of A is lower than that of B.
Question 8 A train starting from rest attains a velocity of 72 km h⁻¹ in 5 minutes. Assuming that the acceleration is uniform, find: (i) the acceleration, and (ii) the distance travelled by the train for attaining this velocity. [Exam Favorite]
Answer:
================================================================================
NUMERICAL SOLUTION (TRAIN ACCELERATION & DISTANCE):
--------------------------------------------------------------------------------
GIVEN DATA:
• Initial Velocity (u) = 0 m/s (Starts from rest)
• Final Velocity (v) = 72 km/h = 72 × (5/18) = 20 m/s
• Time (t) = 5 minutes = 5 × 60 = 300 s
STEP 1: Calculating Acceleration (a)
Formula: a = (v - u) / t
a = (20 - 0) / 300 = 20 / 300 = 1 / 15 m/s² ≈ 0.067 m/s²
STEP 2: Calculating Distance Travelled (s)
Using Third Equation of Motion: v² - u² = 2 · a · s
(20)² - (0)² = 2 × (1/15) × s
400 = (2/15) × s
s = (400 × 15) / 2 = 6000 / 2 = 3000 m = 3 km
FINAL ANSWER:
• Acceleration of Train = 1/15 m/s² (0.067 m/s²)
• Distance Travelled = 3000 m (3 km)
================================================================================
Question 9 A car accelerates uniformly from 18 km h⁻¹ to 36 km h⁻¹ in 5 s. Calculate: (i) the acceleration, and (ii) the distance covered by the car in that time. [Exam Favorite]
Answer:
================================================================================
NUMERICAL SOLUTION (CAR ACCELERATION & DISPLACEMENT):
--------------------------------------------------------------------------------
GIVEN DATA:
• Initial Velocity (u) = 18 km/h = 18 × (5/18) = 5 m/s
• Final Velocity (v) = 36 km/h = 36 × (5/18) = 10 m/s
• Time Taken (t) = 5 s
STEP 1: Calculating Acceleration (a)
Formula: a = (v - u) / t
a = (10 - 5) / 5 = 5 / 5 = 1 m/s²
STEP 2: Calculating Distance Covered (s)
Formula: s = u · t + 1/2 · a · t²
s = (5 × 5) + 1/2 × (1) × (5)²
s = 25 + 1/2 × 25 = 25 + 12.5 = 37.5 m
FINAL ANSWER:
• Acceleration of Car = 1 m/s²
• Distance Covered = 37.5 m
================================================================================
Question 10 The brakes applied to a car produce an acceleration of 6 m s⁻² in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time. [Exam Favorite]
Answer:
================================================================================ NUMERICAL SOLUTION (BRAKING RETARDATION): -------------------------------------------------------------------------------- GIVEN DATA: • Acceleration (a) = -6 m/s² (Opposite to motion) • Time to stop (t) = 2 s • Final Velocity (v) = 0 m/s (Car stops) STEP 1: Finding Initial Velocity (u) Using First Equation of Motion: v = u + a · t 0 = u + (-6) × 2 0 = u - 12 u = 12 m/s STEP 2: Calculating Distance Travelled (s) Using Second Equation of Motion: s = u · t + 1/2 · a · t² s = (12 × 2) + 1/2 × (-6) × (2)² s = 24 - 3 × 4 = 24 - 12 = 12 m FINAL ANSWER: The car travels a distance of 12 m before coming to a complete stop. ================================================================================
Question 11 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 at the end of 2 minutes 20 s? [Exam Favorite]
Answer:
================================================================================ NUMERICAL SOLUTION (CIRCULAR TRACK KINEMATICS): -------------------------------------------------------------------------------- GIVEN DATA: • Diameter of track (d) = 200 m ===> Radius (r) = 100 m • Time for 1 round = 40 s • Total Time (T) = 2 min 20 s = (2 × 60) + 20 = 140 s STEP 1: Number of Rounds Completed (n) n = Total Time / Time for 1 round = 140 / 40 = 3.5 rounds (The athlete completes 3 full rounds and exactly half a round). STEP 2: Total Distance Covered Distance for 1 full round = Circumference = 2 · π · r Distance = 3.5 × (2 × 22/7 × 100) Distance = 3.5 × 2 × 3.1416 × 100 = 7 × 314.16 = 2200 m (or 2.2 km) STEP 3: Net Displacement After 3.5 rounds, the athlete is at the diametrically opposite end from the start. Displacement = Diameter of the circle = 200 m FINAL ANSWER: • Total Distance Covered = 2200 m (2.2 km) • Net Displacement = 200 m (Across the diameter) ================================================================================
Question 12 An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the Earth. [Exam Favorite]
Answer:
================================================================================
NUMERICAL SOLUTION (ORBITAL SPEED):
--------------------------------------------------------------------------------
GIVEN DATA:
• Orbital Radius (r) = 42250 km = 42,250,000 m
• Time Period (T) = 24 hours = 24 × 3600 s = 86400 s
CALCULATION:
Formula: Speed (v) = Total Distance / Time = (2 · π · r) / T
v = (2 × 3.1416 × 42250 km) / 24 h
v = 265464.58 / 24 = 11061.02 km/h
In SI Units (km/s):
v = 265464.58 km / 86400 s = 3.0725 km/s (or 3072.5 m/s)
FINAL ANSWER:
The orbital speed of the satellite is 3.07 km/s (or 11061 km/h).
================================================================================
Question 13 A car moving on a straight road with velocity 54 km h⁻¹ slows down to 18 km h⁻¹ with constant deceleration over a distance of 100 m. Calculate: (i) its acceleration, and (ii) the additional distance required for the car to come to a complete stop if deceleration remains constant. [Exam Favorite]
Answer:
================================================================================ NUMERICAL SOLUTION (TWO-STAGE DECELERATION): -------------------------------------------------------------------------------- GIVEN DATA (STAGE 1): • u₁ = 54 km/h = 54 × (5/18) = 15 m/s • v₁ = 18 km/h = 18 × (5/18) = 5 m/s • Distance (s₁) = 100 m STEP 1: Calculating Acceleration (a) Using Third Equation of Motion: v₁² - u₁² = 2 · a · s₁ (5)² - (15)² = 2 × a × 100 25 - 225 = 200 × a -200 = 200 × a a = -1 m/s² (Retardation of 1 m/s²) STEP 2: Calculating Additional Distance to Stop (Stage 2) • Initial velocity for Stage 2 (u₂) = 5 m/s • Final velocity (v₂) = 0 m/s • Acceleration (a) = -1 m/s² v₂² - u₂² = 2 · a · s₂ (0)² - (5)² = 2 × (-1) × s₂ -25 = -2 × s₂ s₂ = 25 / 2 = 12.5 m FINAL ANSWER: • Acceleration of the Car = -1 m/s² (Retardation = 1 m/s²) • Additional Stopping Distance = 12.5 m ================================================================================
Question 14 Can an object kept on the Earth be considered to be at rest and in motion at the same time? Explain with reference to coordinate frames. [Exam Favorite]
Answer: Yes, rest and motion are strictly relative terms and depend entirely on the chosen frame of reference (reference point):
- With respect to a Frame on Earth: A book resting on a table is at Rest relative to the floor, walls, and an observer standing inside the room because its spatial coordinates do not change with time.
- With respect to a Frame in Outer Space: To an astronaut observing from the Moon or a space station, the same book is in Rapid Motion (moving at approximately 30\text{ km/s} as Earth orbits the Sun, plus rotating with Earth’s diurnal spin).
- Thus, absolute rest or absolute motion does not exist in the universe.
Question 15 (A) Two cars A and B start moving with constant acceleration from rest in a straight line. Car A attains a velocity of 5 m s⁻¹ in 5 s. Car B attains a velocity of 3 m s⁻¹ in 10 s. Plot velocity-time graphs for both cars and calculate displacement in their respective intervals. (B) A clock has a minute hand of length 7 cm. Calculate: (i) distance, (ii) displacement, (iii) average speed, and (iv) average velocity of the tip of the minute hand from 6:00 PM to 7:30 PM. [Exam Favorite]
Answer:
================================================================================ PART A: VELOCITY-TIME GRAPH & DISPLACEMENT OF CARS A & B: -------------------------------------------------------------------------------- 1. Accelerations: • Car A: a_A = (5 - 0) / 5 = 1 m/s² • Car B: a_B = (3 - 0) / 10 = 0.3 m/s² 2. Displacement from Area Under v-t Graph (Triangle): • Displacement of Car A (0 to 5 s) = 1/2 × Base × Height = 1/2 × 5 s × 5 m/s = 12.5 m • Displacement of Car B (0 to 10 s) = 1/2 × Base × Height = 1/2 × 10 s × 3 m/s = 15.0 m ================================================================================ PART B: CLOCK MINUTE HAND KINEMATICS (6:00 PM TO 7:30 PM): -------------------------------------------------------------------------------- GIVEN DATA: • Radius (r) = Length of minute hand = 7 cm • Time Interval = 6:00 PM to 7:30 PM = 90 minutes = 90 × 60 = 5400 s • Revolutions completed = 90 min / 60 min = 1.5 revolutions 1. Distance Travelled by Tip: Distance = 1.5 × (2 · π · r) = 1.5 × (2 × 22/7 × 7 cm) = 1.5 × 44 cm = 66 cm 2. Net Displacement of Tip: At 6:00 PM, tip is at '12' (Top). At 7:30 PM, tip is at '6' (Bottom). Displacement = Straight-line distance across diameter = 2r = 2 × 7 = 14 cm (Downward) 3. Average Speed: v_av = Distance / Time = 66 cm / 90 min = 0.733 cm/min (or 0.0122 cm/s) 4. Average Velocity: v_av = Displacement / Time = 14 cm / 90 min = 0.156 cm/min (Downward) ================================================================================
Frequently Asked Questions (FAQs) – Class 9 Science Chapter 4
Question 1: Define uniform motion and non-uniform motion. [Exam Favorite] Answer:
- Uniform Motion: When an object covers equal distances in equal intervals of time, however small these time intervals may be, its motion is uniform (velocity is constant, acceleration is zero).
- Non-Uniform Motion: When an object covers unequal distances in equal intervals of time (or variable speeds/directions), its motion is non-uniform (acceleration is non-zero).
Question 2: What does the odometer and speedometer of an automobile measure? [Exam Favorite] Answer:
- Odometer: Records and displays the total path length (distance travelled) in kilometers.
- Speedometer: Measures and indicates the instantaneous speed of the vehicle at that specific moment in kilometers per hour (km/h).
Question 3: Derive the first equation of motion (v = u + at) from the definition of acceleration. [Exam Favorite] Answer: By definition, acceleration is the rate of change of velocity: a = \frac{v – u}{t} \implies a \cdot t = v – u \implies v = u + at.
Question 4: What is meant by Uniform Circular Motion? Why is it accelerated? [Exam Favorite] Answer: When an object moves along a circular path with a constant speed, its motion is called uniform circular motion. It is continuously accelerated because its direction of motion changes at every single point along the circumference, requiring a continuous inward centripetal acceleration.
Question 5: What is the nature of the distance-time graph for non-uniform motion? [Exam Favorite] Answer: The distance-time graph for non-uniform motion is a curved non-linear line, indicating that the rate of distance covered varies across successive time intervals.
Question 6: How can you find the displacement from a velocity-time (v-t) graph? [Exam Favorite] Answer: The magnitude of displacement is numerically equal to the total area enclosed under the velocity-time graph curve and the time axis over the specified time interval.
Question 7: What is negative acceleration called? Give an everyday example. [Exam Favorite] Answer: Negative acceleration is termed Retardation or Deceleration. An everyday example is a driver applying brakes to a moving car, causing its velocity to decrease from 60\text{ km/h} to 0\text{ km/h}.
Question 8: Can displacement be greater than distance? [Exam Favorite] Answer: No. Displacement is the straight-line shortest path between two points; therefore, the magnitude of displacement is always less than or equal to the total distance travelled (\vert{}\text{Displacement}\vert{} \le \text{Distance}).
Question 9: Under what conditions are the equations of motion (v = u + at, etc.) applicable? [Exam Favorite] Answer: The three kinematic equations of motion are valid only when an object moves along a straight line with constant (uniform) acceleration. They cannot be applied directly to non-uniform acceleration.
Question 10: Convert 90\text{ km/h} into \text{m/s} and 25\text{ m/s} into \text{km/h}. [Exam Favorite] Answer:
- 90\text{ km/h} = 90 \times \frac{5}{18} = 5 \times 5 = \mathbf{25\text{ m/s}}.
- 25\text{ m/s} = 25 \times \frac{18}{5} = 5 \times 18 = \mathbf{90\text{ km/h}}.
Question 11: What is a reference point (origin)? Why is it needed? [Exam Favorite] Answer: A reference point is a fixed origin relative to which the spatial position, distance, and motion of an object are measured and described. Without a reference point, motion cannot be defined.
Question 12: An object is dropped freely from the top of a building. What is its initial velocity and acceleration? [Exam Favorite] Answer:
- Initial Velocity: u = 0\text{ m/s} (starts from rest).
- Acceleration: a = +g = +9.8\text{ m/s}^2 (downward acceleration due to gravity).
Question 13: What does a horizontal line parallel to the time axis on a v-t graph represent? [Exam Favorite] Answer: It represents that the object is moving with constant velocity (uniform motion), meaning acceleration is strictly zero (a = 0).
Question 14: What is the direction of velocity and acceleration in uniform circular motion? [Exam Favorite] Answer:
- Velocity: Directed tangentially to the circular path at any given point.
- Acceleration (Centripetal): Directed radially inward toward the center of the circle, perpendicular to velocity (\theta = 90^\circ).
Question 15: Why does a runner lean inward while running around a sharp circular track bend? [Exam Favorite] Answer: Leaning inward generates an inward horizontal component of the normal ground reaction force, which provides the necessary centripetal force required to negotiate the circular curve without skidding outward.
Mastering the NCERT Solutions for Class 9 Science Chapter 4 (Exploration), “Describing Motion Around Us”, equips students with the kinematic equations, graphical slope/area analysis, and reaction-braking calculations required for top performance in CBSE physics evaluations. Review the numerical box solutions, the master formula sheet, and the 15 high-yield FAQs above to secure full marks in your examinations.