CG SCERT (Chhattisgarh Board) Class 9 Science Solutions | 2026-27 Official Edition
Complete 100% textbook-grounded CG Board Class 9 Science Chapter 4 Solutions: Motion (गति) in Bilingual Format (English & Hindi Medium).
Set 1 – Q1
Differences Between Speed and Velocity
📝 Question:
Write down the differences between speed and velocity.
💡 Solution:
| Property | Speed | Velocity |
|---|---|---|
| Definition | The distance covered by an object per unit time is called speed. | The displacement of an object per unit time in a given direction is called velocity. |
| Quantity Type | It is a scalar quantity (has magnitude only). | It is a vector quantity (has both magnitude and direction). |
| Value | Speed is always positive or zero; it cannot be negative. | Velocity can be positive, negative, or zero depending on displacement. |
| Formula | Speed = (Distance)/(Time) | Velocity = (Displacement)/(Time) |
Set 1 – Q2
Condition for Average Velocity = Average Speed
📝 Question:
In what situation will the average velocity of an object be equal to its average speed?
💡 Solution:
The average velocity of an object will be equal to its average speed when the object moves along a straight line path in a single direction without changing its direction of motion. In this situation, the magnitude of total displacement is exactly equal to the total distance covered.
Set 1 – Q3
Velocity-Time Graph for Uniform Motion
📝 Question:
Draw a velocity-time graph for a uniform motion.
💡 Solution:
• In uniform motion, the velocity of an object remains constant with time.
• Therefore, the velocity-time graph is a straight line parallel to the time axis (x-axis).
• Graph Description: Plot Time (t) on the x-axis and Velocity (v) on the y-axis. Draw a horizontal straight line at a height corresponding to constant velocity v.
• Therefore, the velocity-time graph is a straight line parallel to the time axis (x-axis).
• Graph Description: Plot Time (t) on the x-axis and Velocity (v) on the y-axis. Draw a horizontal straight line at a height corresponding to constant velocity v.
Set 2 – Q1
Velocity-Time Graph for Uniform Deceleration
📝 Question:
Draw a velocity-time graph for a uniform deceleration motion.
💡 Solution:
• In uniform deceleration (retardation), the velocity of the object decreases uniformly with time.
• The velocity-time graph is a straight line sloping downwards from a higher velocity value on the y-axis toward the time axis (x-axis).
• The velocity-time graph is a straight line sloping downwards from a higher velocity value on the y-axis toward the time axis (x-axis).
Set 2 – Q2
Car Acceleration & Distance Numerical
📝 Question:
A car increases its velocity from 18 km/h to 36 km/h in 5 sec, at a constant acceleration. Find its acceleration and distance covered by it.
💡 Solution:
• Step 1: Convert velocities to SI units (m/s):
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 interval (t) = 5 s
• Step 2: Calculate Acceleration (a):
a = (v – u)/(t) = (10 – 5)/(5) = 5/5 = 1 m/s2
• Step 3: Calculate Distance Covered (s):
s = ut + 1/2at2 = (5 × 5) + 1/2 × 1 × (5)2 = 25 + 12.5 = 37.5 m
• Conclusion: Acceleration = 1 m/s2, Distance covered = 37.5 m.
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 interval (t) = 5 s
• Step 2: Calculate Acceleration (a):
a = (v – u)/(t) = (10 – 5)/(5) = 5/5 = 1 m/s2
• Step 3: Calculate Distance Covered (s):
s = ut + 1/2at2 = (5 × 5) + 1/2 × 1 × (5)2 = 25 + 12.5 = 37.5 m
• Conclusion: Acceleration = 1 m/s2, Distance covered = 37.5 m.
Set 2 – Q3
Braking Deceleration & Stopping Time Numerical
📝 Question:
A car is moving at a constant velocity of 36 km/h. On applying brakes it stops at a distance of 10 m. Calculate deceleration and time taken by the car to stop.
💡 Solution:
• Given Values: Initial velocity u = 36 km/h = 36 × 5/18 = 10 m/s, Final velocity v = 0 m/s (stops), Distance s = 10 m
• Calculate Deceleration (a): Using v2 – u2 = 2as
02 – (10)2 = 2 × a × 10 ⇒ -100 = 20a ⇒ a = -5 m/s2
Magnitude of deceleration = 5 m/s2
• Calculate Time Taken (t): Using v = u + at
0 = 10 + (-5)t ⇒ 5t = 10 ⇒ t = 2 s
• Conclusion: Deceleration = 5 m/s2, Time taken = 2 s.
• Calculate Deceleration (a): Using v2 – u2 = 2as
02 – (10)2 = 2 × a × 10 ⇒ -100 = 20a ⇒ a = -5 m/s2
Magnitude of deceleration = 5 m/s2
• Calculate Time Taken (t): Using v = u + at
0 = 10 + (-5)t ⇒ 5t = 10 ⇒ t = 2 s
• Conclusion: Deceleration = 5 m/s2, Time taken = 2 s.
MCQ 1
Multiple Choice Question (i)
📝 Question:
(i) If distance covered is s = at + 1/2bt2, the acceleration is:
💡 Solution:
Correct Option: (b) b
Explanation: Comparing s = at + 1/2bt2 with s = ut + 1/2At2, coefficient of 1/2t2 is acceleration A = b.
Explanation: Comparing s = at + 1/2bt2 with s = ut + 1/2At2, coefficient of 1/2t2 is acceleration A = b.
Options: (a) a | (b) b | (c) 2b | (d) a + b
MCQ 2
Multiple Choice Question (ii)
📝 Question:
(ii) The area covered by velocity-time graph on the time axis is:
💡 Solution:
Correct Option: (a) m
Explanation: Area under v-t graph represents displacement/distance whose SI unit is meter (m).
Explanation: Area under v-t graph represents displacement/distance whose SI unit is meter (m).
Options: (a) m | (b) m/s | (c) m/s2 | (d) None of these
MCQ 3
Multiple Choice Question (iii)
📝 Question:
(iii) Distance-time graph for two bodies A and B is given. Which body has more velocity?
💡 Solution:
Correct Option: (b) B
Explanation: Slope of distance-time graph equals velocity. Line B has a steeper slope, so body B has greater velocity.
Explanation: Slope of distance-time graph equals velocity. Line B has a steeper slope, so body B has greater velocity.
Options: (a) A | (b) B | (c) A and B | (d) None of these
MCQ 4
Multiple Choice Question (iv)
📝 Question:
(iv) Which one of the following is an example of uniform circular motion:
💡 Solution:
Correct Option: (d) Motion of a racing car along circular track
Explanation: A racing car moving at a constant speed along a circular track is an example of uniform circular motion.
Explanation: A racing car moving at a constant speed along a circular track is an example of uniform circular motion.
Options: (a) Motion of earth around the sun | (b) Motion of a toy train along a linear path | (c) Motion of the second’s hand of a clock | (d) Motion of a racing car along circular track
Blank 1
Fill in the Blank (i)
📝 Question:
(i) A body is moving at a constant velocity, its acceleration will be ……………………………
💡 Solution:
Answer: Zero
Reason: Since velocity is constant (v – u = 0), acceleration a = 0.
Reason: Since velocity is constant (v – u = 0), acceleration a = 0.
Blank 2
Fill in the Blank (ii)
📝 Question:
(ii) The direction of velocity of a body depends on ……………………
💡 Solution:
Answer: Displacement (विस्थापन की दिशा)
Reason: Velocity is defined as displacement per unit time, so its direction is always along displacement.
Reason: Velocity is defined as displacement per unit time, so its direction is always along displacement.
Blank 3
Fill in the Blank (iii)
📝 Question:
(iii) If the speed of an object is 72 km/h, its speed in m/s is ……………………………
💡 Solution:
Answer: 20 m/s
Reason: Calculation: 72 × 5/18 = 20 m/s.
Reason: Calculation: 72 × 5/18 = 20 m/s.
Blank 4
Fill in the Blank (iv)
📝 Question:
(iv) A person covers one round of a circular track of 2 m radius in 2 s. What will be his displacement after 8 seconds?
💡 Solution:
Answer: Zero
Reason: Number of complete rounds in 8 s = 8 / 2 = 4. The person returns to the exact starting point, so displacement is 0.
Reason: Number of complete rounds in 8 s = 8 / 2 = 4. The person returns to the exact starting point, so displacement is 0.
Exercise Q3
Uniform Circular Motion Definition & Examples
📝 Question:
What is the meaning of a uniform circular motion? Give its two examples.
💡 Solution:
• Definition: When an object moves along a circular path with a constant speed, its motion is called uniform circular motion. Although speed remains constant, the direction of motion changes continuously at every point, making it an accelerated motion.
• Examples:
1. Motion of the tip of the second’s hand of a clock.
2. Motion of an artificial satellite orbiting the Earth at a constant speed.
• Examples:
1. Motion of the tip of the second’s hand of a clock.
2. Motion of an artificial satellite orbiting the Earth at a constant speed.
Exercise Q4
Motion of a Body Around the Sun
📝 Question:
A body moves around the sun at a constant speed. Is its motion a uniform motion or a non-uniform motion?
💡 Solution:
• In terms of speed, its motion is uniform because it covers equal distance in equal intervals of time.
• However, in terms of velocity, its motion is non-uniform (accelerated motion). This is because the direction of motion continuously changes along the curved path around the Sun, resulting in a continuously changing velocity vector.
• However, in terms of velocity, its motion is non-uniform (accelerated motion). This is because the direction of motion continuously changes along the curved path around the Sun, resulting in a continuously changing velocity vector.
Exercise Q5
Displacement-Time Graph Velocity Analysis
📝 Question:
The displacement-time graph for an object is given. What result can you find out for its velocity?
💡 Solution:
1. Section A to B: Straight line sloping upwards ⇒ Represents uniform velocity (constant positive velocity).
2. Section B to C: Horizontal straight line (parallel to time axis) ⇒ Position does not change with time, object is at rest (velocity = 0).
3. Section C to D: Steeper straight line sloping upwards ⇒ Represents higher uniform velocity than in section A to B.
2. Section B to C: Horizontal straight line (parallel to time axis) ⇒ Position does not change with time, object is at rest (velocity = 0).
3. Section C to D: Steeper straight line sloping upwards ⇒ Represents higher uniform velocity than in section A to B.
Exercise Q6
Position-Time Graph Parallel to Time Axis
📝 Question:
The position-time graph for a given object is a straight line parallel to its time axis. What can you say about its motion?
💡 Solution:
• When the position-time graph is a straight line parallel to the time axis, the position of the object does not change as time passes.
• Therefore, the object is stationary (at rest) and its velocity is zero.
• Therefore, the object is stationary (at rest) and its velocity is zero.
Exercise Q7
Uniform and Non-Uniform Motion
📝 Question:
Define uniform motion and a non-uniform motion also state two examples of each.
💡 Solution:
• 1. Uniform Motion: When an object covers equal distances in equal intervals of time, its motion is called uniform motion.
Examples: (i) Hands of a clock. (ii) Car moving at constant 40 km/h on a straight highway.
• 2. Non-Uniform Motion: When an object covers unequal distances in equal intervals of time, its motion is called non-uniform motion.
Examples: (i) A bus moving through crowded city traffic. (ii) A freely falling stone from a height.
Examples: (i) Hands of a clock. (ii) Car moving at constant 40 km/h on a straight highway.
• 2. Non-Uniform Motion: When an object covers unequal distances in equal intervals of time, its motion is called non-uniform motion.
Examples: (i) A bus moving through crowded city traffic. (ii) A freely falling stone from a height.
Exercise Q8
Three Equations of Motion & Symbols
📝 Question:
Write all the three equations of motion and explain the symbols used in them.
💡 Solution:
• Three Equations of Motion:
1. First Equation: v = u + at
2. Second Equation: s = ut + 1/2at2
3. Third Equation: v2 – u2 = 2as
• Meaning of Symbols:
u = Initial velocity, v = Final velocity, a = Uniform acceleration, t = Time interval, s = Distance/Displacement.
1. First Equation: v = u + at
2. Second Equation: s = ut + 1/2at2
3. Third Equation: v2 – u2 = 2as
• Meaning of Symbols:
u = Initial velocity, v = Final velocity, a = Uniform acceleration, t = Time interval, s = Distance/Displacement.
Exercise Q9
Athlete Circular Track Distance Numerical
📝 Question:
An athlete covers one round of a circular track of 7 m radius in 44 sec. How much distance will he cover in 1 minute? (Book Answer: 60 m)
💡 Solution:
• Step 1: Circumference of track: Radius r = 7 m ⇒ Distance in 1 round = 2πr = 2 × (22/7) × 7 = 44 m
• Step 2: Speed of athlete: Speed = (Distance)/(Time) = 44 m / 44 s = 1 m/s
• Step 3: Distance in 1 minute (60 s): Distance s = Speed × Time = 1 m/s × 60 s = 60 m
• Conclusion: The athlete will cover 60 m in 1 minute.
• Step 2: Speed of athlete: Speed = (Distance)/(Time) = 44 m / 44 s = 1 m/s
• Step 3: Distance in 1 minute (60 s): Distance s = Speed × Time = 1 m/s × 60 s = 60 m
• Conclusion: The athlete will cover 60 m in 1 minute.
Exercise Q10
Derivation of Second Equation of Motion
📝 Question:
Using the graphical method derive the second equation of motion s = ut + 1/2at2.
💡 Solution:
• Let an object move with initial velocity u at t = 0 with uniform acceleration a. After time t, its velocity becomes v.
• On the velocity-time graph, initial velocity OA = u, final velocity BD = v, and time OD = t.
• The total distance s covered equals the area under graph OABD:
s = Area of rectangle OACD + Area of triangle ABC
• Area of rectangle OACD = OA × OD = u × t = ut
• Area of triangle ABC = 1/2 × Base × Height = 1/2 × AC × BC = 1/2 × t × (v – u)
• Since v – u = at (from first equation), Area of triangle ABC = 1/2 × t × (at) = 1/2at2
• Adding both areas: s = ut + 1/2at2 (Hence proved).
• On the velocity-time graph, initial velocity OA = u, final velocity BD = v, and time OD = t.
• The total distance s covered equals the area under graph OABD:
s = Area of rectangle OACD + Area of triangle ABC
• Area of rectangle OACD = OA × OD = u × t = ut
• Area of triangle ABC = 1/2 × Base × Height = 1/2 × AC × BC = 1/2 × t × (v – u)
• Since v – u = at (from first equation), Area of triangle ABC = 1/2 × t × (at) = 1/2at2
• Adding both areas: s = ut + 1/2at2 (Hence proved).
Exercise Q11
Average Speed of Journey Numerical
📝 Question:
Neeraj reaches school in a car at an average speed of 20 km/h. While returning, his speed becomes 30 km/h due to lesser crowd. What is the average speed of Neeraj’s car for the whole journey? (Book Answer: 24 km/h)
💡 Solution:
• Step 1: Distance: Let distance from home to school be x km.
• Step 2: Times: Time going t1 = x/20 h, Time returning t2 = x/30 h
Total time t = x/20 + x/30 = (3x + 2x)/60 = 5x/60 = x/12 h
• Step 3: Average Speed: Total distance = x + x = 2x km
Average Speed = (Total Distance)/(Total Time) = (2x)/(x/12) = 2x × 12/x = 24 km/h
• Conclusion: Average speed for the whole journey = 24 km/h.
• Step 2: Times: Time going t1 = x/20 h, Time returning t2 = x/30 h
Total time t = x/20 + x/30 = (3x + 2x)/60 = 5x/60 = x/12 h
• Step 3: Average Speed: Total distance = x + x = 2x km
Average Speed = (Total Distance)/(Total Time) = (2x)/(x/12) = 2x × 12/x = 24 km/h
• Conclusion: Average speed for the whole journey = 24 km/h.
Exercise Q12
Freely Falling Ball Speed & Time Numerical
📝 Question:
A ball is dropped from the top of a 20 m high tower. If its velocity is increasing with an acceleration of 10 m/s2 then at what speed will the ball hit the ground? How much time will it take to hit the ground? (Book Answer: 20 m/s, 2 s)
💡 Solution:
• Given: Initial velocity u = 0 m/s, Height s = 20 m, Acceleration a = 10 m/s2
• Final Velocity (v): Using v2 – u2 = 2as ⇒ v2 – 0 = 2 × 10 × 20 = 400 ⇒ v = 20 m/s
• Time Taken (t): Using v = u + at ⇒ 20 = 0 + 10t ⇒ t = 20/10 = 2 s
• Conclusion: Impact speed = 20 m/s, Time taken = 2 s.
• Final Velocity (v): Using v2 – u2 = 2as ⇒ v2 – 0 = 2 × 10 × 20 = 400 ⇒ v = 20 m/s
• Time Taken (t): Using v = u + at ⇒ 20 = 0 + 10t ⇒ t = 20/10 = 2 s
• Conclusion: Impact speed = 20 m/s, Time taken = 2 s.
Exercise Q13
Velocity-Time Table, Acceleration & Distance
📝 Question:
A car is moving on a straight road at a uniform acceleration. The velocity of the car at different times is given in the table. Draw its velocity-time graph, calculate its acceleration, and find out the distance covered by it in 30 seconds.
💡 Solution:
| Time t (s) | 0 | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|---|
| Speed v (m/s) | 5 | 10 | 15 | 20 | 25 | 30 |
• Acceleration (a): a = (v2 – v1)/(t2 – t1) = (10 – 5)/(10 – 0) = 5/10 = 0.5 m/s2
• Distance in 30 seconds (s):
s = ut + 1/2at2 = (5 × 30) + 1/2 × 0.5 × (30)2 = 150 + 225 = 375 m
• Conclusion: Acceleration = 0.5 m/s2, Distance covered = 375 m.
• Distance in 30 seconds (s):
s = ut + 1/2at2 = (5 × 30) + 1/2 × 0.5 × (30)2 = 150 + 225 = 375 m
• Conclusion: Acceleration = 0.5 m/s2, Distance covered = 375 m.
Exercise Q14
Position-Time Graph Segment Speeds
📝 Question:
In the figure, the position-time graph for an object at different times is shown. Calculate the speed of the object: (i) A to B, (ii) B to C, (iii) C to D.
💡 Solution:
• Coordinates: A (0 s, 0 cm), B (5 s, 3 cm), C (7 s, 3 cm), D (9 s, 9 cm)
• (i) Speed from A to B: v = (3 – 0)/(5 – 0) = 3/5 = 0.6 cm/s
• (ii) Speed from B to C: v = (3 – 3)/(7 – 5) = 0/2 = 0 cm/s (Object at rest)
• (iii) Speed from C to D: v = (9 – 3)/(9 – 7) = 6/2 = 3 cm/s
• (i) Speed from A to B: v = (3 – 0)/(5 – 0) = 3/5 = 0.6 cm/s
• (ii) Speed from B to C: v = (3 – 3)/(7 – 5) = 0/2 = 0 cm/s (Object at rest)
• (iii) Speed from C to D: v = (9 – 3)/(9 – 7) = 6/2 = 3 cm/s
Exercise Q15
Train Acceleration & Distance Numerical
📝 Question:
A train starts moving from rest and acquires a velocity of 72 km/h in 5 minutes. If the train moves at a constant acceleration, calculate: (i) Acceleration (Book Answer: 0.06 m/s2), (ii) Distance travelled by the train to acquire this velocity (Book Answer: 3000 m).
💡 Solution:
• Given: u = 0 m/s, v = 72 km/h = 72 × 5/18 = 20 m/s, t = 5 min = 5 × 60 = 300 s
• (i) Acceleration (a): a = (v – u)/(t) = (20 – 0)/(300) = 1/15 ≈ 0.06 m/s2
• (ii) Distance (s): Using v2 – u2 = 2as ⇒ (20)2 – 0 = 2 × (1/15) × s
400 = (2/15)s ⇒ s = (400 × 15)/2 = 3000 m (or 3 km)
• Conclusion: Acceleration = 0.06 m/s2, Distance travelled = 3000 m.
• (i) Acceleration (a): a = (v – u)/(t) = (20 – 0)/(300) = 1/15 ≈ 0.06 m/s2
• (ii) Distance (s): Using v2 – u2 = 2as ⇒ (20)2 – 0 = 2 × (1/15) × s
400 = (2/15)s ⇒ s = (400 × 15)/2 = 3000 m (or 3 km)
• Conclusion: Acceleration = 0.06 m/s2, Distance travelled = 3000 m.