Class 9 Science Ch 7 Motion introduces one of the most important concepts in physics. Motion is a part of everyday life — from a moving car on the road to planets revolving around the sun. In this chapter, students learn how to describe motion using distance, displacement, speed, velocity, and acceleration. The chapter also explains graphical representation of motion and equations of motion through solved examples and activities.
This study guide is based on the NCERT Class 9 Science textbook (CBSE curriculum) and is equally helpful for international students who want a clear understanding of basic physics concepts. With detailed notes, solved questions, and practical examples, learners can strengthen their foundation for higher studies and competitive exams.
Class 9 Science Chapter 7 – Motion (Detailed Concepts)
Introduction
Motion is one of the most basic and important concepts in physics. Every object around us is either at rest or in motion with respect to some observer. A moving car, a flying bird, a rotating fan, or the planets moving around the sun are all examples of motion. In this chapter, we will study how to describe motion, measure it, and represent it mathematically and graphically.
1. Rest and Motion
- An object is said to be at rest if it does not change its position with time relative to its surroundings.
- An object is said to be in motion if it changes its position with time relative to its surroundings.
- Motion is always relative.
- Example: A passenger sitting inside a moving bus is at rest with respect to other passengers but in motion with respect to a tree outside the bus.
2. Types of Motion
- Linear or Rectilinear Motion Motion along a straight line. Example: A car moving on a straight road.
- Circular Motion – Motion of an object along a circular path.
Example: Motion of the moon around the earth. - Oscillatory Motion – To and fro motion about a fixed point.
Example: Motion of a pendulum. - Rotational Motion – Motion of a body about its own axis.
Example: Rotation of the earth on its axis
3. Distance and Displacement
- Distance: The actual path length covered by an object.. It is a scalar quantity (has magnitude only)..It is always positive.
- Displacement: The shortest straight-line distance between the initial and final position of the object in a specific direction.. It is a vector quantity (has both magnitude and direction).
- Can be positive, negative, or zero.
- Example: If a person walks 5 m east and then 5 m west,
- Displacement = 0 .. Distance = 10 m
👉 Key Point: Distance tells how much ground is covered, while displacement tells how far and in which direction.
4. Speed and Velocity
Speed: Distance travelled per unit time. Scalar quantity. Formula:
Speed = Distance ÷ Time
Velocity: Displacement per unit time in a given direction. Vector quantity. Formula:
Velocity = Displacement ÷ Time
👉 Key Point: Two objects can have the same speed but different velocities if they move in different directions.
5. Types of Speed and Velocity
- Uniform Speed: When an object covers equal distances in equal intervals of time.
- Non-Uniform Speed: When an object covers unequal distances in equal intervals of time.
- Uniform Velocity: Displacement increases equally in equal intervals of time in a fixed direction.
- Non-Uniform Velocity: Displacement changes unequally in equal intervals of time or direction keeps changing.
6. Acceleration
- Acceleration is the rate of change of velocity with respect to time.
- Formula: a = (v – u) ÷ t
- u = initial velocity
- v = final velocity
- t = time
- Positive Acceleration: When velocity increases.
- Negative Acceleration (Retardation/Deceleration): When velocity decreases.
Example: A car increases velocity from 20 m/s to 30 m/s in 5 seconds.
Acceleration = (30 – 20) ÷ 5 = 2 m/s².
7. Uniform and Non-Uniform Motion
- Uniform Motion: Equal distances in equal intervals of time.
- Non-Uniform Motion: Unequal distances in equal intervals of time.
👉 Key Point: Most real-life motion is non-uniform, like vehicles in traffic.
8. Graphical Representation of Motion
Distance–Time Graph:
- Straight line → uniform speed.
- Curved line → non-uniform speed.
- Slope = speed.
Velocity–Time Graph:
- Straight horizontal line → uniform velocity.
- Sloping line → uniformly accelerated motion.
- Slope = acceleration.
Area under the graph = displacement.
👉 Key Point: Graphs are very useful for solving numerical problems.
9. Equations of Uniformly Accelerated Motion
These equations are derived using graphs.
- v = u + at
- s = ut + ½at²
- v² – u² = 2as
(where s = displacement, a = acceleration, t = time)
Graphical Derivation of Equations of Motion
We use a velocity–time (v–t) graph for a body moving with uniform acceleration.
- Initial velocity = u
- Final velocity after time t= v
- Acceleration = a
- Displacement in time t= s
On the v–t graph, the line rises straight from (0,u) to (t,v)
The area under this line gives the displacement.
First Equation of Motion
v=u+at
Proof:
From the definition of acceleration,
a=Change in velocity/Time taken =(v−u)/t
10. Average Speed and Average Velocity
- Average Speed = Total Distance ÷ Total Time.
- Average Velocity = Total Displacement ÷ Total Time.
Example: A car goes 100 km east and then returns 100 km west in 4 hours.
- Total Distance = 200 km → Average Speed = 200 ÷ 4 = 50 km/h.
- Displacement = 0 → Average Velocity = 0.
👉 Key Point: Average speed can never be zero but average velocity can be zero.
11. Uniform Circular Motion
When an object moves along a circular path with constant speed, it is called uniform circular motion.
Speed remains constant but velocity changes continuously due to change in direction.
Hence, uniform circular motion is accelerated motion.
Examples:
Motion of a stone tied to a string and whirled in a circle.
Motion of the earth around the sun.
👉 Key Point: Even at constant speed, circular motion involves acceleration.
Summary of Key Points
- Motion is relative – depends on the observer.
- Distance is scalar, Displacement is vector.
- Speed is scalar, Velocity is vector.
- Acceleration is the rate of change of velocity.
- Graphs help visualize and calculate motion.
- Equations of motion are essential for solving problems.
- Circular motion is accelerated motion even at constant speed.
Class 9 Science Ch 7 Motion – Textbook Solution
Intext Question -Page 74
1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer: An object can have zero displacement even when it has moved a distance and shown some displacement. This takes place when the final position of an object coincides with its own initial position. Let’s say, if a person moves around circular area and comes back to the place from where he started then the displacement will be zero.
2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
Answer:
3. 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: None of the above statements is actually true for displacement.
First statement is completely false because displacement can be zero in case of a circular path. Second statement is also false because displacement is less than or equal to the distance travelled by the object in case of motion in opposite direction.
Intext Question -Page 76
1. Distinguish between speed and velocity.
Answer:
| Speed | Velocity |
| Speed is the distance travelled by an object in a given interval of time. | Velocity is the displacement shown by an object in a given interval of time. |
| Speed = Distance / Time | Velocity = Displacement / Time |
| Speed is scalar quantity and has only magnitude. | Velocity is vector quantity and has both magnitude and direction. |
2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Answer: The magnitude of average velocity of an object is equal to its average speed, only in one condition when an object is moving in a straight line.
3. What does the odometer of an automobile measure?
Answer: The odometer of an automobile measures the distance covered by a vehicle or an automobile.
4. What does the path of an object look like when it is in uniform motion?
Answer: An object with uniform motion has a straight line path.
5. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108 m s-1.
Answer
Intext Question -Page 77
1. When will you say a body is in (i) uniform acceleration? (ii) Non-uniform acceleration?
Answer:
(i) A body is said to be in uniform acceleration if it travels in a straight line and its velocity increases or decreases by equal amounts in equal time intervals.
(ii) A body is said to be in non-uniform acceleration if the rate of change of its velocity is not constant, that is differs in different time intervals.
Intext Question -Page 81
1. What is the nature of the distance – ‘time graphs for uniform and non-uniform motion of an object?
Answer: When the motion is uniform, the distance time graph is a straight line with a slope.
When the motion is non-uniform, the distance time graph is not a straight line. It can be any kind of curve.
2. What can you say about the motion of an object whose distance – time graph is a straight line parallel to the time axis?
Answer: If distance time graph is a straight line parallel to the time axis, the body is said to be at rest.
3. What can you say about the motion of an object if its speed – ‘time graph is a straight line parallel to the time axis?
Answer: If speed time graph is a straight line parallel to the time axis, the object is said to be moving uniformly.
4. What is the quantity which is measured by the area occupied below the velocity -time graph?
Answer: The area under the velocity-time graph gives the distance covered by an object.
Intext Question -Page 82
1. A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
Answer:
2. A train is travelling at a speed of 90 km h-1. Brakes are applied so as to produce a uniform acceleration of ?0.5 m s-2. Find how far the train will go before it is brought to rest.
Answer:
3. A trolley, while going down an inclined plane, has an acceleration of 2 cm s-2. What will be its velocity 3 s after the start?
Answer:
4. A racing car has a uniform acceleration of 4 m s – ′2. What distance will it cover in 10 s after start?
Answer:
5. A stone is thrown in a vertically upward direction with a velocity of 5 m s-1. If the acceleration of the stone during its motion is 10 m s-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Answer:
EXERCISE QUESTIONS Pg 85
1. 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?
Answer:
After 3 full rounds the runner is at the start. The additional half round puts the runner at the point diametrically opposite the start. The straight-line distance between two opposite points on a circle = the diameter.
2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Answer:
3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h-1. On his return trip along the same route, there is less traffic and the average speed is 30 km h-1. What is the average speed for Abdul’s trip?
Answer: Let the one-way distance be S
4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s-2 for 8.0 s. How far does the boat travel during this time?
Answer:
5. A driver of a car travelling at 52 km h-1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h-1 in another car applies his 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?
Answer:
6. Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answers 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?
Answer: (a) Which of the three is travelling the fastest?
Speed on a distance–time graph = slope of the line (distance / time).
The steepest line is B (it rises most rapidly), so B is travelling the fastest.
(b) Are all three ever at the same point on the road?
Yes. All three graphs meet at a common point — they all pass through the same point on the graph at about t≈1.2 hour,distance≈8 km.
So at that instant all three are at the same place.
(c) How far has C travelled when B passes A?
“B passes A” means the time when the B and A lines cross. Reading the graph, B meets A at about t≈1.5 h
and at that moment the corresponding distance on C is about C has travelled ≈11 km
(d) How far has B travelled by the time it passes C?
“B passes C” is the time when B and C intersect. From the graph the B–C intersection occurs at about t≈1.2 h
and the distance at that moment is about B has travelled ≈8 km.
7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground? After what time will it strike the ground?
Answer
8. The speed-time graph for a car is shown is Fig. 8.12.
(a) Find out how far the car travels in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
Answer: a)
(b) The part of the graph in red color between time 6 s to 10 s represents uniform motion of the car.
9. State which of the following situations are possible and give an example for each of these:
(a) an object with a constant acceleration but with zero velocity.
(b) An object moving in a certain direction with acceleration in the perpendicular direction.
Answer: (a) An object with a constant acceleration but with zero velocity
- Yes, possible.
At an instant, an object can have zero velocity but still experience constant acceleration. - Example: When a ball is thrown vertically upwards, at the highest point its velocity becomes zero, but the acceleration due to gravity (9.8m/s2 downward) is still constant.
(b) An object moving in a certain direction with acceleration in the perpendicular direction
Example: A satellite moving around Earth in a circular orbit has velocity along the orbit, but acceleration is directed toward Earth (center of orbit).
Yes, possible.
This is the case in uniform circular motion. The object moves along the tangent to the circle (say eastward), while the acceleration (centripetal) always acts toward the center of the circle (perpendicular to velocity).
10. 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.
Answer:
Motion – Worksheet
A. Multiple Choice Questions (MCQs)
- Which of the following is a vector quantity?
a) Distance
b) Speed
c) Velocity
d) Time - An object moves 4 km east and then 3 km west. The displacement is:
a) 7 km
b) 1 km
c) 12 km
d) 0 km - Which graph shows uniform speed?
a) Straight line distance–time graph
b) Curved distance–time graph
c) Horizontal velocity–time graph
d) Both (a) and (c) - In a velocity–time graph, the area under the graph represents:
a) Speed
b) Acceleration
c) Displacement
d) Distance
B. Short Answer Questions
- Define displacement. How is it different from distance?
- Write the three equations of uniformly accelerated motion.
- Why is uniform circular motion considered accelerated motion?
- State two examples each of:
(i) Oscillatory motion
(ii) Rotational motion
C. Numerical Problems
- A car starts from rest and accelerates at 2 m/s2. Find its velocity after 5 s.
- A train moving with uniform acceleration attains a velocity of 72 km/h in 5 minutes. If its initial velocity was 18 km/h, find the acceleration.
- A body moving with an initial velocity of 10 m/s comes to rest after travelling 25 m under uniform retardation. Find the acceleration and time taken to stop.
D. Diagram Questions
- Draw a distance–time graph for:
(i) Uniform motion
(ii) Non-uniform motion - Draw a velocity–time graph for:
(i) Uniform velocity
(ii) Uniformly accelerated motion
Indicate how to find acceleration and displacement from the graph.
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