CLASS 9 SCIENCE

Describing Motion

Complete Chapter Overview, Concepts, Formulas & Examples

1. Introduction to Motion

Motion is one of the most common phenomena in our surroundings. Cars move on roads, birds fly in the sky, planets revolve around the Sun, and people walk from one place to another.

In this chapter, we learn how to describe the motion of an object using quantities such as distance, displacement, speed, velocity and acceleration.

To describe motion scientifically, we need to know the position of an object and how its position changes with time.

Key Idea:

Motion means a change in the position of an object with respect to a reference point as time passes.

2. What is Motion?

An object is said to be in motion if its position changes with time relative to a reference point.

Reference Point

A reference point is a fixed point with respect to which we observe the position or motion of an object.

Example:

Suppose you are sitting inside a moving bus. You are at rest with respect to another passenger sitting beside you, but you are in motion with respect to a person standing on the road.

Important:

Motion is relative. An object can be at rest relative to one reference point and in motion relative to another.

3. Distance and Displacement

Distance

Distance is the total length of the actual path travelled by an object from its initial position to its final position.

Distance

Type: Scalar quantity

Unit: metre (m)

It has magnitude only.

Displacement

Type: Vector quantity

Unit: metre (m)

It has magnitude and direction.

Displacement

Displacement is the shortest distance between the initial and final positions of an object, measured in a particular direction.

Remember:

Distance depends on the actual path travelled, whereas displacement depends only on the initial and final positions.

Example:

A student walks 3 metres east and then 4 metres north. The total distance travelled is:

Distance = 3 + 4 = 7 m

The magnitude of displacement is:

Displacement = √(3² + 4²) = 5 m

Distance vs Displacement

Distance	Displacement
Actual path travelled	Shortest path between initial and final positions
Scalar quantity	Vector quantity
Has magnitude only	Has magnitude and direction
Always positive or zero	Can be positive, negative or zero depending on direction

4. Speed

Speed tells us how fast an object is moving. It is defined as the distance travelled by an object per unit time.

Speed = Distance / Time SI Unit: metre per second (m/s)

Average Speed

Average speed is the ratio of the total distance travelled to the total time taken.

Average Speed = Total Distance / Total Time

Uniform Speed

If an object covers equal distances in equal intervals of time, it is said to have uniform speed.

Non-uniform Speed

If an object covers unequal distances in equal intervals of time, its speed is non-uniform.

Numerical Example:

A car travels 120 km in 3 hours. Find its average speed.

Average Speed = 120 / 3

Average Speed = 40 km/h

5. Velocity

Velocity is the rate of change of displacement with time.

Velocity = Displacement / Time SI Unit: metre per second (m/s)

Velocity is a vector quantity because it has both magnitude and direction.

Difference Between Speed and Velocity

Speed	Velocity
Based on distance	Based on displacement
Scalar quantity	Vector quantity
Direction is not required	Direction is required
Cannot be negative	Can change depending on direction

Exam Tip:

Speed tells us how fast an object moves, while velocity tells us how fast and in which direction the object moves.

6. Acceleration

Acceleration is the rate of change of velocity with respect to time.

a = (v – u) / t a = acceleration, u = initial velocity, v = final velocity, t = time

The SI unit of acceleration is m/s².

Positive Acceleration

When the velocity of an object increases with time, its acceleration is positive.

Negative Acceleration

When the velocity decreases with time, the acceleration is negative. This is commonly called deceleration or retardation.

Example:

A car increases its velocity from 10 m/s to 30 m/s in 5 seconds. Find its acceleration.

a = (v – u) / t

a = (30 – 10) / 5

a = 4 m/s²

7. Uniform and Non-uniform Motion

Uniform Motion

An object is said to be in uniform motion when it covers equal distances in equal intervals of time.

Example: A train travelling at a constant speed in a straight line.

Non-uniform Motion

An object is in non-uniform motion when its speed or direction changes with time.

Example: A car moving through city traffic.

8. Graphical Representation of Motion

Graphs are useful for studying how the position, distance and velocity of an object change with time.

Distance-Time Graph

A distance-time graph shows the relationship between distance travelled and time taken.

Important Rule:

The slope of a distance-time graph gives the speed of the object.

Speed = Slope of distance-time graph

Velocity-Time Graph

A velocity-time graph represents the relationship between velocity and time.

Important Rules:

The slope of a velocity-time graph gives acceleration.
The area under a velocity-time graph gives displacement.

9. Equations of Motion

When an object moves with uniform acceleration, we can use three important equations known as the equations of motion.

v = u + at First equation of motion

s = ut + ½at² Second equation of motion

v² = u² + 2as Third equation of motion

Meaning of Symbols

Symbol	Meaning	SI Unit
u	Initial velocity	m/s
v	Final velocity	m/s
a	Acceleration	m/s²
t	Time	s
s	Displacement	m

Important:

These equations are applicable when the acceleration of the object is uniform.

10. Uniform Circular Motion

When an object moves along a circular path with constant speed, the motion is called uniform circular motion.

Examples:

Motion of the hands of a clock.
A satellite revolving around Earth.
A stone tied to a string and rotated in a circle.
Motion of a point on a rotating fan.

Although the speed may remain constant, the direction of motion continuously changes. Since velocity depends on both speed and direction, the velocity changes continuously.

Therefore:

An object moving with constant speed in a circular path is still accelerating because its direction of velocity changes continuously.

11. Important Formulas at a Glance

Speed

Speed = Distance / Time

Velocity

Velocity = Displacement / Time

Acceleration

a = (v – u) / t

First Equation

v = u + at

Second Equation

s = ut + ½at²

Third Equation

v² = u² + 2as

12. Chapter Summary

Motion is the change in position of an object with time relative to a reference point.
Distance is the total path travelled by an object.
Displacement is the shortest distance between the initial and final positions in a particular direction.
Speed is the distance travelled per unit time.
Velocity is the displacement travelled per unit time.
Acceleration is the rate of change of velocity.
The slope of a distance-time graph represents speed.
The slope of a velocity-time graph represents acceleration.
The area under a velocity-time graph represents displacement.
The equations of motion are applicable to objects moving with uniform acceleration.
An object moving in a circle at constant speed is accelerating because its direction continuously changes.

13. Exam Preparation Tips

For Theory Questions

Learn the definitions of distance, displacement, speed, velocity and acceleration.
Understand the difference between scalar and vector quantities.
Remember the differences between distance and displacement.
Understand why circular motion involves acceleration.

For Numerical Problems

Write the values given in the question.
Convert all quantities into SI units when required.
Choose the correct formula.
Substitute the values carefully.
Always write the final answer with the correct unit.

Class 9 New Science Book 2026 Ch4 Describing Motion – Textbook Answers

Solved Questions & Numericals from New Text Book

The following questions help you understand the concepts of distance, displacement, velocity, acceleration and equations of motion.

Question 1

Distance and Displacement

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?

Solution

Distance between home and shop = 250 m

Movement of father:

Home → 250 m → Shop → 250 m → Home → 250 m → Shop → 250 m → Home

Total distance travelled:

Distance = 250 + 250 + 250 + 250 Distance = 1000 m

Therefore, the total distance travelled = 1000 m.

Since he finally returned home, his initial and final positions are the same.

Displacement = 0 m

Question 2

Distance and Displacement on Different Floors

A student runs from the ground floor to the fourth floor of a school building to collect a book and then comes down to their classroom on the second floor. If the height of each floor is 3 m, find:

the total vertical distance travelled, and
their displacement from the starting point.

Solution

Height of each floor = 3 m

Step 1: Ground Floor → Fourth Floor

Number of floors crossed = 4

Distance upward = 4 × 3 Distance upward = 12 m

Step 2: Fourth Floor → Second Floor

Number of floors moved down = 2

Distance downward = 2 × 3 Distance downward = 6 m

(i) Total Vertical Distance

Total Distance = 12 + 6 Total Distance = 18 m

Therefore, the total vertical distance travelled = 18 m.

(ii) Displacement from Starting Point

Starting point = Ground floor
Final point = Second floor

The student is finally 2 floors above the starting point.

Displacement = 2 × 3 Displacement = 6 m upward

Displacement = 6 m upward

Question 3

Can an Object Accelerate at Constant Speed?

A girl is riding her scooter and finds that its speedometer reading is constant. Is it possible for her scooter to be accelerating and if so, how?

Solution

Yes, it is possible.

Acceleration occurs whenever there is a change in velocity.

Velocity can change when:

the speed changes, or
the direction of motion changes.

If the girl turns the scooter along a curved road, the direction of motion changes continuously.

Therefore, the velocity of the scooter changes even though its speed remains constant.

Conclusion: A scooter can accelerate even when its speed is constant, if its direction of motion changes.

Question 4

Acceleration and Distance Travelled

A car starts from rest and its velocity reaches 24 m/s in 6 s. Find the average acceleration and the distance travelled in these 6 seconds.

Solution

Given:

Initial velocity, u = 0 m/s
Final velocity, v = 24 m/s
Time, t = 6 s

Step 1: Calculate Acceleration

We use the formula:

a = (v − u) / t

Substituting the values:

a = (24 − 0) / 6
a = 24 / 6
a = 4 m/s²

Step 2: Calculate Distance

We use the equation:

s = ut + ½at²

Substituting the values:

s = (0)(6) + ½(4)(6²)
s = 0 + ½ × 4 × 36
s = 72 m

Average acceleration = 4 m/s²

Distance travelled = 72 m

Question 5

Finding Acceleration and Time

A motorbike moving with an initial velocity of 28 m/s and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and the time taken to come to a stop.

Solution

Given:

Initial velocity, u = 28 m/s
Final velocity, v = 0 m/s
Distance travelled, s = 98 m

Step 1: Calculate Acceleration

We use the third equation of motion:

v² = u² + 2as

Substituting the values:

0² = 28² + 2(a)(98)
0 = 784 + 196a
196a = −784
a = −4 m/s²

The negative sign indicates that the motorbike is slowing down.

Step 2: Calculate Time

We use the first equation of motion:

v = u + at

Substituting the values:

0 = 28 + (−4)t
4t = 28
t = 7 s

Acceleration = −4 m/s²

Time taken to stop = 7 s

Quick Revision of These Questions

Question	Main Concept	Final Answer
1	Distance & Displacement	Distance = 1000 m, Displacement = 0 m
2	Distance & Displacement	Distance = 18 m, Displacement = 6 m upward
3	Acceleration & Direction	Yes, changing direction causes acceleration
4	Acceleration & Distance	Acceleration = 4 m/s², Distance = 72 m
5	Equations of Motion	Acceleration = −4 m/s², Time = 7 s

QUESTION 6

Do Objects A and B Ever Have Equal Velocity?

Fig. 4.27 shows a position–time graph of two objects A and B that are moving along parallel tracks in the same direction.

Do objects A and B ever have equal velocity? Justify your answer.

Fig. 4.27 — Position–Time Graph

Fig. 4.27: Position–time graph of objects A and B

✓ Answer

No. Objects A and B do not have equal velocity at any time.

🔑 Why?

The velocity of an object is equal to the slope of its position–time graph.

Velocity = Change in position ÷ Change in time

In the given graph, the straight lines representing objects A and B have different slopes.

The line for object A is steeper than the line for object B. Therefore, object A has a greater velocity than object B.

The two lines intersect at one point. At this point, both objects have the same position at the same instant. However, their slopes are still different.

⚠️ Important

Same position does NOT mean same velocity.

The intersection of the two lines tells us that A and B are at the same position at that instant. Equal velocity would require the two position–time graphs to have the same slope.

✓

Final Answer:

Objects A and B never have equal velocity because the slopes of their position–time graphs are different.

QUESTION 7

Position–Time Graph: Choose the Correct Option(s)

A graph in 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 distance 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 10 s time interval is greater than that of B since B’s speed is lower than A’s in some segments.

Fig. 4.28 — Position–Time Graph

Fig. 4.28: Position–time graph for objects A and B

✓ Correct Option

Correct answer: (i) only

🔑 Key Concept

Average velocity depends on the total displacement and the total time.

Average velocity = Displacement ÷ Time

Average speed depends on the total distance travelled and the total time.

Average speed = Distance ÷ Time

✓ (i) Correct

Both A and B start from the same initial position and reach the same final position at 10 s. Therefore, their displacements are equal.

Since both take the same time of 10 s, their average velocities are equal.

Average velocity = Displacement ÷ Time

✗ (ii) Incorrect

Although both objects take the same time and have the same initial and final positions, this statement is not the correct reason for their average speeds being equal.

However, from the graph both objects move continuously in the same direction without reversing. Therefore, the distance travelled by each object is equal to its displacement.

Hence, their average speeds are also equal. The option’s wording “both cover equal distance in equal time” is true for this graph, so (ii) is also correct.

✗ (iii) Incorrect

Object A does not cover a shorter distance than B. Both objects start at the same position and finish at the same position, and neither reverses its direction.

Therefore, both cover the same total distance. So their average speeds are equal, not different.

✗ (iv) Incorrect

The fact that B’s speed is lower than A’s during some portions of the journey does not mean that A has a greater average speed.

Average speed depends on the total distance travelled over the total time. Since both cover the same distance in the same 10 s, their average speeds are equal.

✓

Final Answer: (i) and (ii)

Both objects have the same average velocity because they have the same displacement in the same time. They also have the same average speed because they travel the same total distance in the same time.

QUESTION 8

Calculate the Total Distance Travelled by the Car

A car starts from rest and accelerates uniformly to 20 m s⁻¹ in 5 seconds.

It then travels at 20 m s⁻¹ for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds.

Find the total distance travelled by the car.

Velocity–Time Graph for the Motion

Velocity–time graph: The area under the graph represents the distance travelled.

✓ Solution

Step 1 — Distance during acceleration

The car starts from rest:

u = 0 m s⁻¹ v = 20 m s⁻¹ t = 5 s

Since acceleration is uniform, the average velocity is:

Average velocity =
(u + v) ÷ 2

= (0 + 20) ÷ 2
= 10 m s⁻¹

Therefore,

Distance = Average velocity × Time
= 10 × 5
= 50 m

Step 2 — Distance at constant velocity

The car travels at a constant velocity of 20 m s⁻¹ for 10 s.

Distance = Velocity × Time
= 20 × 10
= 200 m

Step 3 — Distance while braking

During braking, the car slows uniformly from 20 m s⁻¹ to 0 m s⁻¹ in 6 s.

Average velocity =
(u + v) ÷ 2

= (20 + 0) ÷ 2
= 10 m s⁻¹

Distance = Average velocity × Time
= 10 × 6
= 60 m

🎯 Total Distance Travelled

Total distance = 50 + 200 + 60

= 310 m

✓

Final Answer:

The total distance travelled by the car is 310 metres.

💡 Important Note

If your accompanying figure shows a car speed limit of 50 km/h and a truck speed limit of 40 km/h, those speed-limit values are not used in this particular calculation. The question specifically gives the car’s velocities and times as 20 m s⁻¹, 5 s, 10 s and 6 s.

QUESTION 9

Motion with Acceleration, Constant Speed and Deceleration

A car starts from rest and accelerates uniformly to 20 m s⁻¹ in 5 seconds.

It then travels at constant speed for 10 seconds before decelerating uniformly to rest in 5 seconds.

Find:

The acceleration
The deceleration
The total distance travelled
The total time of the journey

Velocity–Time Graph

The area under a velocity–time graph represents the distance travelled.

Solution

① Acceleration

Given:

u = 0 m s⁻¹, v = 20 m s⁻¹, t = 5 s

a = (v − u) / t

a = (20 − 0) / 5
a = 4 m s⁻²

② Deceleration

During deceleration:

u = 20 m s⁻¹, v = 0 m s⁻¹, t = 5 s

a = (v − u) / t

a = (0 − 20) / 5
a = −4 m s⁻²

Therefore, the deceleration is:

4 m s⁻²

③ Total Distance Travelled

We divide the journey into three parts.

Part 1: Acceleration

Distance = ½(u + v)t
= ½(0 + 20) × 5
= 50 m

Part 2: Constant Speed

Distance = vt
= 20 × 10
= 200 m

Part 3: Deceleration

Distance = ½(u + v)t
= ½(20 + 0) × 5
= 50 m

Total distance = 50 + 200 + 50
= 300 m

④ Total Time of Journey

Total time = 5 + 10 + 5
= 20 s

Final Answers

(a) Acceleration = 4 m s⁻²

(b) Deceleration = 4 m s⁻²

(d) Total time = 20 s

QUESTION 10

Can the Bus Stop Before the Obstacle?

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?

Solution

① Convert the Speed into m/s

36 km h⁻¹ = 36 × 5/18

v = 10 m s⁻¹

② Distance Travelled During Reaction Time

During the reaction time, the driver has not yet applied the brakes. Therefore, the bus continues at 10 m/s.

Distance = Speed × Time

= 10 × 0.5
= 5 m

So, the bus travels 5 m before the brakes are applied.

③ Distance Travelled While Braking

Using:

v² = u² + 2as

At the stopping point:

v = 0, u = 10 m s⁻¹, a = −2.5 m s⁻²

0 = 10² + 2(−2.5)s

0 = 100 − 5s

5s = 100
s = 20 m

④ Total Stopping Distance

Total stopping distance = Reaction distance + Braking distance

= 5 + 20
= 25 m

Distance available 30 m

Distance required to stop 25 m

✓

Yes, the bus will stop before the obstacle.

The bus needs only 25 m to come to rest, while the obstacle is 30 m away.

Therefore, the bus will stop 5 m before the obstacle.

QUESTION 11

Is an Object on Earth at Rest or in Motion?

A student said, “The Earth moves around the Sun”.

In this context, discuss whether an object kept on the Earth can be considered to be at rest.

Answer

🔑 Motion is Relative

An object can be considered to be at rest or in motion depending on the reference point chosen.

🌍

With respect to Earth

An object kept on the Earth does not change its position with respect to the Earth.

It is at REST.

☀️

With respect to the Sun

The Earth moves around the Sun. Therefore, an object on Earth also changes its position with respect to the Sun.

It is in MOTION.

✓

Final Answer

Yes, an object kept on the Earth can be considered to be at rest with respect to the Earth. However, the same object is considered to be in motion with respect to the Sun because the Earth is moving around the Sun.

Therefore, rest and motion are relative concepts.

12.The velocity-time graph from 0 s to 120 s for a cyclist is shown in Fig. 4.30. Shade the areas (in different colours) representing the displacement of the cyclist
(i) while cyclist is moving with constant velocity.
(ii) when the velocity of cyclist is decreasing.
Also, calculate the displacement and average acceleration in the 120 s time interval.

Solution

From the graph:

From $0$ to $20$ s, velocity increases from $0$ to $3$ m s $^{-1}$ −1.
From $20$ s to $100$ s, velocity remains constant at $3$ m s $^{-1}$ −1.
From $100$ s to $120$ s, velocity decreases from $3$ m s $^{-1}$ −1 to $2$ m s $^{-1}$ −1.

(i) Area representing displacement during constant velocity

The cyclist moves with constant velocity from $20$ 20 s to $100$ 100 s.

The displacement is represented by the rectangular area under the graph between:

time = $20$ s to $100$ s
velocity = $3$ m s $^{-1}$

(ii) Area representing displacement when velocity is decreasing

The velocity decreases from $100$ s to $120$ s.

The displacement is represented by the trapezium-shaped area under the graph between:

time = $100$ 100 s to $120$ 120 s

Total displacement

Total displacement = Total area under the velocity-time graph

1. First triangular area (0–20 s)

Using: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ $A_1 = \frac{1}{2} \times 20 \times 3$ $A_1 = 30 \text{ m}$

2. Rectangular area (20–100 s)

Using: $\text{Area} = \text{length} \times \text{breadth}$ $A_2 = (100-20)\times 3$ $A_2 = 80 \times 3$ $A_2 = 240 \text{ m}$

3. Trapezium area (100–120 s)

Using: $\text{Area} = \frac{1}{2}(a+b)h$ $A_3 = \frac{1}{2}(3+2)\times 20$ $A_3 = \frac{1}{2}\times 5 \times 20$ $A_3 = 50 \text{ m}$

Total displacement

$\text{Displacement} = A_1 + A_2 + A_3$ $= 30 + 240 + 50$ $= 320 \text{ m}$

Average acceleration

Using:

$a=\frac{v-u}{t}$

Initial velocity, $u = 0$ m s $^{-1}$
Final velocity, $v = 2$ m s $^{-1}$
Total time, $t = 120$ s $a = \frac{2-0}{120}$ $a = \frac{1}{60}$ $a \approx 0.017 \text{ m s}^{-2}$

13.A girl is preparing for her first marathon by running on a straight road. She uses a smartwatch to calculate her running speed at different intervals. The graph (Fig. 4.31) depicts her velocity versus time. Estimate the running distance based on the graph.

Solution

The distance travelled is equal to the area under the velocity-time graph.

From the graph, the approximate values are:

Time (h)	Velocity (km h $^{-1}$ −1)
0	6.5
1	7.5
2	7.5
3	6.5
4	5.5
5	5.5

We calculate the area interval by interval using the trapezium formula: $\text{Area}=\frac{1}{2}(a+b)h$

where:

$a$ and $b$ are velocities,
$h$ is the time interval.

1. From 0 h to 1 h

$A_1=\frac{1}{2}(6.5+7.5)\times1$ $A_1=7 \text{ km}$

2. From 1 h to 2 h

$A_2=\frac{1}{2}(7.5+7.5)\times1$ $A_2=7.5 \text{ km}$

3. From 2 h to 3 h

$A_3=\frac{1}{2}(7.5+6.5)\times1$ $A_3=7 \text{ km}$

4. From 3 h to 4 h

$A_4=\frac{1}{2}(6.5+5.5)\times1$ $A_4=6 \text{ km}$

5. From 4 h to 5 h

$A_5=\frac{1}{2}(5.5+5.5)\times1$ $A_5=5.5 \text{ km}$

Total running distance

$\text{Distance}=A_1+A_2+A_3+A_4+A_5$ $=7+7.5+7+6+5.5$ $=33 \text{ km}$

Question 14

Question:

On entering a state highway, a car continues to move with a constant velocity of 6 m s⁻¹ for 2 minutes and then accelerates with a constant acceleration of 1 m s⁻² for 6 seconds.

Find the displacement of the car on the state highway in the 2 min 6 s time interval by drawing a velocity-time graph for its motion.

Given Information

Initial constant velocity = 6 m/s
Time at constant velocity = 2 min = 120 s
Acceleration = 1 m/s²
Time of acceleration = 6 s

Step 1: Find the Final Velocity

During the second part of the motion, the car accelerates uniformly. We use:

v = u + at

v = 6 + (1 × 6)

v = 12 m/s

Therefore, after 6 seconds of acceleration, the velocity becomes 12 m/s.

Step 2: Draw the Velocity-Time Graph

For the first 120 seconds, velocity remains constant at 6 m/s. During the next 6 seconds, velocity increases uniformly from 6 m/s to 12 m/s.

Step 3: Calculate Displacement from the Graph

The displacement is equal to the area under the velocity-time graph.

The graph consists of:

A rectangle for the first 120 seconds.
A trapezium for the last 6 seconds.

Part A: Displacement in the First 120 s

Area of rectangle = Length × Breadth

s₁ = 120 × 6

s₁ = 720 m

Part B: Displacement in the Last 6 s

The area under the velocity-time graph during acceleration is a trapezium.

Area of trapezium = ½ × (Sum of parallel sides) × Height

s₂ = ½ × (6 + 12) × 6

s₂ = ½ × 18 × 6

s₂ = 54 m

Step 4: Total Displacement

Total displacement = s₁ + s₂

= 720 + 54

= 774 m

✅ Final Answer: Displacement of the car = 774 m

Quick Summary

Time Interval	Graph Area	Displacement
0–120 s	Rectangle	720 m
120–126 s	Trapezium	54 m
Total		774 m

⭐ Important Concept:
The area under a velocity-time graph gives the displacement. Here, the total area is the area of the rectangle plus the area of the trapezium.

QUESTION 16

Motion of the Tip of a Minute Hand

Question:

Rohan studies science from 6 PM to 7:30 PM at home. Consider the tip of the minute’s hand of the wall clock. During the given time interval, what is its:

distance travelled,
displacement,
speed, and
velocity.

The length of the minute’s hand is 7 cm.

Given

Length of minute hand (radius), r = 7 cm
Initial time = 6:00 PM
Final time = 7:30 PM

Time interval:

Time = 90 minutes

Important Concept

The minute hand of a clock completes one complete revolution in 60 minutes.

Therefore, in 90 minutes, the minute hand completes:

Number of revolutions = 90 / 60 = 1.5 revolutions

(i) Distance Travelled

The tip of the minute hand moves along a circular path. Therefore, distance travelled is the circumference of the circle multiplied by the number of revolutions.

Circumference of the circular path:

Circumference = 2πr

Since the minute hand makes 1.5 revolutions:

Distance = 2πr × 1.5

= 2 × (22/7) × 7 × 1.5

= 44 × 1.5

= 66 cm

Answer: Distance travelled = 66 cm

(ii) Displacement

From 6:00 PM to 7:30 PM, the minute hand completes 1.5 revolutions.

After 1 complete revolution, the minute hand returns to its original position. After another half revolution, it reaches the position opposite to the starting point.

Therefore, the displacement is equal to the diameter of the circle.

Displacement = Diameter = 2r

Displacement = 2 × 7

= 14 cm

Answer: Displacement = 14 cm

(iii) Speed

Speed is defined as the distance travelled per unit time.

Speed = Distance / Time

Distance = 66 cm

Time = 90 minutes

Speed = 66 / 90

Speed = 0.733 cm/min

Answer: Speed = 0.733 cm/min

In SI units:

66 cm = 0.66 m

90 minutes = 5400 s

Speed = 0.66 / 5400

Speed = 1.22 × 10⁻⁴ m/s

(iv) Velocity

Average velocity is defined as displacement divided by the total time taken.

Average Velocity = Displacement / Time

Displacement = 14 cm

Time = 90 minutes

Average Velocity = 14 / 90

Average Velocity = 0.1556 cm/min

Average Velocity ≈ 0.156 cm/min

Answer: Average velocity ≈ 0.156 cm/min

In SI units:

Displacement = 14 cm = 0.14 m

Time = 5400 s

Average Velocity = 0.14 / 5400

Average Velocity ≈ 2.59 × 10⁻⁵ m/s

Final Answer

Distance travelled 66 cm

Displacement 14 cm

Average Speed 0.733 cm/min

Average Velocity 0.156 cm/min

💡 Remember

Distance depends on the actual path travelled, whereas displacement depends only on the initial and final positions.

Similarly, speed is based on distance, while average velocity is based on displacement.

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Class 9 Science

Class 9 Updated New Science Book 2026 Solutions

Explore chapter-wise solutions, concepts, explanations, and study resources for the Class 9 Updated New Science Book 2026.

📖 Class 9 Science Chapter 1 — Exploration → 🔬 Class 9 Science Chapter 2 — The Cell → 🧬 Class 9 Science Chapter 3 — Tissues in Action → 🏃 Class 9 Science Chapter 4 — Describing Motion → 🧪 Class 9 Science Chapter 5 — Exploring Mixtures → ⚡ Class 9 — How Forces Affect Motion → ⚛️ Class 9 — Journey Inside Atom → 🧬 Class 9 Science Chapter 11 — Reproduction →

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🏛️ Official NCERT Textbooks

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Visit Official NCERT Website ↗

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Class 9 New Science Book 2026 Ch4 Describing Motion

Motion is one of the most important concepts in science that helps us understand how objects move around us. In Class 9 New Science Book 2026 Ch4 Describing Motion, students learn about distance, displacement, speed, velocity, acceleration, and different types of motion in a simple and practical way.This chapter explains motion using real-life examples, activities, graphs, and numerical problems. The solutions provided here will help students understand concepts clearly and prepare well for exams and assignments.