NCERT Solutions for Class 9 Science Exploration Chapter 4

Motion is the change in the position of an object with time when observed from a chosen reference point. Distance, displacement, velocity, acceleration, graphs, and equations help describe motion in straight and circular paths.

NCERT Solutions for Class 9 Science Exploration Chapter 4 help students understand Describing Motion Around Us, a Physics chapter from the 2026-27 Class 9 Science textbook. This chapter explains how motion is described using reference point, position, distance, displacement, average speed, average velocity, average acceleration, graphs, and kinematic equations. 

CBSE students should practise this chapter carefully because it includes formula-based numericals, graph interpretation, reasoning questions, and daily-life applications like braking distance and safe driving. These NCERT Class 9 Science Solutions cover Pause and Ponder questions, graph-based questions, and all Revise, Reflect, Refine exercise answers from Class 9 Science Exploration Chapter 4.

Key Takeaways

• Motion: An object is in motion when its position changes with time relative to a reference point.
• Main difference: Distance is the total path covered, while displacement is the net change in position.
• Graph skill: Position-time and velocity-time graphs show velocity, acceleration, and displacement.
• Formula focus: Kinematic equations apply only when acceleration is constant.

NCERT Solutions for Class 9 Science Exploration Chapter 4 Structure 2026

NCERT Solutions for Class 9 Science Exploration Chapter 4 In-Text Questions

Class 9 Science Exploration Chapter 4 uses real-life motion examples to explain Physics quantities. These answers focus on direction, magnitude, units, graphs, and equations.

Q1. When will the displacement of the athlete be zero?

Question: 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?

Answer: The displacement will be zero when the athlete returns to the starting point.

Explanation:
Displacement is the net change in position. If final position is the same as initial position, displacement becomes zero.

If the athlete starts from O, goes to A, and returns to O, then:

$Displacement = 0 , m$

The total distance travelled will not be zero. It will be the full length of the path covered while going and returning.

For example:

$Total , distance = OA + AO$

If $OA = 100 , m$,

$Total , distance = 100 + 100 = 200 , m$

Q2. Fuel used by a vehicle depends on distance or displacement?

Question: Fuel used up in a vehicle depends on which of the following?
(i) Total distance travelled
(ii) Displacement

Answer: Fuel used by a vehicle depends on total distance travelled.

Explanation:
Fuel is consumed while the vehicle moves along the actual path. It does not depend only on the shortest distance between starting and ending points.

If a car moves 10 km away and comes back to the starting point, its displacement is zero. Still, fuel is used for the full 20 km journey.

So, fuel consumption depends on distance travelled, not displacement.

Q3. Is the motion of a ball on an inclined track a straight-line motion?

Question: A ball rolls down an inclined track. Is its motion a straight-line motion? Can it be depicted using a horizontal line? Are distance and displacement equal?

Answer: Yes, the motion is straight-line motion if the ball rolls along a straight inclined track.

Explanation:
The ball moves along a straight path, even though the track is inclined. Its motion can be represented using a horizontal number line for convenience.

The horizontal line does not show the actual slope. It only represents position along the straight track.

If the ball moves from O to A, B, C, and D without turning back, then:

$Magnitude , of , displacement = Total , distance , travelled$

So, distance and displacement are equal at A, B, C, and D.

Q4. Find average speed and average velocity during a road trip.

Question: During a family road trip, you drive 200 km north in three hours. Afterwards, you drive 200 km south in two hours. Find the average speed and average velocity for your entire trip.

Answer: The average speed is $80 , km , h^{-1}$ and the average velocity is $0 , km , h^{-1}$.

Explanation:
Total distance travelled:

$200 , km + 200 , km = 400 , km$

Total time taken:

$3 , h + 2 , h = 5 , h$

Average speed:

$Average , speed = \frac{Total , distance}{Total , time}$

$Average , speed = \frac{400}{5} = 80 , km , h^{-1}$

Displacement is zero because the car returns to the starting point.

$Average , velocity = \frac{Displacement}{Time}$

$Average , velocity = \frac{0}{5} = 0 , km , h^{-1}$

Q5. Conditions for average speed and average velocity

Question: Under what conditions is the magnitude of average velocity equal to average speed? When is average velocity zero while average speed is not zero?

Answer: The magnitude of average velocity equals average speed when the object moves in one direction without turning back.

Explanation:
In such motion:

$Distance , travelled = Magnitude , of , displacement$

So:

$Average , speed = Magnitude , of , average , velocity$

The magnitude of average velocity is zero when displacement is zero. This happens when the object returns to its starting point.

Example:

A swimmer goes from one end of a pool to the other and returns to the starting point.

$Displacement = 0$

$Average , velocity = 0$

The average speed is not zero because distance was travelled.

NCERT Solutions for Class 9 Science Exploration Chapter 4 Exercise Questions

The Revise, Reflect, Refine section of Describing Motion Around Us Class 9 includes numericals, graph interpretation, and reasoning-based questions. These Class 9 Science Chapter 4 exercise solutions use formulas in a copyable format.

Q1. Father goes to a shop and returns home twice. Find distance and displacement.

Question: My father went to a shop from home located 250 m away. He came home to take a cloth bag, went to the shop again, bought provisions, and came back home. Find total distance and displacement.

Answer: The total distance travelled is $1000 , m$ and displacement is $0 , m$.

Explanation:
One trip from home to shop:

$250 , m$

Total path:

Home to shop = $250 , m$

Shop to home = $250 , m$

Home to shop again = $250 , m$

Shop to home again = $250 , m$

Total distance:

$250 + 250 + 250 + 250 = 1000 , m$

Final position is home, which is the starting point.

$Displacement = 0 , m$

Q2. Student runs from ground floor to fourth floor and then comes down to second floor.

Question: A student runs from the ground floor to the fourth floor and then comes down to the classroom on the second floor. Height of each floor is 3 m. Find total vertical distance and displacement.

Answer: The total vertical distance is $18 , m$ and displacement is $6 , m$ upward.

Explanation:
Distance from ground floor to fourth floor:

$4 \times 3 = 12 , m$

Distance from fourth floor to second floor:

$(4 - 2) \times 3 = 6 , m$

Total vertical distance:

$12 + 6 = 18 , m$

Final position is second floor.

$Displacement = 2 \times 3 = 6 , m$ upward

Q3. Can a scooter accelerate if speedometer reading is constant?

Answer: Yes, the scooter can accelerate even if the speedometer reading is constant.

Explanation:
The speedometer shows the magnitude of velocity, which is speed. Acceleration happens when velocity changes.

Velocity changes if speed changes or direction changes.

If the scooter moves on a curved road at constant speed, its direction changes continuously. So, it has acceleration due to change in direction.

This idea is important in uniform circular motion Class 9.

Q4. A car starts from rest and reaches 24 m/s in 6 s. Find acceleration and distance.

Answer: The average acceleration is $4 , m , s^{-2}$ and distance travelled is $72 , m$.

Explanation:
Given:

$u = 0 , m , s^{-1}$

$v = 24 , m , s^{-1}$

$t = 6 , s$

Average acceleration:

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

$a = \frac{24 - 0}{6} = 4 , m , s^{-2}$

Distance travelled:

$s = ut + \frac{1}{2}at^2$

$s = 0 \times 6 + \frac{1}{2} \times 4 \times 6^2$

$s = 2 \times 36 = 72 , m$

Q5. A motorbike stops after travelling 98 m. Find acceleration and time.

Question: A motorbike moving with initial velocity $28 , m , s^{-1}$ and constant acceleration stops after travelling $98 , m$. Find acceleration and time taken.

Answer: The acceleration is $-4 , m , s^{-2}$ and time taken is $7 , s$.

Explanation:
Given:

$u = 28 , m , s^{-1}$

$v = 0 , m , s^{-1}$

$s = 98 , m$

Use:

$v^2 = u^2 + 2as$

$0^2 = 28^2 + 2 \times a \times 98$

$0 = 784 + 196a$

$a = -4 , m , s^{-2}$

Now use:

$v = u + at$

$0 = 28 + (-4)t$

$4t = 28$

$t = 7 , s$

The negative acceleration means the motorbike slows down.

Q6. Do objects A and B ever have equal velocity in the position-time graph?

Answer: Objects A and B have equal velocity if their position-time graphs have equal slopes at any point or interval.

Explanation:
In a position-time graph Class 9 question, velocity is represented by slope.

If both graphs are straight and parallel, their velocities are equal. If the slopes are different, their velocities are different.

From Fig. 4.27, if the lines of A and B are not parallel, their velocities are not equal. If they intersect, that only means equal position at that instant, not equal velocity.

Equal velocity depends on equal slope, not equal position.

Q7. Choose correct options from the position-time graph of A and B.

Answer: The correct options are (i) and (iii).

Explanation:
Both objects have the same initial and final positions in 10 seconds. So, their displacements are equal.

Average velocity:

$Average , velocity = \frac{Displacement}{Time}$

Since displacement and time are the same, average velocities are equal.

Object B covers a longer path because its position changes more during the journey. So, average speed of A is lower than that of B.

Thus, option (i) and option (iii) are correct.

Q8. Truck slows from 54 km/h to 36 km/h in 36 s. Find distance.

Answer: The truck travels $450 , m$ while slowing down.

Explanation:
Convert velocities:

$54 , km , h^{-1} = 15 , m , s^{-1}$

$36 , km , h^{-1} = 10 , m , s^{-1}$

Given:

$u = 15 , m , s^{-1}$

$v = 10 , m , s^{-1}$

$t = 36 , s$

For constant acceleration, average velocity:

$Average , velocity = \frac{u + v}{2}$

$Average , velocity = \frac{15 + 10}{2} = 12.5 , m , s^{-1}$

Distance:

$s = Average , velocity \times time$

$s = 12.5 \times 36 = 450 , m$

Q9. Car accelerates, moves uniformly, then stops. Find total distance.

Question: A car starts from rest, reaches $20 , m , s^{-1}$ in 5 s, travels at $20 , m , s^{-1}$ for 10 s, and stops in 6 s. Find total distance.

Answer: The total distance travelled is $310 , m$.

Explanation:
First part:

$u = 0 , m , s^{-1}$

$v = 20 , m , s^{-1}$

$t = 5 , s$

$s_1 = \frac{u + v}{2} \times t$

$s_1 = \frac{0 + 20}{2} \times 5 = 50 , m$

Second part:

$s_2 = vt$

$s_2 = 20 \times 10 = 200 , m$

Third part:

$u = 20 , m , s^{-1}$

$v = 0 , m , s^{-1}$

$t = 6 , s$

$s_3 = \frac{u + v}{2} \times t$

$s_3 = \frac{20 + 0}{2} \times 6 = 60 , m$

Total distance:

$s = s_1 + s_2 + s_3$

$s = 50 + 200 + 60 = 310 , m$

Q10. Will the bus stop before the obstacle?

Question: A bus travels at $36 , km , h^{-1}$ and sees an obstacle 30 m ahead. The driver reacts after 0.5 s. Braking acceleration is $2.5 , m , s^{-2}$. Will it stop before the obstacle?

Answer: Yes, the bus will stop before reaching the obstacle.

Explanation:
Convert speed:

$36 , km , h^{-1} = 10 , m , s^{-1}$

Distance travelled during reaction time:

$s_1 = vt$

$s_1 = 10 \times 0.5 = 5 , m$

Distance covered after brakes are applied:

Given:

$u = 10 , m , s^{-1}$

$v = 0 , m , s^{-1}$

$a = -2.5 , m , s^{-2}$

Use:

$v^2 = u^2 + 2as$

$0 = 10^2 + 2 \times (-2.5) \times s$

$0 = 100 - 5s$

$s = 20 , m$

Total stopping distance:

$5 + 20 = 25 , m$

The obstacle is 30 m ahead.

$25 , m < 30 , m$

So, the bus stops before the obstacle.

Q11. Can an object kept on Earth be considered at rest?

Answer: Yes, an object kept on Earth can be considered at rest relative to the Earth.

Explanation:
Rest and motion depend on the chosen reference point. If the Earth is taken as the reference point, the object’s position does not change.

So, the object is at rest with respect to Earth.

However, Earth moves around the Sun. So, the same object is in motion with respect to the Sun.

This shows that rest and motion are relative.

Q12. Use the velocity-time graph of a cyclist to calculate displacement and acceleration.

Answer: The displacement is found from the area under the velocity-time graph, and average acceleration is found from change in velocity divided by time.

Explanation:
From Fig. 4.30, the motion has three parts:

1. Velocity increases from $0$ to $6 , m , s^{-1}$ in 40 s.
2. Velocity remains constant at $6 , m , s^{-1}$ from 40 s to 80 s.
3. Velocity decreases from $6 , m , s^{-1}$ to $0$ from 80 s to 120 s.

Displacement from 0 to 40 s:

$s_1 = \frac{1}{2} \times 40 \times 6 = 120 , m$

Displacement from 40 s to 80 s:

$s_2 = 40 \times 6 = 240 , m$

Displacement from 80 s to 120 s:

$s_3 = \frac{1}{2} \times 40 \times 6 = 120 , m$

Total displacement:

$s = 120 + 240 + 120 = 480 , m$

Average acceleration over 120 s:

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

$a = \frac{0 - 0}{120} = 0 , m , s^{-2}$

The displacement is $480 , m$ and average acceleration over the full interval is $0 , m , s^{-2}$.

Q13. Estimate the distance run from the velocity-time graph.

Answer: The distance is estimated from the area under the velocity-time graph.

Explanation:
From Fig. 4.31, velocity is shown in $km , h^{-1}$ and time is shown in hours.

The graph forms a triangle with:

Base = $6 , h$

Height = $7.5 , km , h^{-1}$

Distance:

$Distance = Area , under , velocity-time , graph$

$Distance = \frac{1}{2} \times base \times height$

$Distance = \frac{1}{2} \times 6 \times 7.5$

$Distance = 22.5 , km$

So, the girl ran approximately $22.5 , km$.

Q14. Find displacement using velocity-time graph.

Question: A car moves with constant velocity $6 , m , s^{-1}$ for 2 minutes and then accelerates at $1 , m , s^{-2}$ for 6 s. Find displacement in 2 min 6 s.

Answer: The total displacement is $774 , m$.

Explanation:
First part:

$t_1 = 2 , min = 120 , s$

$v = 6 , m , s^{-1}$

$s_1 = vt$

$s_1 = 6 \times 120 = 720 , m$

Second part:

$u = 6 , m , s^{-1}$

$a = 1 , m , s^{-2}$

$t_2 = 6 , s$

$s_2 = ut + \frac{1}{2}at^2$

$s_2 = 6 \times 6 + \frac{1}{2} \times 1 \times 6^2$

$s_2 = 36 + 18 = 54 , m$

Total displacement:

$s = s_1 + s_2$

$s = 720 + 54 = 774 , m$

Q15. Compare velocity-time graphs of cars A and B.

Question: Car A reaches $5 , m , s^{-1}$ in 5 s. Car B reaches $3 , m , s^{-1}$ in 10 s. Both start from rest. Find displacement in the mentioned time intervals.

Answer: Car A travels $12.5 , m$ and Car B travels $15 , m$.

Explanation:
For Car A:

$u = 0 , m , s^{-1}$

$v = 5 , m , s^{-1}$

$t = 5 , s$

Displacement is area under velocity-time graph:

$s_A = \frac{1}{2} \times 5 \times 5$

$s_A = 12.5 , m$

For Car B:

$u = 0 , m , s^{-1}$

$v = 3 , m , s^{-1}$

$t = 10 , s$

$s_B = \frac{1}{2} \times 10 \times 3$

$s_B = 15 , m$

Accelerations:

$a_A = \frac{5 - 0}{5} = 1 , m , s^{-2}$

$a_B = \frac{3 - 0}{10} = 0.3 , m , s^{-2}$

Car A has higher acceleration, but Car B travels slightly more distance during its longer time interval.

Q16. Motion of the tip of the minute hand from 6 PM to 7:30 PM

Question: Rohan studies from 6 PM to 7:30 PM. Consider the tip of the minute hand of a wall clock. Find distance, displacement, speed, and velocity. Length of minute hand is 7 cm.

Answer: Distance travelled is $66 , cm$, displacement is $14 , cm$ downward, speed is $0.0122 , cm , s^{-1}$, and velocity is $0.00259 , cm , s^{-1}$ downward.

Explanation:
Time interval:

$6:00 , PM , to , 7:30 , PM = 90 , min$

In 60 minutes, the minute hand completes one revolution.

In 90 minutes, it completes:

$1.5 , revolutions$

Radius:

$r = 7 , cm$

Distance in one revolution:

$2\pi r = 2 \times \frac{22}{7} \times 7 = 44 , cm$

Distance in 1.5 revolutions:

$Distance = 1.5 \times 44 = 66 , cm$

At 6:00, the tip is at 12. At 7:30, the tip is at 6.

Displacement is the straight-line distance from 12 to 6:

$Displacement = diameter = 2r = 14 , cm$ downward

Time in seconds:

$90 , min = 90 \times 60 = 5400 , s$

Speed:

$Speed = \frac{Distance}{Time}$

$Speed = \frac{66}{5400} = 0.0122 , cm , s^{-1}$

Velocity:

$Velocity = \frac{Displacement}{Time}$

$Velocity = \frac{14}{5400} = 0.00259 , cm , s^{-1}$ downward

NCERT Solutions for Class 9 Science Exploration

Topics Covered in NCERT Solutions for Class 9 Science Exploration Chapter 4

Class 9 Science Exploration Chapter 4 covers motion in a straight line and uniform circular motion. The chapter builds the foundation for later Physics topics like force, gravitation, and energy.

• Motion and rest
• Reference point
• Position of an object
• Motion in a straight line Class 9
• Distance and displacement Class 9
• Magnitude and direction
• Scalars and vectors
• Average speed and average velocity Class 9
• Uniform and non-uniform motion
• Rate of change
• Average acceleration Class 9
• Acceleration during speeding up and slowing down
• Acceleration due to gravitational force
• Position-time graph Class 9
• Velocity-time graph Class 9
• Slope of a graph
• Area under a velocity-time graph
• Kinematic equations Class 9
• Braking distance and road safety
• Motion in a plane
• Uniform circular motion Class 9
• Tangent to a circular path
• Acceleration due to change in direction

Important Formulas in NCERT Solutions for Class 9 Science Exploration Chapter 4

Class 9 Science Chapter 4 solutions use formulas for speed, velocity, acceleration, graph area, and uniformly accelerated motion. These formulas should be written with correct SI units.

Important Concepts in Describing Motion Around Us Class 9

Describing Motion Around Us Class 9 teaches students to describe motion using measurable quantities. The chapter uses both numbers and graphs to make motion easier to compare.