NCERT Solutions for Class 9 Science Chapter 4 Describing Motion Around Us

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1.0Download NCERT Class 9 Science Chapter 4 Describing Motion Around Us

Chapter 4: Describing Motion Around Us introduces the concepts of motion, distance, displacement, speed, and velocity, helping students understand how objects move and how motion can be measured and represented.

NCERT Solutions Class 9 Science Chapter 4

2.0Learning Outcomes – NCERT Class 9 Science Chapter 4, Describing Motion Around Us

Differentiate between distance and displacement, and explain why displacement can be zero even when distance travelled is not.
Calculate average speed and average velocity, and identify the conditions under which they are equal.
Define acceleration and distinguish between uniform and non-uniform acceleration.
Interpret position-time and velocity-time graphs to determine speed, velocity, and acceleration.
Derive and apply the three kinematic equations of motion to solve numerical problems.
Analyse real-life motion scenarios, such as braking distance and free fall, using graphical and equation-based methods.
Understand uniform circular motion and explain why it involves constant acceleration despite constant speed.
Recognise that motion is relative to a chosen reference point or frame of reference.

3.0Detailed NCERT Class 9 Science Chapter 4 Solutions

1. How much distance should we maintain from the truck ahead to avoid a collision if it suddenly applies the brakes? Does this distance depend upon the speed with which we are moving?

Solution

To avoid a collision, we should maintain a safe distance from the truck ahead so that we have enough time to stop if it suddenly applies the brakes. Yes, this distance depends on the speed of our vehicle. As the speed increases, the stopping distance also increases. Therefore, we should keep a larger distance at higher speeds to ensure safety.

2. 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?

Solution

Athlete running back and forth on a straight track The displacement of the athlete will be zero when he returns to his starting point. In that case, the total distance travelled is not zero; it is the actual length of the path covered during the entire journey.

3. Fuel used up in a vehicle depends on which of the following? Justify your answer.
(i) Total distance travelled
(ii) Displacement

Solution

Fuel used up in a vehicle depends on: (i) Total distance travelled
Fuel is consumed according to the actual path covered by the vehicle.
It does not depend on displacement because displacement only considers the shortest distance between initial and final positions.

4. A ball rolls down an inclined track as shown in figure. Is its motion, a straight line motion? Assuming the starting point of the ball ( 0 ) to be the origin, can its motion from 0 to D be depicted using a horizontal line as shown in figure? Are the values of total distance travelled and magnitude of displacement from O equal or different at positions $A, B, C$ and D ?

A ball rolling down an inclined track

Solution

Distances from 0:

At A $= 40 cm$
At $B = 40 + 10 = 50 cm$
At $C = 40 + 10 + 20 = 70 cm$
At $D = 40 + 10 + 20 + 30 = 100 cm$

Since the ball moves in a straight line from O:

Total distance travelled = Magnitude of displacement at each position. Therefore:
At A: Distance = Displacement = 40 cm
At B: Distance $=$ Displacement $= 50 cm$
At C: Distance = Displacement = 70 cm
At D: Distance = Displacement = 100 cm

So, the values are equal at all positions A, B, C and D.

5. 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.

Solution

Average speed and average velocity

Distance travelled north $= 200 km$
Distance travelled south $= 200 km$
Total distance $= 200 + 200 = 400 km$
Total time $= 3 + 2 = 5 h$

Average Speed $=$ Total distance ÷ Total time $= 400 \div 5 = 80 km / h$ Since the final position is the same as the starting position: Displacement $= 0 km$ Average Velocity $=$ Displacement ÷ Total time $= 0 \div 5 = 0 km / h$

Average Speed $= 80 km / h$
Average Velocity $= 0 km / h$

6. Under what condition(s) is the (i) magnitude of average velocity of an object equal to its average speed? (ii) magnitude of average velocity of an object zero while its average speed is not zero?

Solution

Conditions (i) When an object moves in a straight line in one direction without changing direction. (ii) When an object returns to its starting point, so displacement is zero but distance travelled is not zero. For example, completing one round of a circular track.

7. 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 $= 250 m$ Total journey: Home to shop $= 250 m$ Shop to home $= 250 m$ Home to shop $= 250 m$ Shop to home $= 250 m$ Total distance travelled $= 250 + 250 + 250 + 250$ $= 1000 m$ Displacement $=$ Final position - Initial position = 0 Total distance $= 1000 m$ Displacement $= 0 m$

8. 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: (i) the total vertical distance travelled, and (ii) their displacement from the starting point.

Solution

Height of each floor $= 3 m$ Distance from ground floor to fourth floor $= 4 \times 3 = 12 m$

Distance from fourth floor to second floor $= 2 \times 3 = 6 m$

Total distance travelled $= 12 + 6 = 18 m$ Displacement = Final position - Initial position $=$ height of second floor $= 2 \times 3 = 6 m$ upward

9. 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 depends on change in velocity. Velocity changes if speed or direction changes. Even if speed is constant, a change in direction causes acceleration.

Example: Motion in a circular path.

10. 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.

Solution

Given: Initial velocity, $u = 0$ Final velocity, $v = 24 m / s$ Time, $t = 6 s$ Using formula: $v = u + at$ $24 = 0 + a \times 6$ $a = 4 m / s^{2}$ Distance travelled: $s = ut + \frac{1}{2} at^{2}$ $s = 0 + \frac{1}{2} \times 4 \times 36$ $s = 72 m$ Acceleration $= 4 m / s^{2}$ Distance $= 72 m$

11. A motorbike moving with initial velocity $28 m s^{- 1}$ 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: $u = 28 m / s$ $v = 0$ $s = 98 m$ Using formula: $v^{2} = u^{2} + 2 as$ $0 = (28)^{2} + 2 a \times 98$ $0 = 784 + 196 a$ $a = - 4 m / s^{2}$ Now, $v = u + at$ $0 = 28 - 4 t$ $t = 7 s$ Acceleration $= - 4 m / s^{2}$ Time $= 7 s$

12. Figure shows a position-time graph of two objects A and B that are moving along the parallel tracks in the same direction. Do objects A and B ever have equal velocity? Justify your answer.

Solution

Velocity is given by the slope of the position-time graph.

At the point where both lines have the same slope, their velocities are equal.

The graph A and B are straight lines which means both A and B have constant velocity but since the slopes are different, there velocities are different.

Although the two graphs intersect at about $t = 5 s$ (meaning they are at the same position at that instant.

But their slopes are not equal : → Therefore, objects A and B never have equal velocities because their positiontime graphs have different slopes throughout.

13. A graph in figure 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.

Solution

Two objects A and B move from 0 s to 10 s . Understanding the graph Both A and B start from the same position. Both reach the same final position at 10 s . Object A has a straight line (uniform motion).

Object B has a curved line (non-uniform motion), meaning its speed changes. (i) Average velocity

Average velocity = Displacement / Time Since both objects start and end at the same positions, their displacement is equal. Time interval is also same ( 10 s ). So, Average velocity of $A =$ Average velocity of $B$ Statement (i) is correct (ii) Average speed

Average speed = Total distance / Time Object A moves in a straight line, so distance = displacement. Object B follows a curved path in the graph, meaning it covers more distance than A.

The distances covered by A and B are not equal, therefore their average speeds are not equal, provided time taken is same

So, distances are not equal. Statement (ii) is incorrect (iii) Comparison of distances

From the graph, B's curve lies below and then rises steeply, indicating longer path travelled. Object A covers less distance than B. So, average speed of A is less than that of $B$ .

Statement (iii) is correct (iv) Speed comparison in parts

Although B may be slower than A at some intervals, overall, it covers more total distance.

So, average speed of A is not greater. Statement (iv) is incorrect Correct options are (i) and (iii)

14. A truck driver driving at the speed of 54 $km h^{- 1}$ notices a road sign with a speed limit of $40 km h^{- 3}$ Figure for trucks. He slows down to $36 km h^{- 1}$ in 36 s . What was the distance travelled by him during this time? Assume the acceleration to be constant while slowing down.

Solution

Given: Initial speed $= 54 km / h$ Final speed $= 36 km / h$ Time $= 36 s$ Step 1: Convert units $54 km / h = 15 m / s$ $36 km / h = 10 m / s$ Step 2: Use formula Distance $=$ average velocity × time Average velocity = $\frac{( initial velocity + final velocity )}{2}$ $= \frac{( 15 + 10 )}{2} = 12.5 m / s$ Step 3: Calculate distance Distance $= 12.5 \times 36$ $= 450 m$ Since acceleration is constant, velocity changes uniformly.

So, we can use average velocity to find distance.

Distance travelled $= 450 m$

15. A car starts from rest and accelerates uniformly to $20 ms^{- 1}$ in 5 seconds. It then travels at $20 ms^{- 1}$ for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.

Solution

Given: 1st: acceleration 2nd: constant velocity 3rd: retardation 1st: Acceleration (0 to $20 m / s$ in 5 s ) Distance $= ($ initial + final velocity $) /2 \times$ time

Distance $= (\frac{u + v}{2}) \times t$ $= (0 + 20) /2 \times 5$ $= 10 \times 5$ $= 50 m$ 2nd: Constant velocity ( $20 m / s$ for 10 s ) Distance $=$ velocity × time $= 20 \times 10$ $= 200 m$ 3rd: Retardation ( $20 m / s$ to 0 in 6 s ) Distance $= (20 + 0) /2 \times 6$ $= 10 \times 6$ $= 60 m$ Total distance $= 50 + 200 + 60$ $= 310 m$ The motion is divided into three parts because velocity changes differently in each phase.

Total distance travelled $= 310 m$

16. 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 $2.5 ms^{- 2}$ Will the bus be able to stop before reaching the obstacle?

Solution

Given: Speed $= 36 km / h$ Distance to obstacle $= 30 m$ Reaction time $= 0.5 s$ Deceleration $= 2.5 m / s^{2}$ Step 1 : Convert speed $36 km / h = 10 m / s$ Step 2 : Distance during reaction time Driver does not apply brakes immediately. Distance $=$ speed × time $= 10 \times 0.5$ $= 5 m$ Step 3: Distance during braking Use equation: $v^{2} = u^{2} + 2 as$ Final velocity $= 0$ $0 = 1 0^{2} - 2 \times 2.5 \times s$ $0 = 100 - 5 s$ 5s = 100 $s = 20 m$ Step 4: Total stopping distance = reaction distance + braking distance $= 5 + 20$ $= 25 m$ Step 5: Compare with obstacle distance Obstacle distance $= 30 m$ Stopping distance $= 25 m$ Since 25 m is less than 30 m , the bus stops before reaching the obstacle. Yes, the bus will stop safely before reaching the obstacle.

17. 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.

Solution

When we say an object is at rest, we must always specify the reference point. If the reference point is the Earth, then the object appears at rest because it is not changing its position relative to the Earth. However, the Earth itself is moving around the Sun and also rotating about its axis. So, with respect to the Sun, the same object is actually in motion. An object can be at rest or in motion depending on the reference frame. Motion is always relative.

18. The velocity-time graph from 0 s to 120 s for a cyclist is shown in figure 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.

Solution

Understanding the graph: From 0 to 20 s : velocity increases from 0 to $3 m / s$ (acceleration)

From 20 to 100 s: velocity is constant at $3 m / s$

From 100 to 120 s: velocity decreases from $3 m / s$ to $2 m / s$ (deceleration) (i) Constant velocity region

From 20 s to 100 s , the graph is a horizontal line.

This shows constant velocity. (ii) Decreasing velocity region

From 100 s to 120 s, the graph slopes downward.

This shows velocity is decreasing. (iii) Displacement

Displacement $=$ area under velocitytime graph

Area 1 (triangle, $0 - 20 s$ ) $= \frac{1}{2} \times$ base × height $= \frac{1}{2} \times 20 \times 3$ $= 30 m$

Area 2 (rectangle, $20 - 100 s$ ) $=$ length × breadth $= 80 \times 3$ $= 240 m$ Area 3 (trapezium, $100 - 120 s$ ) $= \frac{1}{2} \times ($ sum of parallel sides $) \times$ height $= \frac{1}{2} \times (3 + 2) \times 20$ $= 50 m$ Total displacement $= 30 + 240 + 50$ $= 320 m$ (iv) Average acceleration

Average acceleration = (final velocity

initial velocity) / total time $= (2 - 0) /120$ $= 0.0167 m / s^{2}$

19. 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 figure depicts her velocity versus time. Estimate the running distance based on the graph.

Solution

Area under the graph Section:1 From $t = 0$ to $t = 0.5 h$ (Rectangle) Width $= 0.5 - 0 = 0.5$ Height $= 7.0$ Area $_{1} = 0.5 \times 7.0 = 3.5$ Section 2: from $t = 0.5$ to $t = 1.75 h$ (Trapezium) Width $= 1.75 - 0.5 = 1.25$ Parallel heights = 7.0 and 7.5 Area $2 = \frac{1}{2} \times (7.0 + 7.5) \times 1.25 = \frac{1}{2} \times 14.5 \times 1.25 = 9.0625$

Section 3: From $t = 1.75$ to $t = 3 h$ (Rectangle) Width $= 3.0 - 1.75 = 1.25$ Height $= 7.5 km h^{- 1}$ Аrea $3 = 1.25 \times 7.5 = 9.375$ Section 4: From $t = 3.0$ to $t = 4.5$ (Trapezium) Width $= 4.5 - 3.0 = 1.5$ Area $_{4} = \frac{1}{2} \times (7.5 + 7.0) \times 1.5 = \frac{1}{2} \times 14.5 \times 1.5 = 10.875$

Section 5: From $t = 4.5$ to $t = 5.5 h$ (Trapezium) Width $= 5.5 - 4.5 = 1.0$ Parallel heights $= 7.0$ and 6.5 Section 6: From $t = 5.5$ to $t = 6.5 h$ (Rectangle)

Width $= 6.5 - 5.5 = 1.0$ Area $_{6} = 1.0 \times 6.5 = 6.5$ Step 3: Sum the Areas to Find total distance

Total distance $=$ Area $_{1} +$ Area $_{2} +$ Area $_{3} +$ Area $_{4} +$ Area $_{5} +$ Area $_{6}$

Total distance $= 3.5 + 9.0625 + 9.375 + 10.875 + 6.75 + 6.5$

Total Distance $= 46.0625 km$

20. 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 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.

Solution

Given: Velocity $= 6 m / s$ for 2 minutes Acceleration $= 1 m / s^{2}$ for next 6 s Convert time: 2 minutes $= 120 s$ First part (uniform velocity) Displacement $=$ velocity × time $= 6 \times 120$ $= 720 m$ Second part (accelerated motion) Initial velocity $= 6 m / s$ Time $= 6 s$ Displacement: $s = u t + \frac{1}{2} a t^{2}$ $= 6 \times 6 + \frac{1}{2} \times 1 \times 36$ $= 36 + 18$ $= 54 m$ Total displacement $= 720 + 54$ $= 774 m$

21. Two cars A and B start moving with a 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 the velocity-time graphs for both the cars in the same graph. Using the graph, calculate the displacement mentioned in the two time intervals (Hint: Calculate the acceleration in both cases. Then calculate their velocities at five instants of time to plot the graph).

Solution

Since both cars start from rest ( $u = 0$ ) with constant acceleration, we use formula $a = \frac{v - u}{t}$ Car A: $a_{A} = \frac{5 - 0}{5} = 1 m / s^{2}$ Car B: $ab = \frac{3 - 0}{10} = 0.3 m / s^{2}$ Since both equations represent a direct linear relationship ( $v \propto t$ ). both graphs will be straight lines starting from the origin $(0, 0)$ .

Area under the curve of graph A $S_{A} = \frac{1}{2} \times$ base × height $\frac{1}{2} \times 5 \times 5$ $\frac{25}{2} = 12.5 m$

Area under the curve of graph B $S_{B} = \frac{1}{2} \times$ base × height $S_{B} = \frac{1}{2} \times 10 \times 3$ $S_{B} = 15 m$ 2. Velocity at five instants of time

To complete the data for case A at $t = 7.5$ s and $t = 10 s$ $\Rightarrow Att = 7.5 s \Rightarrow vA = 1 \times 7.5 = 7.5 m / s$ $\Rightarrow Att = 10 s \Rightarrow VA = 1 \times 10 = 10 m / s$

Time ( t ) seconds	Velocity of A ( $v_{A} = 1 t$ ) in m/s	Velocity of B $(v_{B} = 0.3 t)$ in m/s
0	0	0
2.5	2.5	0.75
5.0	5.0	1.5
7.5	7.5	2.25
10.0	10.0	3.0

(A)

(B)

22. 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: what is its: (i) distance travelled, (ii) displacement, (iii) speed, and (iv) velocity.

The length of the minute's hand is 7 cm figure.

Solution

Given: Length of minute hand $= 7 cm$ Time = 6:00 PM to 7:30 PM = 1.5 hours (i) Distance travelled

In 1 hour, minute hand completes 1 revolution. So, in 1.5 hours, it completes 1.5 revolutions. Distance $=$ number of revolutions × circumference Circumference $= 2 π r$ $= 2 \times \frac{22}{7} \times 7$ $= 44 cm$ Distance $= 1.5 \times 44 = 66 cm$ (ii) Displacement

After 1.5 revolutions, the hand is opposite to starting point. So, displacement = diameter of circle $= 2 \times 7$ $= 14 cm$ (iii) Speed

Speed = distance / time Time $= 1.5$ hours $= 5400 s$ Speed $= \frac{66}{5400} = 0.0122 cm / s$ (iv) Velocity

Velocity = displacement / time $= \frac{14}{5400}$ $= 0.0026 cm / s$

4.0 Important Key Concepts of NCERT Class 9 Science Chapter 4 – Describing Motion Around Us

Topic	What Students Learn
Rest and Motion	Understand how the position of an object changes with respect to a reference point and distinguish between rest and motion.
Distance and Displacement	Differentiate between distance and displacement and identify their characteristics.
Speed and Velocity	Learn the concepts of speed, average speed, velocity, and average velocity, along with their units.
Acceleration	Understand the meaning of acceleration and calculate it for different types of motion.
Uniform and Non-uniform Motion	Distinguish between uniform motion and non-uniform motion using practical examples.
Equations of Motion	Apply the equations of motion to solve numerical and conceptual problems involving uniformly accelerated motion.
Motion Graphs	Interpret and analyse distance-time and velocity-time graphs to describe the motion of objects.
Applications of Motion	Apply the concepts of motion to solve real-life and textbook problems involving moving objects.

5.0Quick Revision on Class 9 Science Chapter 4 Describing Motion Around Us

6.0Related Study Materials Class 9 Science

Study ALLEN's Class 9 Science study material as per latest NCERT syllabus. Access NCERT Solutions along with NCERT Textbooks, Revision Notes, Sample Papers and Previous Years’ Question Papers to enhance your concepts, revise important topics and prepare with confidence for school and CBSE exams.

CBSE Class 9 Science Syllabus

Class 9 Science Revision Notes

NCERT Textbook for Class 9 Science

CBSE Sample Papers for Class 9 Science

7.0Advantages of Chapter 4 Science Class 9 NCERT Solutions

Explains Distance and Displacement: Helps students clearly differentiate between the two concepts.
Strengthens Motion Concepts: Simplifies speed, velocity, and acceleration with easy examples.
Improves Graph Interpretation: Teaches how to read and analyse position-time and velocity-time graphs.
Clarifies Equations of Motion: Explains the application of kinematic equations in a systematic way.
Develops Understanding of Circular Motion: Introduces the concept of uniform circular motion with clarity.
Builds a Strong Foundation: Prepares students for advanced mechanics in higher classes.

NCERT Solutions for Class 9 Science Chapter 4 Describing Motion Around Us

1.0Download NCERT Class 9 Science Chapter 4 Describing Motion Around Us

NCERT Solutions Class 9 Science Chapter 4

2.0Learning Outcomes – NCERT Class 9 Science Chapter 4, Describing Motion Around Us

Differentiate between distance and displacement, and explain why displacement can be zero even when distance travelled is not.
Calculate average speed and average velocity, and identify the conditions under which they are equal.
Define acceleration and distinguish between uniform and non-uniform acceleration.
Interpret position-time and velocity-time graphs to determine speed, velocity, and acceleration.
Derive and apply the three kinematic equations of motion to solve numerical problems.
Analyse real-life motion scenarios, such as braking distance and free fall, using graphical and equation-based methods.
Understand uniform circular motion and explain why it involves constant acceleration despite constant speed.
Recognise that motion is relative to a chosen reference point or frame of reference.

3.0Detailed NCERT Class 9 Science Chapter 4 Solutions

1. How much distance should we maintain from the truck ahead to avoid a collision if it suddenly applies the brakes? Does this distance depend upon the speed with which we are moving?

Solution

2. 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?

Solution

3. Fuel used up in a vehicle depends on which of the following? Justify your answer.
(i) Total distance travelled
(ii) Displacement

Solution

A ball rolling down an inclined track

Solution

Distances from 0:

At A $= 40 cm$
At $B = 40 + 10 = 50 cm$
At $C = 40 + 10 + 20 = 70 cm$
At $D = 40 + 10 + 20 + 30 = 100 cm$

Since the ball moves in a straight line from O:

Total distance travelled = Magnitude of displacement at each position. Therefore:
At A: Distance = Displacement = 40 cm
At B: Distance $=$ Displacement $= 50 cm$
At C: Distance = Displacement = 70 cm
At D: Distance = Displacement = 100 cm

So, the values are equal at all positions A, B, C and D.

5. 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.

Solution

Average speed and average velocity

Distance travelled north $= 200 km$
Distance travelled south $= 200 km$
Total distance $= 200 + 200 = 400 km$
Total time $= 3 + 2 = 5 h$

Average Speed $= 80 km / h$
Average Velocity $= 0 km / h$

Solution

Height of each floor $= 3 m$ Distance from ground floor to fourth floor $= 4 \times 3 = 12 m$

Distance from fourth floor to second floor $= 2 \times 3 = 6 m$

Total distance travelled $= 12 + 6 = 18 m$ Displacement = Final position - Initial position $=$ height of second floor $= 2 \times 3 = 6 m$ upward

9. 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 depends on change in velocity. Velocity changes if speed or direction changes. Even if speed is constant, a change in direction causes acceleration.

Example: Motion in a circular path.

10. 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.

Solution

12. Figure shows a position-time graph of two objects A and B that are moving along the parallel tracks in the same direction. Do objects A and B ever have equal velocity? Justify your answer.

Solution

Velocity is given by the slope of the position-time graph.

At the point where both lines have the same slope, their velocities are equal.

The graph A and B are straight lines which means both A and B have constant velocity but since the slopes are different, there velocities are different.

Although the two graphs intersect at about $t = 5 s$ (meaning they are at the same position at that instant.

But their slopes are not equal : → Therefore, objects A and B never have equal velocities because their positiontime graphs have different slopes throughout.

Solution

Object B has a curved line (non-uniform motion), meaning its speed changes. (i) Average velocity

Average speed = Total distance / Time Object A moves in a straight line, so distance = displacement. Object B follows a curved path in the graph, meaning it covers more distance than A.

The distances covered by A and B are not equal, therefore their average speeds are not equal, provided time taken is same

So, distances are not equal. Statement (ii) is incorrect (iii) Comparison of distances

From the graph, B's curve lies below and then rises steeply, indicating longer path travelled. Object A covers less distance than B. So, average speed of A is less than that of $B$ .

Statement (iii) is correct (iv) Speed comparison in parts

Although B may be slower than A at some intervals, overall, it covers more total distance.

So, average speed of A is not greater. Statement (iv) is incorrect Correct options are (i) and (iii)

Solution

So, we can use average velocity to find distance.

Distance travelled $= 450 m$

Solution

Given: 1st: acceleration 2nd: constant velocity 3rd: retardation 1st: Acceleration (0 to $20 m / s$ in 5 s ) Distance $= ($ initial + final velocity $) /2 \times$ time

Total distance travelled $= 310 m$

Solution

17. 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.

Solution

Understanding the graph: From 0 to 20 s : velocity increases from 0 to $3 m / s$ (acceleration)

From 20 to 100 s: velocity is constant at $3 m / s$

From 100 to 120 s: velocity decreases from $3 m / s$ to $2 m / s$ (deceleration) (i) Constant velocity region

From 20 s to 100 s , the graph is a horizontal line.

This shows constant velocity. (ii) Decreasing velocity region

From 100 s to 120 s, the graph slopes downward.

This shows velocity is decreasing. (iii) Displacement

Displacement $=$ area under velocitytime graph

Area 1 (triangle, $0 - 20 s$ ) $= \frac{1}{2} \times$ base × height $= \frac{1}{2} \times 20 \times 3$ $= 30 m$

Average acceleration = (final velocity

initial velocity) / total time $= (2 - 0) /120$ $= 0.0167 m / s^{2}$

Solution

Section 5: From $t = 4.5$ to $t = 5.5 h$ (Trapezium) Width $= 5.5 - 4.5 = 1.0$ Parallel heights $= 7.0$ and 6.5 Section 6: From $t = 5.5$ to $t = 6.5 h$ (Rectangle)

Width $= 6.5 - 5.5 = 1.0$ Area $_{6} = 1.0 \times 6.5 = 6.5$ Step 3: Sum the Areas to Find total distance

Total distance $=$ Area $_{1} +$ Area $_{2} +$ Area $_{3} +$ Area $_{4} +$ Area $_{5} +$ Area $_{6}$

Total distance $= 3.5 + 9.0625 + 9.375 + 10.875 + 6.75 + 6.5$

Total Distance $= 46.0625 km$

20. 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 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.

Solution

Area under the curve of graph A $S_{A} = \frac{1}{2} \times$ base × height $\frac{1}{2} \times 5 \times 5$ $\frac{25}{2} = 12.5 m$

Area under the curve of graph B $S_{B} = \frac{1}{2} \times$ base × height $S_{B} = \frac{1}{2} \times 10 \times 3$ $S_{B} = 15 m$ 2. Velocity at five instants of time

To complete the data for case A at $t = 7.5$ s and $t = 10 s$ $\Rightarrow Att = 7.5 s \Rightarrow vA = 1 \times 7.5 = 7.5 m / s$ $\Rightarrow Att = 10 s \Rightarrow VA = 1 \times 10 = 10 m / s$

Time ( t ) seconds	Velocity of A ( $v_{A} = 1 t$ ) in m/s	Velocity of B $(v_{B} = 0.3 t)$ in m/s
0	0	0
2.5	2.5	0.75
5.0	5.0	1.5
7.5	7.5	2.25
10.0	10.0	3.0

(A)

(B)

The length of the minute's hand is 7 cm figure.

Solution

Given: Length of minute hand $= 7 cm$ Time = 6:00 PM to 7:30 PM = 1.5 hours (i) Distance travelled

After 1.5 revolutions, the hand is opposite to starting point. So, displacement = diameter of circle $= 2 \times 7$ $= 14 cm$ (iii) Speed

Speed = distance / time Time $= 1.5$ hours $= 5400 s$ Speed $= \frac{66}{5400} = 0.0122 cm / s$ (iv) Velocity

Velocity = displacement / time $= \frac{14}{5400}$ $= 0.0026 cm / s$

4.0 Important Key Concepts of NCERT Class 9 Science Chapter 4 – Describing Motion Around Us

Topic	What Students Learn
Rest and Motion	Understand how the position of an object changes with respect to a reference point and distinguish between rest and motion.
Distance and Displacement	Differentiate between distance and displacement and identify their characteristics.
Speed and Velocity	Learn the concepts of speed, average speed, velocity, and average velocity, along with their units.
Acceleration	Understand the meaning of acceleration and calculate it for different types of motion.
Uniform and Non-uniform Motion	Distinguish between uniform motion and non-uniform motion using practical examples.
Equations of Motion	Apply the equations of motion to solve numerical and conceptual problems involving uniformly accelerated motion.
Motion Graphs	Interpret and analyse distance-time and velocity-time graphs to describe the motion of objects.
Applications of Motion	Apply the concepts of motion to solve real-life and textbook problems involving moving objects.

5.0Quick Revision on Class 9 Science Chapter 4 Describing Motion Around Us

6.0Related Study Materials Class 9 Science

CBSE Class 9 Science Syllabus

Class 9 Science Revision Notes

NCERT Textbook for Class 9 Science

CBSE Sample Papers for Class 9 Science

7.0Advantages of Chapter 4 Science Class 9 NCERT Solutions

Explains Distance and Displacement: Helps students clearly differentiate between the two concepts.
Strengthens Motion Concepts: Simplifies speed, velocity, and acceleration with easy examples.
Improves Graph Interpretation: Teaches how to read and analyse position-time and velocity-time graphs.
Clarifies Equations of Motion: Explains the application of kinematic equations in a systematic way.
Develops Understanding of Circular Motion: Introduces the concept of uniform circular motion with clarity.
Builds a Strong Foundation: Prepares students for advanced mechanics in higher classes.

Frequently Asked Questions

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Can I use the NCERT Solutions for Class 9 Science Chapter 4 for exam preparation?

What is the difference between distance and displacement?

How is velocity different from speed?

What is uniform motion?

Why are motion graphs important?

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NCERT Solutions for Class 9 Science Chapter 4 Describing Motion Around Us

1.0Download NCERT Class 9 Science Chapter 4 Describing Motion Around Us

2.0Learning Outcomes – NCERT Class 9 Science Chapter 4, Describing Motion Around Us

3.0Detailed NCERT Class 9 Science Chapter 4 Solutions

4.0 Important Key Concepts of NCERT Class 9 Science Chapter 4 – Describing Motion Around Us

5.0Quick Revision on Class 9 Science Chapter 4 Describing Motion Around Us

6.0Related Study Materials Class 9 Science

7.0Advantages of Chapter 4 Science Class 9 NCERT Solutions

Frequently Asked Questions

Do the NCERT Solutions for Class 9 Science Chapter 4 cover all textbook questions?

Can I use the NCERT Solutions for Class 9 Science Chapter 4 for exam preparation?

What is the difference between distance and displacement?

How is velocity different from speed?

What is uniform motion?

Why are motion graphs important?

Join ALLEN!

NCERT Solutions for Class 9 Science Chapter 4 Describing Motion Around Us

1.0Download NCERT Class 9 Science Chapter 4 Describing Motion Around Us

2.0Learning Outcomes – NCERT Class 9 Science Chapter 4, Describing Motion Around Us

3.0Detailed NCERT Class 9 Science Chapter 4 Solutions

4.0 Important Key Concepts of NCERT Class 9 Science Chapter 4 – Describing Motion Around Us

5.0Quick Revision on Class 9 Science Chapter 4 Describing Motion Around Us

6.0Related Study Materials Class 9 Science

7.0Advantages of Chapter 4 Science Class 9 NCERT Solutions