class 9 science motion case study questions

Case Study Questions Class 9 Science Motion

Case study questions class 9 science chapter 8 motion.

CBSE Class 9 Case Study Questions Science Motion. Important Case Study Questions for Class 9 Exam. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Motion.

At Case Study Questions there will given a Paragraph. In where some Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks or 4 marks.

CBSE Case Study Questions Class 9 Science – Motion

(1) Distance and displacement are two quantities that seem to mean the same but are different with different meanings and definitions. Distance is the measure of “how much distance an object has covered during its motion” while displacement refers to the measure of“how far the abject actually from initial place.” using this data answer following questions.

(i)Which of the following relation is always true when object moves in straight line

(a)distance is always equal to displacement

(b)distance is always greater than or equal to displacement

(c)distance is always lesser than or equal to displacement

(d)none of the above

(ii) Kapiltravels 20 km to North but then come back to South for 40 km to pick up a friend. Whatis kapil’s total distance?

(d) none of the above

(iii) Rahul travels 20 km to East but then come back to West for 10 km. Find displacement.

(iv) Define distance and displacement of particle.

(v) Write difference between distance and displacement

Answer key -1

(iv) Distance and displacement are two quantities that seem to mean the same but are different with different meanings and definitions. Distance is the measure of “how much distance an object has covered during its motion” while displacement refers to the measure of “how far the abject actually from initial place.”

(v) difference between distance and displacement is given by

(2 ) Answer the following questions by observing following diagram

(i ) What is distance and displacement when particle moves from point A to B?

(a)distance is equal to displacement

(b)distance is greater than and equal to displacement

(c)distance is lesser than and equal to displacement

(ii)What is Displacement when particle moves from point A to D?

(iii) What is Displacement when particle moves from point A to C through A-B-C?

(c) 10√2 km

(iv) Find distance covered when particle moves in path ABCDA i.e. starts from A and ends at A?

(v) Find Displacement covered when particle moves in path ABCDA i.e. starts from A and ends at A?

ANSWER KEY-2

(3) The speed of an object is the distance covered per unit time,and velocity is the displacement per unit time.To specify the speed of anobject, we require only its magnitude while Velocity is the speed of an object moving in adefinite direction.

(i) S.I unit of speed is

(d) none of these

(ii) Which of the following is true

(a)speed is scalar

(b)velocity is vector

(c)both a and b

(d)none of these

(iii) To specify speed we require

(a)magnitude

(b)direction

(c)both magnitude and direction

(iv) Define speed and velocity of  particle.

(v)Differentiate between speed and velocity.

Answer key-3

(iv)The speed of an object is the distance covered by object per unit time.Velocity is defined as the displacement by particle per unit time.

(v) Difference between speed and velocity.

(4) The speed of an object need not be constant. In most cases, objects will be in non-uniform

motion. Thereforewe describe the rate of motion of such objects in terms of their average speed. The average speed of an object is obtained by dividing the total distance travelled by the total time taken. That is, average speed =total distance travelled / total time taken. Answer the following.

(i)An object travels 20 m in 5 s and then another 20 m in 5s. What is the average speed in m/s of the object?

(d) None of these

(ii) An object travels 20 m in 5 s and then another 40 m in 5s. What is the average speed in m/s of the object?

(d)None of these

(iii)A man starts walking from a point P on a circular field of radius 7 km and after 1 hour later he comes to same point P after one complete round. find his speed. (take pi=22/7)

(a) 30km/hr

(b) 40km/hr

(c) 44km/hr

(d) 33km/hr

(iv) A man travelled on square field of side 10m .he completed one round of field by taking time 2s, 3s 1s and 2s respectively for each side. Find his average speed.

(v) Define average speed.write its formula and SI unit

Answer key-4

(v) The average speed of an object is obtained by dividing the total distance travelled by the total time taken. That is, average speed =total distance travelled / total time taken.Its SI unit is given by  m/s.

(5) We know that the circumference of a circle of radius r is given by . If the body takes

t seconds to go once around the circular path of radius r, the speed v is given by

When an object moves in a circular path with uniform speed, its motion is called

uniform circular motion. Refer the paragraph and answer the following questions

(i) What happens when a body is moving with constant speed?

(a)acceleration is non uniform

(b)velocity is uniform

(c)velocity is changing

(ii) Cyclist on circular track with constant speed is example of

(a)uniform circular motion

(b)nonuniform circular motion

(iii) Which of the following changes whenbody performsuniform circular motion?

(b)direction of velocity

(iv) Define uniform circular motion. Give 2 examples of uniform circular motion

(v) Give the name of device used for measurement of speed of rotation

Answer key -5

(iv) When an object moves in a circular path with uniform speed, its motion is calleduniform circular motion. Rotation of hour hand of clock and motion of fan at constant speed are examples of uniform circular motion.

(v) Odometers and speedometers are two of the many instruments used to measure speed of rotation.

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• Class 9 Science Case...

Class 9 Science Case Study Questions

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If you are wondering how to solve class 9 science case study questions, then myCBSEguide is the best platform to choose. With the help of our well-trained and experienced faculty, we provide solved examples and detailed explanations for the recently added Class 9 Science case study questions.

You can find a wide range of solved case studies on myCBSEguide, covering various topics and concepts. Class 9 Science case studies are designed to help you understand the application of various concepts in real-life situations.

The rationale behind Science

Science is crucial for Class 9 students’ cognitive, emotional, and psychomotor development. It encourages curiosity, inventiveness, objectivity, and aesthetic sense.

In the upper primary stage, students should be given a variety of opportunities to engage with scientific processes such as observing, recording observations, drawing, tabulating, plotting graphs, and so on, whereas in the secondary stage, abstraction and quantitative reasoning should take a more prominent role in science teaching and learning. As a result, the concept of atoms and molecules as matter’s building units, as well as Newton’s law of gravitation, emerges.

Science is important because it allows Class 9 Science students to understand the world around us. It helps to find out how things work and to find solutions to problems at the Class 9 Science level. Science is also a source of enjoyment for many people. It can be a hobby, a career, or a source of intellectual stimulation.

Case study questions in Class 9 Science

The inclusion of case study questions in Class 9 science CBSE is a great way to engage students in critical thinking and problem-solving. By working through real-world scenarios, Class 9 Science students will be better prepared to tackle challenges they may face in their future studies and careers. Class 9 Science Case study questions also promote higher-order thinking skills, such as analysis and synthesis. In addition, case study questions can help to foster creativity and innovation in students. As per the recent pattern of the Class 9 Science examination, a few questions based on case studies/passages will be included in the CBSE Class 9 Science Paper. There will be a paragraph presented, followed by questions based on it.

Examples of Class 9 science class case study questions

Class 9 science case study questions have been prepared by myCBSEguide’s qualified teachers. Class 9 case study questions are meant to evaluate students’ knowledge and comprehension of the material. They are not intended to be difficult, but they will require you to think critically about the material. We hope you find Class 9 science case study questions beneficial and that they assist you in your exam preparation.

The following are a few examples of Class 9 science case study questions.

Class 9 science case study question 1

• due to its high compressibility
• large volumes of a gas can be compressed into a small cylinder
• transported easily
• all of these
• shape, volume
• volume, shape
• shape, size
• size, shape
• the presence of dissolved carbon dioxide in water
• the presence of dissolved oxygen in the water
• the presence of dissolved Nitrogen in the water
• liquid particles move freely
• liquid have greater space between each other
• both (a) and (b)
• none of these
• Only gases behave like fluids
• Gases and solids behave like fluids
• Gases and liquids behave like fluids
• Only liquids are fluids

Answer Key:

• (d) all of these
• (a) shape, volume
• (b) the presence of dissolved oxygen in the water
• (c) both (a) and (b)
• (c) Gases and liquids behave like fluids

Class 9 science case study question 2

• 12/32 times
• 18 g of O 2
• 18 g of CO 2
• 18 g of CH 4
• 1 g of CO 2
• 1 g of CH 4 CH 4
• 2 moles of H2O
• 20 moles of water
• 6.022  ×  1023 molecules of water
• 1.2044  ×  1025 molecules of water
• (I) and (IV)
• (II) and (III)
• (II) and (IV)
• Sulphate molecule
• Ozone molecule
• Phosphorus molecule
• Methane molecule
• (c) 8/3 times
• (d) 18g of CH ​​​​​4
• (c) 1g of H ​​​​​​2
• (d) (II) and (IV)
• (c) phosphorus molecule

Class 9 science case study question 3

• collenchyma
• chlorenchyma
• It performs photosynthesis
• It helps the aquatic plant to float
• It provides mechanical support
• Sclerenchyma
• Collenchyma
• Epithelial tissue
• Parenchyma tissues have intercellular spaces.
• Collenchymatous tissues are irregularly thickened at corners.
• Apical and intercalary meristems are permanent tissues.
• Meristematic tissues, in its early stage, lack vacuoles, muscles
• (I) and (II)
• (III) and (I)
• Transpiration
• Provides mechanical support
• Provides strength to the plant parts
• None of these
• (a) Collenchyma
• (b) help aquatic plant to float
• (b) Sclerenchyma
• (d) Only (III)
• (c) provide strength to plant parts

Cracking Class 9 Science Case Study Questions

There is no one definitive answer to Class 9 Science case study questions. Every case study is unique and will necessitate a unique strategy. There are, nevertheless, certain general guidelines to follow while answering case study questions.

• To begin, double-check that you understand the Class 9 science case study questions. Make sure you understand what is being asked by reading it carefully. If you’re unclear, seek clarification from your teacher or tutor.
• It’s critical to read the Class 9 Science case study material thoroughly once you’ve grasped the question. This will provide you with a thorough understanding of the problem as well as the various potential solutions.
• Brainstorming potential solutions with classmates or other students might also be beneficial. This might provide you with multiple viewpoints on the situation and assist you in determining the best solution.
• Finally, make sure your answer is presented simply and concisely. Make sure you clarify your rationale and back up your claim with evidence.

A look at the Class 9 Science Syllabus

The CBSE class 9 science syllabus provides a strong foundation for students who want to pursue a career in science. The topics are chosen in such a way that they build on the concepts learned in the previous classes and provide a strong foundation for further studies in science. The table below lists the topics covered in the Class 9 Science syllabus of the Central Board of Secondary Education (CBSE). As can be seen, the Class 9 science syllabus is divided into three sections: Physics, Chemistry and Biology. Each section contains a number of topics that Class 9 science students must study during the course.

CBSE Class 9 Science (Code No. 086)

Theme: Materials Unit I: Matter-Nature and Behaviour Definition of matter; solid, liquid and gas; characteristics – shape, volume, density; change of state-melting (absorption of heat), freezing, evaporation (cooling by evaporation), condensation, sublimation. Nature of matter:  Elements, compounds and mixtures. Heterogeneous and homogenous mixtures, colloids and suspensions. Particle nature and their basic units:  Atoms and molecules, Law of constant proportions, Atomic and molecular masses. Mole concept: Relationship of mole to mass of the particles and numbers. Structure of atoms:  Electrons, protons and neutrons, valency, the chemical formula of common compounds. Isotopes and Isobars.

Theme: The World of the Living Unit II: Organization in the Living World Cell – Basic Unit of life:  Cell as a basic unit of life; prokaryotic and eukaryotic cells, multicellular organisms; cell membrane and cell wall, cell organelles and cell inclusions; chloroplast, mitochondria, vacuoles, endoplasmic reticulum, Golgi apparatus; nucleus, chromosomes – basic structure, number. Tissues, Organs, Organ System, Organism: Structure and functions of animal and plant tissues (only four types of tissues in animals; Meristematic and Permanent tissues in plants).

Theme: Moving Things, People and Ideas Unit III: Motion, Force and Work Motion:  Distance and displacement, velocity; uniform and non-uniform motion along a straight line; acceleration, distance-time and velocity-time graphs for uniform motion and uniformly accelerated motion, derivation of equations of motion by graphical method; elementary idea of uniform circular motion. Force and Newton’s laws:  Force and Motion, Newton’s Laws of Motion, Action and Reaction forces, Inertia of a body, Inertia and mass, Momentum, Force and Acceleration. Elementary idea of conservation of Momentum. Gravitation:  Gravitation; Universal Law of Gravitation, Force of Gravitation of the earth (gravity), Acceleration due to Gravity; Mass and Weight; Free fall. Floatation:  Thrust and Pressure. Archimedes’ Principle; Buoyancy. Work, energy and power:  Work done by a Force, Energy, power; Kinetic and Potential energy; Law of conservation of energy. Sound:  Nature of sound and its propagation in various media, speed of sound, range of hearing in humans; ultrasound; reflection of sound; echo.

Theme: Food Unit IV: Food Production Plant and animal breeding and selection for quality improvement and management; Use of fertilizers and manures; Protection from pests and diseases; Organic farming.

PRESCRIBED BOOKS:

• Science-Textbook for class IX-NCERT Publication
• Assessment of Practical Skills in Science-Class IX – CBSE Publication
• Laboratory Manual-Science-Class IX, NCERT Publication
• Exemplar Problems Class IX – NCERT Publication

myCBSEguide: A true helper

There are numerous advantages to using myCBSEguide to achieve the highest results in Class 9 Science.

• myCBSEguide offers high-quality study materials that cover all of the topics in the Class 9 Science curriculum.
• myCBSEguide provides practice questions and mock examinations to assist students in the best possible preparation for their exams.
• On our myCBSEguide app, you’ll find a variety of solved Class 9 Science case study questions covering a variety of topics and concepts. These case studies are intended to help you understand how certain principles are applied in real-world settings
• myCBSEguide is that the study material and practice problems are developed by a team of specialists who are always accessible to assist students with any questions they may have. As a result, students may be confident that they will receive the finest possible assistance and support when studying for their exams.

So, if you’re seeking the most effective strategy to study for your Class 9 Science examinations, myCBSEguide is the place to go!

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Chapter 8 Class 9 - Motion

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In this chapter, we will learn

What is Motion and What are the Different Types of Motion

What is meaning of Distance and Displacement

What is Uniform Motion and Non Uniform Motion - with graphs

What is Speed

How do we calculate Average Speed

What is Velocity

How do we calculate Average Velocity

What is Acceleration  

First Equation of Motion - Explanation and Derivation

Second Equation of Motion - Explanation and Derivation

Third Equation of Motion - Explanation and Derivation

What is Uniform Circular Motion

Distance Time Graph - and finding velocity from it

Speed Time Graph or Velocity Time Graph - and how to find acceleration and distance from it

Graphical Derivation of Equations of Motion

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• NCERT Solutions
• NCERT Class 9
• NCERT 9 Science
• Chapter 8: Motion

NCERT Solutions for Class 9 Science Chapter 8: Motion

Ncert solutions class 9 science chapter 8 – free pdf download.

* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 7.

NCERT Solutions for Class 9 Science Chapter 8 Motion is designed with the intention of clarifying doubts and concepts easily. Class 9 NCERT Solutions in Science is a beneficial reference and guide that helps students clear doubts instantly in an effective way. 

Download Exclusively Curated Chapter Notes for Class 9 Science Chapter – 8 Motion

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NCERT Solutions for Class 9 Science approaches students in a student-friendly way and is loaded with questions, activities, and exercises that are CBSE exam and competitive exam-oriented. NCERT Solutions for Class 9 Science is the contribution of our faculty, having vast teaching experience. It is developed keeping in mind the concept-based approach along with the precise answering method for CBSE examinations. Refer to NCERT Solutions for Class 9 for best scores in CBSE and competitive exams. It is a detailed and well-structured solution for a solid grip on the concept-based learning experience. NCERT for Class 9 Science Solutions is made available in both web and PDF format for ease of access.

• Chapter 1 Matter in Our Surroundings
• Chapter 2 Is Matter Around Us Pure?
• Chapter 3 Atoms and Molecules
• Chapter 4 Structure of the Atom
• Chapter 5 The Fundamental Unit of Life
• Chapter 6 Tissues
• Chapter 7 Diversity in Living Organisms
• Chapter 9 Force And Laws Of Motion
• Chapter 10 Gravitation
• Chapter 11 Work and Energy
• Chapter 12 Sound
• Chapter 13 Why Do We Fall Ill?
• Chapter 14 Natural Resources
• Chapter 15 Improvement in Food Resources

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Access Answers of Science NCERT class 9 Chapter 8: Motion  (All intext and exercise questions solved)

Intext Questions – 1   Page: 100

1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Yes, an object which has moved through a distance can have zero displacement if it comes back to its initial position.

Example: If a person jogs in a circular park which is circular and completes one round. His initial and final position is the same.

Hence, his displacement is zero.

2. A farmer moves along the boundary of a square field of side 10m 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?

Side of the given square field = 10m

Hence, the perimeter of a square = 40 m

Time taken by the farmer  to cover the boundary of 40 m = 40 s

So, in 1 s, the farmer covers a distance of 1 m

Distance covered by the farmer in 2 min 20 sec = 1 x 140 = 140 m

The total number of rotations taken by the farmer to cover a distance of 140 m = total distance/perimeter

At this point, let us say the farmer is at point B from the origin O

Therefore,  from Pythagoras theorem, the displacement s = √(10 2 +10 2 )

s =  10 √ 2

s = 14.14 m

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.

Neither of the statements is true.

(a) Given statement is false because the displacement of an object which travels a certain distance and comes back to its initial position is zero.

(b) Given statement is false because the displacement of an object can be equal to, but never greater than the distance travelled.

Intext Questions – 2   Page: 102

1. Distinguish between speed and velocity.

2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Since average speed is the total distance travelled in a time frame and velocity is the total displacement in the time frame, the magnitude of average velocity and average speed will be the same when the total distance travelled is equal to the displacement.

3. What does the odometer of an automobile measure?

An odometer, or odograph, is a device that measures the distance travelled by an automobile based on the perimeter of the wheel as the wheel rotates.

4. What does the path of an object look like when it is in uniform motion?

The path of an object in uniform motion is a straight line.

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 × 10 8 m/s.

Given that the signal travels in a straight line, the distance between the spaceship and the ground station is equal to the total distance travelled by the signal.

5 minutes = 5*60 seconds = 300 seconds.

Speed of the signal = 3 × 10 8 m/s.

Therefore, total distance = (3 × 10 8 m/s) * 300s

= 9*10 10 meters.

Intext Questions – 3   Page: 103

1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?

Uniform Acceleration:  When an object is travelling in a straight line with an increase in velocity at equal intervals of time, then the object is said to be in uniform acceleration.

The free-falling of an object is an example of uniform acceleration.

Non-Uniform Acceleration:  When an object is travelling with an increase in velocity but not at equal intervals of time is known as non-uniform acceleration.

Bus moving or leaving from the bus stop is an example of non-uniform acceleration.

2. A bus decreases its speed from 80 km h –1 to 60 km h –1 in 5 s. Find the acceleration of the bus.

Given, the initial velocity (u) = 80km/hour = 80000m/3600s= 22.22 m.s -1

The final velocity (v) = 60km/hour = 60000m/3600s= 16.66 m.s -1

Time frame, t = 5 seconds.

Therefore, acceleration (a) =(v-u)/t = (16.66 m.s -1 – 22.22 m.s -1 )/5s

= -1.112 m.s -2

Therefore, the total acceleration of the bus is -1.112m.s -2 . It can be noted that the negative sign indicates that the velocity of the bus is decreasing.

3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h –1 in 10 minutes. Find its acceleration.

Given parameters

Initial velocity (u) = 0

Final velocity (v) = 40 km/h

v = 40 × (5/18)

v = 11.1111 m/s

Time (t) = 10 minute

t = 60 x 10

Acceleration (a) =?

Consider the formula

11.11 = 0 + a × 600

11,11 = 600 a

a = 11.11/600

a = 0.0185 ms -2

Intext Questions – 4   Page: 107

1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

For uniform motion, the distance-time graph is a straight line. On the other hand, the distance-time graph of an object in non-uniform motion is a curve.

The first graph describes the uniform motion and the second one describes the non-uniform motion.

2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

The distance-time graph can be plotted as follows.

When the slope of the distance-time graph is a straight line parallel to the time axis, the object is at the same position as time passes. That means the object is 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?

The speed-time graph can be plotted as follows.

Since there is no change in the velocity of the object (Y-Axis value) at any point of time (X-axis value), the object is said to be in uniform motion.

4. What is the quantity which is measured by the area occupied below the velocity-time graph?

Considering an object in uniform motion, its velocity-time graph can be represented as follows.

Now, the area below the velocity-time graph is the area of the rectangle OABC, which is given by OA*OC. But OA is the velocity of the object and OC represents time. Therefore, the shaded area can be represented as:

Area under the velocity-time graph = velocity*time.

Substituting the value of velocity as displacement/time in the previous equation, it is found that the area under the velocity-time graph represents the total displacement of the object.

Intext Questions – 5 Page: 109,110

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.

(a) Given, the bus starts from rest. Therefore, initial velocity (u) = 0 m/s

Acceleration (a) = 0.1 m.s -2

Time = 2 minutes = 120 s

Acceleration is given by the equation a=(v-u)/t

Therefore, terminal velocity (v) = (at)+u

= (0.1 m.s -2 * 120 s) + 0 m.s -1

= 12 m.s -1 + 0 m.s -1

Therefore, terminal velocity (v) = 12 m/s

(b) As per the third motion equation, 2as = v 2 – u 2

Since a = 0.1 m.s -2 , v = 12 m.s -1 , u = 0 m.s -1 , and t = 120 s, the following value for s (distance) can be obtained.

Distance, s =(v 2 – u 2 )/2a

=(12 2 – 0 2 )/2(0.1)

Therefore, s = 720 m.

The speed acquired is 12 m.s -1 and the total distance travelled is 720 m.

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.

Given, initial velocity (u) = 90 km/hour = 25 m.s -1

Terminal velocity (v) = 0 m.s -1

Acceleration (a) = -0.5 m.s -2

As per the third motion equation, v 2 -u 2 =2as

Therefore, distance traveled by the train (s) =(v 2 -u 2 )/2a

s = (0 2 -25 2 )/2(-0.5) meters = 625 meters

The train must travel 625 meters at an acceleration of -0.5 ms -2 before it reaches the rest position.

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?

Given, initial velocity (u) = 0 (the trolley begins from the rest position)

Acceleration (a) = 0.02 ms -2

Time (t) = 3s

As per the first motion equation, v=u+at

Therefore, terminal velocity of the trolley (v) = 0 + (0.02 ms -2 )(3s)= 0.06 ms -1

Therefore, the velocity of the trolley after 3 seconds will be 6 cm.s -1

4. A racing car has a uniform acceleration of 4 m s -2 . What distance will it cover in 10 s after start?

Given, the car is initially at rest; initial velocity (u) = 0 ms -1

Acceleration (a) = 4 ms -2

Time period (t) = 10 s

As per the second motion equation, s = ut+1/2 at 2

Therefore, the total distance covered by the car (s) = 0 * 10m + 1/2 (4ms -2 )(10s) 2

= 200 meters

Therefore, the car will cover a distance of 200 meters after 10 seconds.

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?

Given, initial velocity (u) = 5 m/s

Terminal velocity (v) = 0 m/s (since the stone will reach a position of rest at the point of maximum height)

Acceleration = 10 ms -2 in the direction opposite to the trajectory of the stone = -10 ms -2

As per the third motion equation, v 2 – u 2 = 2as

Therefore, the distance travelled by the stone (s) = (0 2 – 5 2 )/ 2(10)

Distance (s) = 1.25 meters

As per the first motion equation, v = u + at

Therefore, time taken by the stone to reach a position of rest (maximum height) = (v – u) /a

=(0-5)/-10 s

Time taken = 0.5 seconds

Therefore, the stone reaches a maximum height of 1.25 meters in a timeframe of 0.5 seconds.

Exercises Page: 112,113

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?

Given, diameter of the track (d) = 200m

Therefore, the circumference of the track (π*d) = 200π meters.

Distance covered in 40 seconds = 200π meters

Distance covered in 1 second = 200π/40

Distance covered in 2minutes and 20 seconds (140 seconds) = 140 * 200π/40 meters

= (140*200*22)/(40* 7) meters = 2200 meters

Number of rounds completed by the athlete in 140 seconds = 140/40 = 3.5

Therefore, the final position of the athlete (with respect to the initial position) is at the opposite end of the circular track. Therefore, the net displacement will be equal to the diameter of the track, which is 200m.

Therefore, the net distance covered by the athlete is 2200 meters and the total displacement of the athlete is 200m.

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?

Given, distance covered from point A to point B = 300 meters

Distance covered from point A to point C = 300m + 100m = 400 meters

Time taken to travel from point A to point B = 2 minutes and 30 seconds = 150 seconds

Time taken to travel from point A to point C = 2 min 30 secs + 1 min = 210 seconds

Displacement from A to B = 300 meters

Displacement from A to C = 300m – 100m = 200 meters

Average speed = total distance travelled/ total time taken

Average velocity = total displacement/ total time taken

Therefore, the average speed while traveling from A to B = 300/150 ms -1 = 2 m/s

Average speed while traveling from A to C = 400/210 ms -1 = 1.9 m/s

Average velocity while traveling from A to B =300/150 ms -1 = 2 m/s

Average velocity while traveling from A to C =200/210 ms -1 = 0.95 m/s

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?

Distance travelled to reach the school = distance travelled to reach home = d (say)

Time taken to reach school = t 1

Time taken to reach home = t 2

therefore, average speed while going to school = total distance travelled/ total time taken = d/t 1 = 20 kmph

Average speed while going home = total distance travelled/ total time taken = d/t 2 = 30 kmph

Therefore, t 1 = d/20 and t 2 = d/30

Now, the average speed for the entire trip is given by total distance travelled/ total time taken

= (d+d)/(t 1 +t 2 )kmph = 2d/(d/20+d/30)kmph

= 2/[(3 + 2)/60]

= 120/5 kmh -1 = 24 kmh -1

Therefore, Abdul’s average speed for the entire trip is 24 kilometers per hour.

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?

Given, initial velocity of the boat = 0 m/s

Acceleration of the boat = 3 ms -2

Time period = 8s

As per the second motion equation, s = ut + 1/2 at 2

Therefore, the total distance travelled by boat in 8 seconds = 0 + 1/2 (3)(8) 2

= 96 meters

Therefore, the motorboat travels a distance of 96 meters in a time frame of 8 seconds.

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?

The speed v/s time graphs for the two cars can be plotted as follows.

The total displacement of each car can be obtained by calculating the area beneath the speed-time graph.

Therefore, displacement of the first car = area of triangle AOB

= (1/2)*(OB)*(OA)

But OB = 5 seconds and OA = 52 km.h -1 = 14.44 m/s

Therefore, the area of the triangle AOB is given by: (1/2)*(5s)*(14.44ms -1 ) = 36 meters

Now, the displacement of the second car is given by the area of the triangle COD

= (1/2)*(OD)*(OC)

But OC = 10 seconds and OC = 3km.h -1 = 0.83 m/s

Therefore, area of triangle COD = (1/2)*(10s)*(0.83ms -1 ) = 4.15 meters

Therefore, the first car is displaced by 36 meters whereas the second car is displaced by 4.15 meters. Therefore, the first car (which was traveling at 52 kmph) travelled farther post the application of brakes.

6. Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer 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?

(a) Since the slope of line B is the greatest, B is travelling at the fastest speed.

(b) Since the three lines do not intersect at a single point, the three objects never meet at the same point on the road.

(c) Since there are 7 unit areas of the graph between 0 and 4 on the Y axis, 1 graph unit equals 4/7 km.

Since the initial point of an object C is 4 graph units away from the origin, Its initial distance from the origin is 4*(4/7)km = 16/7 km

When B passes A, the distance between the origin and C is 8km

Therefore, total distance travelled by C in this time = 8 – (16/7) km = 5.71 km

(d) The distance that object B has covered at the point where it passes C is equal to 9 graph units.

Therefore, total distance travelled by B when it crosses C = 9*(4/7) = 5.14 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?

Given, initial velocity of the ball (u) = 0 (since it began at the rest position)

Distance travelled by the ball (s) = 20m

Acceleration (a) = 10 ms -2

As per the third motion equation,

v 2 – u 2 = 2as

= 2*(10ms -2 )*(20m) + 0

v 2 = 400m 2 s -2

Therefore, v= 20ms -1

The ball hits the ground with a velocity of 20 meters per second.

As per the first motion equation,

Therefore, t = (v-u)/a

= (20-0)ms -1 / 10ms -2

= 2 seconds

Therefore, the ball reaches the ground after 2 seconds.

8. The speed-time graph for a car is shown in Fig. 8.12

(a) Find how far does the car travel 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?

The shaded area represents the displacement of the car over a time period of 4 seconds. It can be calculated as:

(1/2)*4*6 = 12 meters. Therefore the car travels a total of 12 meters in the first four seconds.

(b) Since the speed of the car does not change from the points (x=6) and (x=10), the car is said to be in uniform motion from the 6 th to the 10 th second.

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 with an acceleration but with uniform speed. (c) an object moving in a certain direction with an acceleration in the perpendicular direction.

(a) It is possible; an object thrown up into the air has a constant acceleration due to gravity acting on it. However, when it reaches its maximum height, its velocity is zero.

(b) it is possible; acceleration implies an increase or decrease in speed, and uniform speed implies that the speed does not change over time

Circular motion is an example of an object moving with acceleration but with uniform speed.

An object moving in a circular path with uniform speed is still under acceleration because the velocity changes due to continuous changes in the direction of motion.

(c) It is possible; for an object accelerating in a circular trajectory, the acceleration is perpendicular to the direction followed by the object.

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.

Given, the radius of the orbit = 42250 km

Therefore, circumference of the orbit = 2*π*42250km = 265571.42 km

Time is taken for the orbit = 24 hours

Therefore, speed of the satellite = 11065.4 km.h -1

The satellite orbits the Earth at a speed of 11065.4 kilometers per hour.

NCERT Class 9 Science Chapter 8 explains the concept of motion, and types of motion with relevant examples for a clear understanding of the concept. It explains the causes of phenomena like sunrise, sunset, and changing of the seasons. It helps students understand uniform and non-uniform motion. Distance-time graphs and velocity-time graphs, which are considered important concepts for examination, are explained in an easy way in NCERT Solutions . It describes how the acceleration of an object is the change in velocity per unit time. NCERT Class 9 Science Chapter 8 is covered under Unit III: Motion, Force and Work and can get you maximum marks.

• NCERT Solutions for Class 9 explains motion in terms of distance moved or displacement.
• Uniform and non-uniform motions of objects are explained through the graph and examples.
• Uniform circular motion concept is made understandable in a simple way.
• Problems on acceleration, velocity, and average velocity are also solved.

Key Features of NCERT Solutions for Class 9 Science Chapter 8: Motion

• A simple and easily understandable approach is followed in NCERT Solutions to make students aware of topics.
• Provides complete solutions to all the questions present in the respective NCERT textbooks.
• NCERT Solutions offers detailed answers to all the questions to help students in their preparations.
• These solutions will be useful for CBSE exams, Science Olympiads, and other competitive exams.

Disclaimer:

Dropped Topics –  8.5 Equations of motion by graphical method, 8.5.1 Equation for Velocity–Time Relation, 8.5.2 Equation for Position–Time relation and 8.5.3 Equation for Position– Velocity.

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Case Study Questions of Chapter 9 Force and Laws of Motion PDF Download

Case study Questions on Class 9 Science Chapter 9 are very important to solve for your exam. Class 9 Science Chapter 9 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 9 Science Chapter 9 Force and Laws of Motion

In CBSE Class 9 Science Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Force and Laws of Motion Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 9 Science  Chapter 9 Force and Laws of Motion

Case Study/Passage-Based Questions

Question 1:

The sum of the momentum of the two objects before the collision is equal to the sum of momentum after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum. This statement can alternatively be given as the total momentum of the two objects is unchanged or conserved by the collision. The Law of conservation of momentum is applicable to the system of particles. Answer the following questions.

(i)Law of conservation of momentum is applicable to

(a) A system of particles

(b) Only for 2 particles

(c) Only for 1 particle

(d) None of the above

Answer: (a) A system of particles

(ii) Law of conservation of momentum holds good provided that

(a) There should be external unbalanced force acting on particles

(b) There should not be any external unbalanced force acting on particles

(c) No internal forces acting on particles

Answer: (b) There should not be any external unbalanced force acting on particles

(iii)The total momentum of the two objects when collision occurs is

(a) Changed

(b) Remains conserved

(c) Become zero

Answer: (b) Remains conserved

(iv) State law of conservation of momentum.

Answer: The sum of momentum of the two objects before collision is equal to the sum of momentum after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum. This statement can alternatively be given as the total momentum of the two objects is unchanged or conserved by the collision.

(v) If action and Reaction are equal in magnitude and opposite in direction then why they do not cancel each other?

Answer: Action and Reaction are equal in magnitude and opposite in direction but they do not cancel each other because they are not action on sane object. As these forces are acting on different object hence produces different acceleration and does not cancel each other.

Question 2:

The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and never on the same object. It is important to note that even though the action and reaction forces are always equal in magnitude; these forces may not produce accelerations of equal magnitudes, this is because each force acts on a different object that may have a different mass. The two opposing forces are also known as action and reaction forces. Answer the following questions.

(i) Action reaction forces are always

(a) Equal and in the same direction

(b) Equal and in the opposite direction

(c) Unequal and in the same direction

Answer: (b) Equal and in the opposite direction

(ii) Which of the following is correct about action reaction forces?

(a) They act on different objects

(b) They are equal in magnitude and opposite in direction

(c) Both forces acted on different object simultaneously

(d) All the above

Answer: (d) All the above

(iii) State third law of motion

Answer: The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and neveron the same object.

(iv) Give 5 examples of third law of motion

Answer: Examples of third law of motion are Swimming or rowing a boat. •Static friction while pushing an object. •Walking. •Standing on the ground or sitting on a chair. •The upward thrust of a rocket. •Resting against a wall or tree.

Hope the information shed above regarding Case Study and Passage Based Questions for Class 9 Science Chapter 9 Force and Laws of Motion with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 9 Science Force and Laws of Motion Case Study and passage-based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible

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NCERT Solutions for Class 9 Science Chapter 8 Motion

NCERT Solutions for Class 9 Science (physics) Chapter 8 Motion are given below. In these solutions, we have answered all the intext and exercise questions provided in NCERT class 9 science textbook. Class 9 NCERT Solutions Science Chapter 8 provided in this article are strictly based on the CBSE syllabus and curriculum. Students can easily download these solutions in PDF format for free from our app.

Class 9 Science Chapter 8 Textbook Questions and Answers

INTEXT QUESTIONS

PAGE NO. 100

Question 1: An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Answer: Yes, zero displacement is possible if an object has moved through a distance. Suppose a body is moving in a circular path and starts moving from point A and it returns back at same point A after completing one revolution, then the distance will be equal to its circumference while displacement will be zero.

Question 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?

Given, side of the square field = 10 m Therefore, perimeter = 10 m × 4 = 40 m Farmer moves along the boundary in 40 second Time = 2 minutes 20 second = 2 × 60 + 20 = 140 s Since, in 40 second farmer moves 40 m

Therefore, in 1 second distance covered by farmer = 40 ÷ 40 = 1m.

Therefore, in 140 second distance covered by farmer = 1 × 140 m = 140 m

Thus, after 2 minute 20 second the displacement of farmer will be equal to 14.1 m north east from initial position.

Question 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: (a) Not true

Displacement can become zero when the initial and final position of the object is the same.

(b) Not true

Displacement is the shortest measurable distance between the initial and final positions of an object. It cannot be greater than the magnitude of the distance travelled by an object. However, sometimes, it may be equal to the distance travelled by the object.

PAGE NO. 102

Question 1: Distinguish between speed and velocity

Answer: Speed has only magnitude while velocity has both magnitude and direction. So, speed is a scalar quantity but velocity is a vector quantity.

Question 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 will be equal to its average speed in the condition of uniform velocity in a straight line motion.

Question 3: What does the odometer of an automobile measure?

Answer: In automobiles, odometer is used to measure the distance.

Question 4: What does the path of an object look like when it is in uniform motion?

Answer: In the case of uniform motion, the path of an object will look like a straight line.

Question 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, 3×10 8 ms -1 .

Answer: Here we have, speed = 3 × 10 8 m/s

Time = 5 minute = 5 × 60 s = 300 s

Using, Distance = Speed × Time ⇒ Distance = 3 × 10 8 × 300 m = 900 × 10 8 m = 9.0 × 10 10 m

PAGE NO 103

Question 1: When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?

Answer: (i) A body is said in uniform acceleration when its motion is along a straight line and its velocity changes by equal magnitude in equal interval of time.

(ii) A body is said in non-uniform acceleration when its motion is along a straight line and its velocity changes by unequal magnitude in equal interval of time.

Question 2: A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.

Question 3: A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 minutes. Find its acceleration.

PAGE NO. 107

Question 1: What is the nature of the distance – time graphs for uniform and non-uniform motion of an object?

Answer: (a) The slope of the distance-time graph for an object in uniform motion is a straight line. (b) The slope of the distance-time graph for an object in non-uniform motion is not a straight line.

Question 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: When the slope of distance-time graph is a straight line parallel to time axis, the object is stationary.

Question 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: When the graph of a speed time graph is a straight line parallel to the time axis, the object is moving with constant speed.

Question 4: What is the quantity which is measured by the area occupied below the velocity- time graph?

Answer: The quantity of distance is measured by the area occupied below the velocity time graph.

PAGE NO 109-110

Question 1: A bus starting from rest moves with a uniform acceleration of 0.1ms –2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

Answer:  Here we have, Initial velocity (u) = 0 m/s Acceleration (a) = 0.1ms –2   Time (t) = 2 minute = 120 seconds 

(a) The speed acquired: We know that, v = u + at ⇒ v = 0 + 0.1 × 120 m/s ⇒ v = 12 m/s Thus, the bus will acquire a speed of 12 m/s after 2 minute with the given acceleration.

(b) The distance travelled:

Thus, bus will travel a distance of 720 m in the given time of 2 minute.  

Question 2: A train is travelling at a speed of 90 km/h. 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.

Question 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: Here we have, Initial velocity, u = 0 m/s Acceleration (a) = 2 cm/s 2 = 0.02 m/s 2 Time (t) = 3 s Final velocity, v = ?

We know that, v = u + at            Therefore, v = 0 + 0.02 × 3 m/s           ⇒ v = 0.06 m/s Therefore, the final velocity will be 0.06 m/s after start.

Question 4:  A racing car has a uniform acceleration of 4 m/s 2 . What distance will it cover in 10 s after start?

Thus, racing car will cover a distance of 200 m after start in 10 s with given acceleration.

Question 5: A stone is thrown in a vertically upward direction with a velocity of 5 m/s. 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?

Thus, stone will attain a height of 1.25 m and time taken to attain the height is 0.5 s.

Question 1: An athlete completes one round of circular track of diameter 200 m in 40 sec.  What will be the distance covered and the displacement at the end of 2 minutes 20 sec? 

Answer: Time taken = 2 min 20 sec = 140 sec. Radius, r = 100 m. In 40 sec the athlete complete one round.    

So, in 140 sec the athlete will complete = 140 ÷ 40 = 3.5 round.

At the end of his motion, the athlete will be in the diametrically opposite position.

Displacement = diameter = 200 m.

Question 2: Joseph jogs from one end A to another end B of a straight 300 m road in 2 minutes and 30 sec 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 (b) from A to C?

Answer: For motion from A to B: Distance covered = 300 m = Displacement = 300 m.

Time taken = 150 sec.

(a) We know that,   Average speed  = Total distance covered ÷ Total time taken  = 300 m ÷ 150 sec = 2 ms -1  

Average velocity = Net displacement ÷ time taken = 300 m ÷ 150 sec = 2 ms -1

(b) For motion from A to C:

Distance covered = 300 + 100 = 400 m.

Displacement = AB – CB = 300 – 100 = 200 m.

Time taken = 2.5 min + 1 min = 3.5 min = 3.5 × 60 min = 210 sec.

Therefore,        Average speed = Total distance covered ÷ Total time taken  = 400 ÷ 210 = 1.90 ms -1 .

Average velocity = Net displacement ÷ time taken  = 200 m ÷ 210 sec = 0.952ms -1 . 

Question 3: Abdul, while driving to school, computes the average speed for his trip to be 20 kmh -1 . On his return trip along the same route, there is less traffic and the average speed is 30 km/h. What is the average speed of Abdul’s trip? 

Question 4: A motor boat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms -2 for 8.0 s. How far does the boat travel during this time?

Answer:  Here,      u = 0 m/s                  a = 3 ms -2                t = 8 s

Using,  s = ut + ½ at 2   ⇒ s = 0 × 8 + ½ × 3×8 2 ⇒ s = 96 m.

Question 5: A driver of a car travelling at 52 kmh -1 applies the brakes and accelerates uniformly in the opposite direction. The car stops after 5 s. Another driver going at 34 kmh -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 two cars. Which of the two cars travelled farther after the brakes were applied?  

Answer: In in the following graph, AB and CD are the time graphs for the two cars whose initial speeds are 52 km/h(14.4 m/s) and 34 km/h(8.9 m/s), respectively.

Distance covered by the first car before coming to rest = Area of triangle AOB = ½ × AO × BO = ½ × 52 kmh -1 × 5 s = ½  (52 × 1000 × 1/3600) ms -1 × 5 s = 36.1 m 

Distance covered by the second car before coming to rest = Area of triangle COD = ½ × CO × DO = ½ × 34 km h -1 × 10 s = ½ × (34 × 1000 × 1/3600) ms -1 ×10 s = 47.2 m

Thus, the second car travels farther than the first car after they applied the brakes.

Question 6: Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer 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) B is travelling fastest as he is taking less time to cover more distance. (b) All three are never at the same point on the road. (c) Approximately 6 kms.   [as 8 – 2 = 6] (d) Approximately 7 kms. [as 7 – 0 = 7] 

Question 7: A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10ms -2 , with what velocity will it strike the ground? After what time will it strike the ground?

Answer: Given, initial velocity of the ball (u) = 0 (since it began at the rest position) Distance travelled by the ball (s) = 20m Acceleration (a) = 10 ms -2

As per the third motion equation, v 2 = u 2 +2as ⇒ v 2 = 2 × (10ms ‒2 ) × (20m) + 0 ⇒ v 2  = 400m 2 s ‒2 ⇒ v = 20ms -1

The ball hits the ground with a velocity of 20 meters per second.

As per the first motion equation, t = (v-u)/a = (20-0)ms ‒1  / 10ms ‒2 = 2 seconds

Therefore, the ball reaches the ground after 2 seconds.

Question 8: The speed – time graph for a car is shown in Figure:

(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: Shaded area representing the distance travelled is as follows:

(a) The shaded area represents the displacement of the car over a time period of 4 seconds. It can be calculated as:

Distance = 1/2 × 4 × 6 = 12 m

Therefore, the car travels a total of 12 meters in the first four seconds.

(b) Since the speed of the car does not change from the points (x=6) and (x=10), the car is said to be in uniform motion from the 6 th  to the 10 th  second.

Question 9: State which of the following situations are possible and give an example of each of the following:

(a) an object with a constant acceleration but with zero velocity

(b) an object moving with an acceleration but with uniform speed.

(c) an object moving in a certain direction with an acceleration in the perpendicular direction.

Answer: (a) It is possible; an object thrown up into the air has a constant acceleration due to gravity acting on it. However, when it reaches its maximum height, its velocity is zero.

(b) it is impossible; acceleration implies an increase or decrease in speed, and uniform speed implies that the speed does not change over time

(c) It is possible; for an object accelerating in a circular trajectory, the acceleration is perpendicular to the direction followed by the object.

Question 10: An artificial is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hrs to revolve around the earth.   

Answer: Here, 

Given, radius of the orbit = 42250 km

Therefore, circumference of the orbit = 2 × π × 42250 km = 265571.42 km

Time taken for the orbit = 24 hours

Therefore, speed of the satellite = 265571.42 ÷ 24 = 11065.4 km.h -1

The satellite orbits the Earth at a speed of 11065.4 kilometers per hour.

Class 9 Science NCERT Solutions Chapter 8 Motion

CBSE Class 9 Science NCERT Solutions Chapter 8 helps students to clear their doubts and to score good marks in the board exam. All the questions are solved by experts with a detailed explanation that will help students complete their assignments & homework. Having a good grasp over CBSE NCERT Solutions for Class 9 Science will further help the students in their preparation for board exams and other competitive exams such as NTSE, Olympiad, etc.

NCERT Solutions for Class 9 Science Chapter 8 PDF

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• Important Questions for CBSE Class 9 Science Chapter 8 - Motion

CBSE Class 9 Science Chapter-8 Important Questions - Free PDF Download

Class 9 plays a vital role in the lives of the students. It is the foundation class for the upcoming Board examinations of class 10th. The curriculum and entire academic marking of class 9th are also comparatively more challenging, and the students must learn the subjects well to score fantastic marks in the examinations. Motion chapter in Class 9th Science involves various concepts related to Velocity, distance, and Displacement, requiring more preparation and practice. 

Important questions for Class 9th Science Chapter 8 are essential to help the students get a brief practice during the exam times and score well in the examinations. Vedantu provides precisely designed Class 9th Science Motion Important Questions , based on the previous year papers' research. The expert team prepares the study materials to ensure the best practice resource for the students.

Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students. You can download Class 9 Maths and Class 9 Science NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations.

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Study Important Questions for Class 9 Science Chapter 8 - Motion

Very Short Answer Questions (1 Mark)

1. Which of the following statements is correct?

Both speed and velocity are same

Speed is a scalar and velocity is a vector

Speed is a vector and velocity is scalar

None of these

Ans: b) speed is a scalar and velocity is a vector 

2. What is the slope of the body when it moves with uniform velocity?

May be positive or negative

Ans: b)Zero

3. Which of the following is the position time graph for a body at rest?

4. What does an area velocity time graph give?

Acceleration

Displacement

None of the above

Ans: c) Displacement

5. If a body starts from rest, what can be said about the acceleration of the body?

Positively accelerated

Negative accelerated

Uniform accelerated

Ans: a) Positively accelerated

6. What does the slope of the position-time graph give?

Uniform speed

Both (a) and (c) depending upon the type of graph.

Ans: a) Speed

7. When a body moves uniformly along the circle, then:

Its velocity changes but speed remain the same

Its speed changes but velocity remain the same

Both speed and velocity changes

Both speed and velocity remains the same

Ans: a) Its velocity changes but speed remains the same

8. Which of the following statements is correct?

Speed distance are scalar, velocity and displacement are vector

Speed distance are vector, velocity and displacement are vector

Speed and velocity are scalar, distance and velocity are vector

Speed and velocity are vector, distance and displacement are scalar

Ans: a) Speed distance are scalar, velocity and displacement are vector

9. What does the slope of a velocity-time graph give?

Change in velocity.

Ans: c) Acceleration

10. The displacement  of the body can be-

All of these.

Ans: d) All of these.

11. Which of the following gives both direction and magnitude-

Ans: b) Vector

12. If a moving body comes to rest, then its acceleration is

All of these depending upon initial velocity.

Ans: b) Negative

Short Answer Questions (2 Marks)

1. Distinguish between speed and velocity.

Ans: Speed of a body is the distance travelled by a body as per unit time while velocity is the rate and direction of an object’s movement.

2. Under what condition(s) is the magnitude of the average velocity of an object equal to its average speed?

Ans: If the distance travelled by a body is equal to the displacement, then the magnitude of the average velocity of an object will be equal to its average speed.

3. What does the odometer of an automobile measure?

Ans: The odometer of an automobile is used to measure the distance covered by an automobile.

4. What does the path of an object look like when it is in uniform motion?

Ans: Graphically the path of an object will be linear; it looks like a straight line when it is in uniform motion.

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, \[\mathbf{3}\times \mathbf{108m}{{s}^{-1}}\] 

Ans: The given data is that time is five minutes and speed is$(3\times {{10}^{8}}m{{s}^{-1}})$ 

 \[Distance=Speed\times Time\] 

$\Rightarrow 5\min \times (3\times {{10}^{8}}m{{s}^{-1}})$ 

$\Rightarrow (5\times 60)\sec \times (3\times {{10}^{8}}m{{s}^{-1}})$ 

$\Rightarrow 300\sec \times (3\times {{10}^{8}}m{{s}^{-1}})$ 

$\Rightarrow 900\times {{10}^{8}}m{{s}^{-1}}=9\times {{10}^{10}}m$ 

$\therefore Dis\tan ce=9\times {{10}^{7}}km$ 

6. Which of the following is true for displacement?

It cannot be zero.

Ans: False with respect to the concept of displacement

Its magnitude is greater than the distance travelled by the object.

Ans: False with respect to the concept of displacement.

7. When will you say a body is in

Uniform acceleration?

Ans: When an object travels in a straight line and its velocity changes by equal amount in an equal interval of time, it is said to have uniform acceleration.

Non-uniform acceleration?

Ans:  Non-uniform acceleration is also called variable acceleration. When the velocity of an object changes by unequal amounts in equal intervals of time, it is said to have non-uniform acceleration.

8. A bus decreases its speed from $80km{{h}^{-1}}$ to $60km{{h}^{-1}}$ in \[\mathbf{5s}\]. Find the acceleration of the bus.

Ans: Initial speed of bus (u) $=80km{{h}^{-1}}$ 

$\Rightarrow \dfrac{80\times 1000}{60\times 60\sec }$ 

$\Rightarrow \dfrac{200}{9m{{s}^{-1}}}=22.22m{{s}^{-1}}$  Final speed of bus (v) $=60km{{h}^{-1}}$ 

$\Rightarrow \dfrac{60\times 1000}{60\times 60\sec }=\dfrac{50}{3m{{s}^{-1}}}=16.67m{{s}^{-1}}time(t)=5s$ 

Acceleration (a) $=\dfrac{(v-u)}{t}=\dfrac{(16.67-22.22)}{5}=\dfrac{-5.55}{5}=-1.11m/{{s}^{2}}$ 

$\therefore Acceleration\text{ }\left( a \right)=-1.11m/{{s}^{2}}$ 

9. What is the nature of the distance time graphs for uniform and non-uniform motion of an object?

Ans: If an object has a uniform motion then the nature of distance time graph will be linear, that is it would in a straight line and if it has non-uniform motion then the nature of the distance-time graph will be a curved line.

10. What is the quantity which is measured by the area occupied below the velocity-time graph?

Ans: The area occupied below the velocity-time graph measures the distance moved by any object.

11. A bus starting from rest moves with a uniform acceleration of \[\mathbf{0}.\mathbf{1m}{{\mathbf{s}}^{-2}}\] for \[\mathbf{2}\text{ }\mathbf{minutes}\] . Find

The speed acquired,

Ans: $u=0,a=0.1m{{s}^{-2}},t=2\min =120\sec $ 

$v=u+at=0+0.1\times 120=12m{{s}^{-1}}$ 

\[\mathbf{Speed}\text{ }\mathbf{acquired}=v=12m{{s}^{-1}}\] 

The distance travelled.

Ans: $s=ut+\dfrac{1}{2}a{{t}^{2}}=0\times 120+\dfrac{1}{2}0.1\times {{120}^{2}}=720m$

12. A trolley, while going down an inclined plane, has an acceleration of$2cm{{s}^{-2}}$ . What will be its velocity \[\mathbf{3s}\]after the start?

Ans: Given:$u=0,a=2cm/{{s}^{2}},t=3s$  

$v=u+at=0+2\times 3=6cm/s$ 

13. A racing car has a uniform acceleration of$4m{{s}^{-2}}$ . What distance will it cover in \[\mathbf{10s}\]after start?

Ans: Given:$u=0,a=4m/{{s}^{2}},t=10s$  

$s=ut+\dfrac{1}{2}a{{t}^{2}}$ 

$s==0\times 10+\dfrac{1}{2}\times 4\times {{10}^{2}}$ 

$\therefore s=200m$ 

14. Differentiate between distance and displacement?

Ans: The difference between distance and displacement is as below,

15. Derive mathematically the first equation of motion\[\mathbf{V}=\mathbf{u}+\mathbf{at}\]?

Ans: Acceleration is defined as the rate of change of velocity.

Let V=final velocity; \[{{\mathbf{V}}_{o}}\]= initial velocity, T= time, a = acceleration. 

So by definition of acceleration

$a=\dfrac{V-{{V}_{o}}}{T}$ 

$at=V-{{V}_{o}}$ 

$V={{V}_{o}}+at$ 

 If \[{{V}_{o}}=u=\] initial velocity, then \[\left[ V=\text{ }u+at \right]\] 

16. Calculate the acceleration of a body which starts from rest and travels \[\mathbf{87}.\mathbf{5m}\text{ }\mathbf{5sec}\]?

Ans: Given Data: $u=0$ (starts from rest) u= initial velocity

a=? a=acceleration

\[T=5\sec ,t=time\] 

\[S=\text{ }87.5m\text{ }\left( S=distance \right)\] 

From second equation of motion

$S=ut+\dfrac{1}{2}a{{t}^{2}}$ 

$\Rightarrow 87.5=0+\dfrac{1}{2}a{{t}^{2}}$ 

$\Rightarrow 87.5=\dfrac{1}{2}a{{t}^{2}}\to (i)$ 

$\Rightarrow 87.5\times 2=a\times {{(5)}^{2}}$ 

$\Rightarrow \dfrac{87.5\times 2}{25}=a$ 

$\Rightarrow \dfrac{175.0}{25}=a$ 

$\therefore S=7m/{{s}^{2}}=a$ 

17. Define uniform velocity and uniform acceleration?

Ans: Uniform velocity :- A body is said to move with uniform velocity if equal displacement takes place in equal intervals of time, however small these intervals may be. 

Uniform acceleration :- A body is said to move with uniform acceleration if equal changes in velocity take place in equal intervals of time, however, small intervals may be.

18. A car travels at a speed of \[\mathbf{40km}/\mathbf{hr}\]for two hour and then at \[\mathbf{60km}/\mathbf{hr}\]for three hours. What is the average speed of the car during the entire journey?

Ans:  Given: In first case;${{t}_{1}}=time=2hrs$ 

${{v}_{1}}=speed=40km/hr$ 

${{s}_{1}}=dis\tan ce=speed\times time$ 

${{s}_{1}}=40\times 2=80km$ 

In second case, Given ${{t}_{2}}=time=3hrs$ 

${{v}_{2}}=speed=60km/hr$ 

${{s}_{2}}=dis\tan ce=speed\times time$ 

${{s}_{2}}=60\times 3=180km$ 

The total distance = ${{s}_{1}}+{{s}_{2}}=80+180=260km$ 

Total time, ${{t}_{1}}+{{t}_{2}}=2+3=5hrs$ 

Average speed = $\dfrac{total\text{ }distance}{total\text{ }time}=\dfrac{260}{5}$ 

$\therefore Average\text{ }speed=52km/hr$ .

19. The velocity-time graph of two bodies A and B traveling along the +x direction are given in the figure

Are the bodies moving with uniform acceleration?

Ans: Yes the bodies are moving with uniform acceleration.

Which body is moving with greater acceleration A or B?

Ans: Body A is moving with greater acceleration.

20. Derive the second equation of motion, \[\mathbf{s}=~\mathbf{ut}+~\dfrac{1}{2}a{{t}^{2}}\] numerically?

Ans: Let at time \[\mathbf{t}=\mathbf{0}\] , body has initial velocity = ${{V}_{o}}$ 

At time‘t’, the body has final velocity=V

S=distance traveled in time‘t’

We know, total distance traveled = Average velocity $\times $ time

\[Average\text{ }velocity~~=\dfrac{initial\text{ }velocity+final\text{ }velocity}{2}\] 

$\Rightarrow \dfrac{{{V}_{o}}+V}{2}$

$Total~distance=s=\dfrac{{{V}_{o}}+V}{2}\times t$ 

$\Rightarrow 2s=({{V}_{o}}+V)t\to (i)$ 

Now from first equation of motion, $V={{V}_{o}}+at\to (ii)$ 

Use the value of (V) from (ii) in (i)

$2s=({{V}_{o}}+{{V}_{o}}+at)t$

$2s=2{{V}_{o}}t+\dfrac{1}{2}a{{t}^{2}}$  

$Let,{{V}_{o}}=u$ 

$\Rightarrow s=ut+\dfrac{1}{2}a{{t}^{2}}$ 

21. Calculate the acceleration and distance of the body moving with $5m/{{s}^{2}}$ which comes to rest after traveling for 6sec?

Ans: Acceleration=a=?

Final velocity = V =o (body comes to rest)

Distance = s =?

\[Time\text{ }=t=\text{ }6~sec\] 

From, \[V~=~u~+~at\] 

$O=5+a\times 6$

$-a\times 6=5$ 

$a=\dfrac{-5}{6}m/{{s}^{2}}$ 

${{v}^{2}}-{{u}^{2}}=2as$ 

${{O}^{2}}-25=2\times \dfrac{-5}{6}\times s$ 

$\Rightarrow -25=\dfrac{-5}{3}\times s$ 

$\Rightarrow \dfrac{25\times 3}{5}=s$ 

$\therefore s=15m$ 

22. A body is moving with a velocity of $12m/s$ and it comes to rest in 18m, what was the acceleration?

Ans : Initial velocity\[=u=12m/s\] 

Find velocity =V=0

\[S=distance=18m\] 

A= acceleration =?

From\[~{{3}^{rd}}\]equation of motion;

${{O}^{2}}-{{(12)}^{2}}=2\times a\times 18$ 

$\Rightarrow \dfrac{-144}{36}=a$ 

$a=\dfrac{-144}{36}$ 

$[a=-4m/{{s}^{2}}]$ 

$\Rightarrow 4m/{{s}^{2}}$  

23. A body starts from rest and moves with a uniform acceleration of $4m/{{s}^{2}}$until it travels a distance of $800m$, find the find velocity?

Ans: Initial velocity $=u=0$ 

Final velocity$=v=?$ 

Acceleration$=a=4m/{{s}^{2}}$ 

Distance$=s=800m$ 

${{u}^{2}}-(0)=2\times 4\times 800$ 

$u=80m/s$ 

$\therefore {{u}^{2}}=6400$ 

24. The driver of a car traveling along a straight road with a speed of $72kmph$ observes a sign board which gives the speed limit to be$54kmph$. The signboard is $70m$ahead when the driver applies the brakes calculate the acceleration of the car which will cause the car to pass the signboard at the stated speed limit?

Ans: Initial speed \[=u=72\text{ }km/hr\] 

$\Rightarrow \dfrac{72\times 5}{18}=20m/s$ 

Final speed \[=v=54\text{ }km/hr\] 

$\Rightarrow \dfrac{54\times 5}{18}=15m/s$ 

Distance\[~~=~\mathbf{S}~=~\mathbf{70m}\] 

Now, ${{v}^{2}}-{{u}^{2}}=2as$ 

${{(15)}^{2}}-{{(20)}^{2}}=2\times a\times 70$ 

$\Rightarrow 225-400=140a$ 

$\Rightarrow -175=140a$ 

$[a=-1.25m/{{s}^{2}}]$ 

25. Differentiate between scalars and vectors?

Ans: The difference between scalars and vectors is as below,

Short Answer Questions (3 Marks)

1. An object has moved through a distance. Can it have zero displacements? If yes, support your answer with an example.

Ans: Yes, if an object is moved through a distance it can have zero displacement because displacement of an object is the actual change in its position when it moves from one position to the other position. So if an object travels from point A to B and then returns back to point A again, the total displacement will be zero.

2. A farmer moves along the boundary of a square field of side $10m$in$40s$. What will be the magnitude of displacement of the farmer at the end of$2\min 20\sec $ ?

Ans: Distance covered by farmer in \[\mathbf{40}\text{ }\mathbf{seconds}~=4\times (10)m=40m\] 

Speed of the farmer = distance/time \[=40m/40s=1m/s.\]

Total time given in the question\[=2min20sec=60+60+20=140sec\] 

Since he completes $1round$of the field in \[40\]seconds so in he will complete $3$rounds in \[120\text{ }seconds~\left( 2mins \right)or120m\] distance is covered in\[2min\]. In another \[20sec\]will cover another\[20m\]so total distance covered in \[2min20sec=120+20=140m.\]

\[Displacement=\sqrt{{{10}^{2}}}+{{10}^{2}}\] 

$\Rightarrow \sqrt{200}~=\sqrt{10}\sqrt{2m}~$ (as per diagram)

$\Rightarrow 10\times ~1.414$ 

$\therefore Displacement=~14.14m$ 

3. A train starting from a railway station and moving with uniform acceleration attains a speed$40km{{h}^{-1}}$ in $10\min $. Find its acceleration.

Ans: Since the train starts from rest (railway station) = u = zero

Final velocity of train $=v=40km{{h}^{-1}}$ 

$\Rightarrow \dfrac{(40\times 1000)}{60\times 60m{{s}^{-1}}}=100/9m{{s}^{-1}}$ 

$\Rightarrow 11.11m{{s}^{-1}}$ 

\[\begin{array}{*{35}{l}} time\left( t \right)=10min=10\times 60=600\text{ }seconds \\ ~ \\ \end{array}\]  

\[Since\text{ }a\text{ }=\dfrac{\left( v~u \right)}{t}~=11.11m{{s}^{-1}}/600\sec =0.018m/{{s}^{2}}~\] 

4. What can you say about the motion of an object whose distance time graph is a straight line parallel to the time axis?

Ans: If the object’s distance time graph is a straight line parallel to the time axis indicates that with increasing time the distance of that object is not increasing hence the object is at rest that is not moving.

Distance-time graph of an object at rest

5. What can you say about the motion of an object if its speed time graph is a straight line parallel to the time axis?

Ans: Such a graph indicates that the object is travelling with uniform velocity.

Speed time graph of an object moving with uniform speed

6. A train is travelling at a speed of $90km{{h}^{-1}}$ . Brakes are applied so as to produce a uniform acceleration of$-0.5m{{s}^{-2}}$ . Find how far the train will go before it is brought to rest.

Ans: $u=90km{{h}^{-1}}=\dfrac{(90\times 1000)}{60\times 60}=25m{{s}^{-1}}$ 

\[Given:a=-0.5m{{s}^{-2}},v=0\left( \mathbf{trainis}~\mathbf{brought}~\mathbf{to}~\mathbf{rest} \right)\] 

\[v=u+at=25+(-0.5)\times t\] 

\[\Rightarrow 0=25-0.5x\] 

\[\Rightarrow 0.5t=25,ort=\dfrac{25}{0.5}=50\sec \] 

\[s=ut+\dfrac{1}{2}a{{t}^{2}}=25\times 50+\dfrac{1}{2}\times (-0.5)\times {{50}^{2}}\] 

\[\Rightarrow 1250-625=625m\] 

$\therefore s=625m$ 

7. A stone is thrown in a vertically upward direction with a velocity of$5m{{s}^{-1}}$ . If the acceleration of the stone during its motion is$10m{{s}^{-1}}$ in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Ans: Given:$u=5m{{s}^{-1}},a=-10m{{s}^{-2}}$  

$v=0~~~\left( \mathbf{sinceat}\text{ }\mathbf{maximum}\text{ }\mathbf{height}\text{ }\mathbf{its}\text{ }\mathbf{velocity}\text{ }\mathbf{will}\text{ }\mathbf{be}\text{ }\mathbf{zero} \right)$ 

$v=u+at=5+(-10)\times t$ 

$\Rightarrow 0=5-10t$ $10t=5,or,t=\dfrac{5}{10}=0.5\sec $ 

$S=ut+\dfrac{1}{2}a{{t}^{2}}=5\times 0.5+\dfrac{1}{2}\times (-10)\times {{0.5}^{2}}$ 

$\Rightarrow 2.5-1.25=1.25m$ 

$\therefore S=1.25m$ 

8. Derive the second equation of motion$S=ut+\dfrac{1}{2}a{{t}^{2}}$  graphically?

Ans : Let at time \[\mathbf{T}=\mathbf{0}\] body moves with initial velocity u and at time ‘t’ body has final velocity ‘v’ and at time ‘t’ it covers a distance’s.

\[\mathbf{AC}=\mathbf{v},\text{ }\mathbf{AB}=\mathbf{u},\text{ }\mathbf{OA}=\mathbf{t},\text{ }\mathbf{DB}=\mathbf{OA}=\mathbf{t},\text{ }\mathbf{BC}=\mathbf{AC}-\mathbf{AB}=\mathbf{V}-\mathbf{u}\] 

Area under a \[v-t\]curve gives displacement so,

\[\mathbf{S}=\mathbf{Area}\text{ }\mathbf{of}\text{ }\Delta \mathbf{DBC}+\mathbf{Area}~\mathbf{of}\text{ }\mathbf{rectangle}\text{ }\mathbf{OABD}\to \left( \mathbf{i} \right)\] 

\[Area\text{ }of~\Delta DBC=\dfrac{1}{2}\times Base\times Height\Rightarrow \dfrac{1}{2}\times ~DB\times BC\] 

\[\Rightarrow \dfrac{1}{2}\times t\times (v-u)\to (ii)\] 

\[Area\text{ }of\text{ }rectangle\text{ }OABD\text{ }=\text{ }length\times Breadth\] 

\[\Rightarrow OA\times BA\] 

\[\Rightarrow t\times u\to (iii)\] 

\[S=ut+\dfrac{1}{2}\times t\times (v-u)\] 

\[S=ut+\dfrac{1}{2}\times t\times at(\therefore useV-u=at)\] 

\[\therefore S=ut+\dfrac{1}{2}a{{t}^{2}}\] 

9. A car moving with a certain velocity comes to a halt if the retardation was$5m/{{s}^{2}}$ find the initial velocity of the car?

Ans: \[\mathbf{V}=\mathbf{0}\left( \mathbf{comes}\text{ }\mathbf{to}\text{ }\mathbf{rest} \right)\mathbf{V}=\mathbf{final}\text{ }\mathbf{velocity}\] 

$S=62.5m$ 

$a=-5m/{{s}^{2}}(retardation)$ 

From \[{{3}^{rd}}\] equation of motion,

${{O}^{2}}-{{u}^{2}}=2\times (-5)\times 62.5$ 

$-{{u}^{2}}=-10\times 62.5$ 

${{u}^{2}}=625$

$u=\sqrt{625}[u=25m/s]$

10. Two cars A and B are moving along in a straight line. Car A is moving at a speed of $80kmph$ while car B is moving at a speed of $50kmph$in the same direction, find the magnitude and direction of

Five v the relative of car A with respect to B.

The relative velocity of car B with respect to A.

Ans: Velocity of car A = $80kmph$  

Velocity of Car B = \[-\text{ }50~~kmph\] 

(-ve sign indicates that Car B is moving in the opposite direction to Car A )

The relative velocity of car A with respect to B

velocity of  car A + (- velocity of car B)

$\Rightarrow 80+(-(-50))$ 

$\Rightarrow 80+50$ 

$\Rightarrow +130kmph$

$+130kmph$shows that for a person in car B, car A will appear to move in the same direction with a speed of the sum of their individual speed.

  Relative velocity of car B with respect to A

 $\Rightarrow $ Velocity of car B+ (- velocity of car A)

\[ \Rightarrow -50~~+~\left( -80 \right) \] 

\[ \Rightarrow -130kmph \] 

It shows that car B will appear to move with \[130~~~kmph\] in opposite direction to car A

11. A ball starts from rest and rolls down $16m$down an inclined plane in$4s$

What is the acceleration of the ball? 

Ans:  Given: \[\mathbf{u}=\text{ }\mathbf{initial}\text{ }\mathbf{velocity}\text{ }=\text{ }\mathbf{0}\] (body starts from rest)\[S=\text{ }distance\text{ }=\text{ }16\text{ }m\text{ }\]  

\[T=\text{ }time\text{ }=\text{ }4s\]

From, $s=ut+\dfrac{1}{2}a{{t}^{2}}$ 

$16=0\times t+\dfrac{1}{2}a\times {{\left( 4 \right)}^{2}}$ 

$16=\dfrac{1}{2}\times a\times 16$ 

$\dfrac{16\times 2}{16}=a$ 

$[2m/{{s}^{2}}=a]$

What is the velocity of the ball at the bottom of the incline? 

Ans: From, $v=u+at$ 

$v=0+2\times 4$

$[v=8m/s]$ 

12. Two boys A and B, travel along the same path. The displacement – time graph for their journey is given in the following figure. 

How far down the road has B travelled when A starts the journey?

Ans: When A starts his journey at\[4\text{ }sec\] , B has already covered a distance of \[857m\].

Without calculation, the speed, state who is traveling faster A or B? 

Ans: A travels faster than B because A starts his journey late but crosses B and covers more distance then B in the same time as B 

What is the speed of A? 

Ans:  

\[Speed\text{ }of\text{ }A\text{ }=\text{ }\dfrac{Distance\text{ }covered}{Time\text{ }taken}\] 

Let at \[t\text{ }=12\text{ }min\],\[distance\text{ }covered=3500m\]  

$\Rightarrow \dfrac{3500}{12}=375m/\min $ 

$\therefore Speed\text{ }of\text{ }A=375m/\min $ 

What is the speed of B?

\[Speed\text{ }of\text{ B}=\dfrac{Distance\text{ }covered}{Time\text{ }taken}\] 

\[{{V}_{B}}=\dfrac{3000}{12}\]

\[{{V}_{B}}=214m/\min \] 

Are the speed of A and B uniform?

What does point X on the graph represent?

Ans: X on the graph represents the point at which both A and B are at the same position  

What is the speed of approach of A towards B? What is the speed of separation of A from B?

Ans: Speed of approach of A towards \[B=375\text{ }m/min-214\text{ }m/min\]

\[\Rightarrow 161\text{ }m/min\] 

Speed of separation of A from \[B=161\text{ }m/min\] 

13. A body is dropped from a height of$320m$. The acceleration due to the gravity is $10m/{{s}^{2}}?$

How long does it take to reach the ground? 

Ans:  Given Data: Height = h

\[Distance=s=320m\]  

Acceleration due to gravity $=g=10m/{{s}^{2}}$ 

Initial velocity \[=u=0\] 

From $s=ut+\dfrac{1}{2}a{{t}^{2}}$ 

$h=ut\times \dfrac{1}{2}g{{t}^{2}}$ 

$320=0\times t+\dfrac{1}{2}\times 10\times {{t}^{2}}$ 

$\dfrac{320\times 2}{10}={{t}^{2}}$ 

$64={{t}^{2}}$ 

$t=8\sec $ 

What is the velocity with which it will strike the ground?

Ans: From $v=u+at$ 

$v=0+10\times 8$ 

$v=80m/s$ 

14. Derive third equation of motion${{v}^{2}}-{{u}^{2}}=2as$numerically?

Ans: We know,  

$v=u+at......(i)$ 

$s=ut+\dfrac{1}{2}a{{t}^{2}}......(ii)$

When, v= final velocity 

u= initial velocity

a = acceleration 

t = time 

s = distance 

From equation (i) $t=\dfrac{v-u}{a}$ 

Put the value of t in equation (ii) 

$s=u\times \dfrac{v-u}{a}+\dfrac{1}{2}a\times \dfrac{v-u}{a}$ 

$s=\dfrac{uv-{{u}^{2}}}{a}+\dfrac{1}{2}a\times \dfrac{{{v}^{2}}+{{u}^{2}}-2uv}{{{a}^{2}}}$ 

$s=\dfrac{uv-{{v}^{2}}}{a}+\dfrac{1}{2}\times \dfrac{{{v}^{2}}+{{u}^{2}}-2uv}{a}$ 

$s=\dfrac{2uv-2{{u}^{2}}+{{v}^{2}}+{{u}^{2}}-2vu}{2a}$ 

$s=\dfrac{{{v}^{2}}-{{u}^{2}}}{2a}$ 

$2as={{v}^{2}}-{{u}^{2}}$ 

${{v}^{2}}={{u}^{2}}+2as$ 

16. The velocity-time graph of the runner is given in the graph. 

What is the total distance covered by the runner in$16s$?

Ans:  We know that area under v-t graph gives displacement: 

So, Area = distance = s = area of triangle + area of rectangle \[Area\text{ }of\text{ }triangle=\dfrac{1}{2}\times base\times height\] 

\[\Rightarrow \dfrac{1}{2}\times 6\times 10\] \[\Rightarrow 30m\] 

\[\therefore Area\text{ }of\text{ }triangle=30m\] 

\[Area\text{ }of\text{ }rectangle=length\times breadth\] 

\[\Rightarrow \left( 16-6 \right)\times 10\] 

\[\Rightarrow 10\times 10\] 

\[\Rightarrow 100m\] 

Total area = \[180m\] 

Total distance = \[180m\] 

What is the acceleration of the runner at $t=11s$?

Ans: Since at \[t=11sec\], particles travel with uniform velocity so, there is no change in velocity hence acceleration = zero. 

17. A boy throws a stone upward with a velocity of $60m/s$ . 

How long will it take to reach the maximum height$(g=-10m/{{s}^{2}})$ ?

Ans:\[u=60\text{ }m/s;\text{ g= -10m/}{{\text{s}}^{2}};v=0\]

 The time to reach maximum height is

\[\text{v}=\text{u}+\text{at}=\text{u}+\text{gt}\]

\[0=60-10\text{t}\]

\[\text{t}=\dfrac{60}{10}=6\text{s}\]

What is the maximum height reached by the ball? 

Ans: The maximum height is:

\[\text{v}^2=\text{u}^2+2\text{gs}\]

\[\text{s}=-\dfrac{\text{u}^2}{2\text{g}}=\dfrac{60^2}{2\times 10}\]

\[\Rightarrow 180 \text{m}\]

How long will it take to reach the ground?

Ans: The time to reach the top is equal to the time taken to reach back to the ground. Thus, the time to reach the ground after reaching the top is\[6s\]or the time to reach the ground after throwing is\[~6+6=12s\]. 

18. The displacement x of a particle in meters along the x- axis with time ‘t’ in seconds according to the equation- $X=2m+\left( \dfrac{12m}{s} \right)t$ 

Draw a graph if x versus t for $t=0$ and $t=5\sec $ 

Ans: \[X=20m+\left( 12 \right)\text{ }t\] 

At \[t=0\text{ }\]

\[X=20+120=12\text{ }m\] 

At \[t=1\] 

\[X=20+12=32m\] 

At \[t\text{ }=2\]

\[X=20+24=44m\]  

At \[t=5\] 

\[X=20+125=72m\]

What is the displacement comes out of the particles initially? 

Ans: At \[T=0\] (initially) 

Displacement \[=20m.\] 

What is the slope of the graph obtained? 

\[Slope=\dfrac{{{y}_{2}}-y1}{{{x}_{2}}-{{x}_{1}}}=\dfrac{72-44}{5-2}=\dfrac{28}{3}\] 

$\Rightarrow 9.3m/s$ 

19. The velocity of a body in motion is recorded every second as shown

calculate the –

Acceleration 

Ans: Acceleration =slope of the velocity time graph

$a=\dfrac{{{V}_{2}}-{{V}_{1}}}{{{t}_{2}}-{{t}_{1}}}$ 

$a=\dfrac{54-24}{1-6}=\dfrac{30}{-5}=-6m/{{s}^{2}}$

Distances travelled and draw the graph. 

Ans: Distance $\Rightarrow S=ut+\dfrac{1}{2}a{{t}^{2}}$ 

$\Rightarrow 60\times 10+\dfrac{1}{2}(-6)\times {{(10)}^{2}}$ 

$\Rightarrow 600-300=300m$

20. Draw the graph for uniform retardation – 

Position – time graph 

Velocity – time 

Acceleration- time

21. The displacement – time graph for a body is given. State whether the velocity and acceleration of the body in the region BC, CD, DE and EF are positive, negative or Zero. 

For AB, the curve is upward stopping i.e. slope is increasing so velocity is positive and remains the same so, V= +ve but a = 0.

For BC, the curve has still has +ve slope so, V = +ve but velocity is decreasing with respect to time so, a=negative 

For CD, both velocity and acceleration are Zero because slope is Zero. 

For DE, velocity is the (v is increasing with respect to time) and so is the acceleration is +ve. (v) For EF, velocity is +ve (positive slope of x-t graph) but acceleration is Zero because velocity remains some with time.

22. Derive the third equation of motion$-{{v}^{2}}-{{u}^{2}}=2as$  as graphically? 

Ans: Let at time \[t=0\], the body moves with initial velocity u and time at ‘t’ has final velocity ‘v’ and in time ‘t’ covers a distance ‘s’ 

Area under v-t graph gives displacement

\[\mathbf{S}=\mathbf{Area}\text{ }\mathbf{of}\text{ }\Delta \mathbf{DBC}+\mathbf{Area}~\mathbf{of}\text{ }\mathbf{rectangle}\text{ }\mathbf{OAB}D\] 

$S=\dfrac{1}{2}\times base\times height+length\times breadth$ 

$S=\dfrac{1}{2}\times DB\times BC+OA\times AB$ 

$S=\dfrac{1}{2}\times t\times \left( v-u \right)+t\times u\to (i)$ 

$Now,v-u=at$ 

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

Put the value of ‘t’ in equation (i)

$s=\dfrac{1}{2}\times (v-u)\times \dfrac{v-u}{a}+u\times \left( \dfrac{v-u}{a} \right)$ 

$s=\dfrac{{{(v-u)}^{2}}2u(v-u)}{2a}$ 

$s=\dfrac{{{v}^{2}}+{{u}^{2}}-2uv+2vu-2{{u}^{2}}}{2a}$ 

$2as={{v}^{2}}-{{u}^{2}}$

Third equation of motion

Long Answer Questions (5 Marks)

1. An athlete completes one round of a circular track of diameter $200m$in $40s$. What will be the distance covered and the displacement at the end of $2\min 20\sec ?$ 

Ans: circumference of circular track \[=2\pi r\] 

 $\Rightarrow 2\times \dfrac{22}{7}\times \dfrac{diameter}{2}$ 

$\Rightarrow 2\times \dfrac{22}{7}\times \dfrac{200}{2}=\dfrac{4400}{7}m$

Rounds completed by athlete in \[2min20sec=s=\dfrac{140}{40}=\text{ }3.5\] 

Therefore, total distance covered $=\dfrac{4400}{7}\times 3.5=2200m$ 

since one complete round of circular track needs \[40s\]  so he will complete \[3rounds\text{ }in\text{ }2mins\]and in next \[20s\] he can complete half round Therefore, displacement \[=diameter=200m\].

2. Joseph jogs from one end A to the other end B of a straight $300m$road in $2\min 50\sec $ and then turns around and jogs $100m$back to point C in another$1\min $ . What are Joseph’s average speeds and velocities in jogging 

From A to B 

Ans: \[Distance=300m\] 

\[time=2min30seconds=150\text{ }seconds\] 

Average speed from A to B = average velocity from A to B 

\[\Rightarrow 300m/150s=2m/s\] 

From A to C?

Ans: average speed from A to C \[=\dfrac{\left( 300+100 \right)\text{ }m}{\left( 150+60 \right)\text{ }sec}\] 

$\Rightarrow \dfrac{400m}{210s}=1.90m/s$ 

Displacement from A to C \[=\left( 300100 \right)m=200m\] 

\[time=2min30sec+1min\text{ }=210s\] 

\[velocity\text{ }=\dfrac{\text{ }displacement}{time}=\dfrac{200m}{210s}=0.95m/s\] 

3. Abdul, while driving to school, computes the average speed for his trip to be $20km{{h}^{-1}}$. On his return trip along the same route, there is less traffic and the average speed is $40km{{h}^{-1}}$. What is the average speed for Abdul’s trip? 

Ans. If we suppose that distance from Abdul’s home to school = x km s 

while driving to school:$speed=20km{{h}^{-1}}$ , 

\[velocity\text{ }=\text{ }\dfrac{displacement}{time}\] 

\[20\text{ }=\dfrac{x}{t},or,\text{ }t=\dfrac{x}{20}hr\] 

on his return trip: \[speed=40km{{h}^{-1}},40=\text{ }x/t\] 

or, \[t=\dfrac{x}{40}hr\] 

\[total\text{ }distance\text{ }travelled=\text{ }x+x=2x\] 

\[total\text{ }time\text{ }=\text{ }t\text{ }+\text{ }t=\dfrac{x}{20}+\dfrac{x}{40}=\dfrac{\left( 2x\text{ }+\text{ }x \right)}{40}=\dfrac{3x}{40}hr\] 

average speed for Abdul’s trip \[=\text{ }\dfrac{2x}{\left( \dfrac{3x}{40} \right)\text{ }}=\dfrac{80x}{3x}=26.67km/hr\] 

4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of $3.0m{{s}^{-2}}$ for $8.0s$. How far does the boat travel during this time?

Ans: since the motorboat starts from rest so \[u=0\]

Given data:

$time(t)=8s,a=3m/{{s}^{2}}$  

$distacnce(s)=ut+\dfrac{1}{2}a{{t}^{2}}$

\[distacnce(s)=0+\dfrac{1}{2}\times 3\times {{8}^{2}}\] 

$\therefore distacnce(s)=96m$ 

5. A driver of a car travelling at applies the brakes and accelerates uniformly in the opposite direction. The car stops in$5s$. Another driver going at $3km{{h}^{-1}}$ in another car applies is brakes slowly and stops in$10s$. 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?

As given in the figure below AB (in red line) and CD(in red line) are the Speed-time graph for given two cars with initial speeds \[52km{{h}^{-1}}\] and $3km{{h}^{-1}}$  respectively. Distance Travelled by first car before coming to rest=Area of $\Delta OAB$ 

\[\Rightarrow \left( \dfrac{1}{2} \right)\times OB\times OA\] 

\[\Rightarrow \left( \dfrac{1}{2} \right)\times 5s\times 52km{{h}^{-1}}\] 

\[\Rightarrow \left( \dfrac{1}{2} \right)\times 5\times \left( \dfrac{52\times 1000}{3600} \right)m\] 

\[\Rightarrow \left( \dfrac{1}{2} \right)\times 5\times \left( \dfrac{130}{9} \right)m\] 

\[\Rightarrow \dfrac{325}{9}m\] 

\[\Rightarrow 36.11m\] 

Distance Travelled by second car before coming to rest=Area of $\Delta OCD$ 

\[\Rightarrow \dfrac{1}{2}\times OD\times OA\] 

\[\Rightarrow \dfrac{1}{2}\times 10s\times 3km{{h}^{-1}}\] 

\[\Rightarrow \dfrac{1}{2}\times 10\times \left( \dfrac{3\times 1000}{3600} \right)m\] 

\[\Rightarrow \dfrac{1}{2}\times 10\times \left( \dfrac{5}{6} \right)m\] 

\[\Rightarrow 5\times \left( \dfrac{5}{6} \right)m\] 

\[\Rightarrow \dfrac{25}{6}=4.16m\]

$\therefore $ Clearly the first car will travel farther \[\left( 36.11\text{ }m \right)\] than the first car \[\left( 4.16\text{ }m \right).\] 

6. Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:

Which of the three is travelling the fastest?

Ans: It is clear from graph that B covers more distance in less time. Therefore, B is the fastest.

Are all three ever at the same point on the road? Fig. 8.11 

Ans: All of them never come at the same point at the same time.

How far has C travelled when B passes A? 

Ans: According to graph; each small division shows about\[0.57\text{ }km\] .

A is passing B at point S which is in line with point P (on the distance axis) and shows about \[9.14\text{ }km\] 

Thus, at this point C travels about 

\[\text{9}\text{.14-(0}\text{.57}\times \text{3}\text{.75)km= }9.14\text{ }km\text{ }\text{ }2.1375\text{ }km=7.0025\text{ }km\approx 7km\] 

Thus, when A passes B, C travels about\[7\text{ }km\].

 How far has B travelled by the time it passes C?

Ans: B passes C at point Q at the distance axis which is \[\approx 4km+0.57km\times 2.25=\text{ }5.28\text{ }km\] Therefore, B travelled about \[5.28\text{ }km\] when passes to C.

7. A ball is gently dropped from a height of \[20m\]. If its velocity increases uniformly at the rate of $10m{{s}^{-2}}$, with what velocity will it strike the ground? After what time will it strike the ground? 

Ans:   Let us assume, the final velocity with which ball will strike the ground be 'v' and time it takes to strike the ground be 't' 

Initial Velocity of ball \[u=0\] 

Distance or height of fall \[s=20\text{ }m\] 

Downward acceleration $a=10m{{s}^{-2}}$ 

As we know,${{v}^{2}}={{u}^{2}}-2as$ 

$or,2as={{v}^{2}}-{{u}^{2}}$ 

${{v}^{2}}=2as+{{u}^{2}}$ 

$\Rightarrow 2\times 10\times 20+0$ 

$v=\sqrt{400m{{s}^{-1}}}$

$\therefore $ Final velocity of ball,$v=20m{{s}^{-1}}$ 

 \[t=\text{ }\dfrac{\left( v-u \right)}{a}\] 

$\therefore $Time taken by the ball to strike $=\dfrac{\left( 20-0 \right)}{10}$ 

\[\Rightarrow \dfrac{20}{10}\] 

\[\therefore Time\text{ }taken\text{ }by\text{ }the\text{ }ball\text{ }to\text{ }strike=2\sec \]

8. The speed-time graph for a car is shown is Fig. 8.12.

Find how far does the car travel in the first\[4\] seconds. Shade the area on the graph that represents the distance travelled by the car during the period. 

Distance travelled by car in the \[4\text{ }second\] 

The area under the slope of the speed – time graph gives the distance travelled by an object. In the given graph 

\[56\] full squares and \[12\] half squares come under the area slope for the time of \[4\text{ }second\]. 

Total number of squares \[=56+\dfrac{12}{2}=62\] squares 

The total area of the squares will give the distance travelled by the car in \[4\text{ }second\] . on the time axis, 

\[5\text{ }squares\text{ }=2seconds,\text{ }therefore\text{ }1\text{ }square\text{ }=\dfrac{2}{5}seconds\] 

on speed axis there are \[3\text{ }squares=2m/s\] 

therefore, area of one square $=\dfrac{2}{5}s\times \dfrac{2}{3}m/s=\dfrac{4}{15}m$ 

so area of  \[62\text{ }squares=\dfrac{4}{15}m\times 62=\dfrac{248}{15}m=16.53m\] 

Hence the car travels \[16.53m\] in the first\[4\text{ }seconds\].

Which part of the graph represents uniform motion of the car? 

Ans: The straight line part of graph, from point A to point B represents a uniform motion of 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 

Ans: An object with a constant acceleration can still have the zero velocity. For example, an object which is at rest on the surface of earth will have zero velocity but still being acted upon by the gravitational force of earth with an acceleration of \[9.81\text{ }m{{s}^{-2}}\]towards the center of earth. Hence when an object starts falling freely can have constant acceleration but with zero velocity. 

b) An object moving in a certain direction with acceleration in the perpendicular direction. 

Ans: When an athlete moves with a velocity of constant magnitude along the circular path, the only change in his velocity is due to the change in the direction of motion. Here, the motion of the athlete moving along a circular path is, therefore, an example of an accelerated motion where acceleration is always perpendicular to direction of motion of an object at a given instance. Hence it is possible when an object moves on a circular path. 

10. An artificial satellite is moving in a circular orbit of radius \[42250km\] . Calculate its speed if it takes \[24hrs\]to revolve around the earth. 

Ans. Let us assume An artificial satellite, which is moving in a circular orbit of radius \[42250\text{ }km\] covers a distance 's' as it revolve around earth with speed 'v' in given time 't' of \[24\text{ }hours\] . 

\[\Rightarrow 42250\text{ }km\] 

Radius of circular orbit r 

$\Rightarrow 4225\times 1000m$ Time taken by artificial satellite \[t=\text{ }24\text{ }hours\] 

$\Rightarrow 24\times 60\times 60s$ Distance covered by satellite s=circumference of circular orbit 

\[\Rightarrow 2\pi \text{ }r\] 

∴ Speed of satellite\[v=\dfrac{\left( 2\pi \text{ }r \right)}{t}\] 

$\Rightarrow \dfrac{\left[ 2\times \left( \dfrac{22}{7} \right)\times 42250\times 1000 \right]}{\left( 24\times 60\times 60 \right)}$ 

$\Rightarrow \dfrac{\left( 2\times 22\times 42250\times 1000 \right)}{\left( 7\times 24\times 60\times 60 \right)m{{s}^{-1}}}$ 

$\Rightarrow 3073.74m{{s}^{-1}}$ 

$\therefore Speed=3.073km/s$

11. The position of a body at different times are recorded in the table given below: 

Draw the displacement time graph for the above data? 

What is the slope of the graph? 

 \[Slope\text{ }of\text{ }the\text{ }graph=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\] 

\[\Rightarrow \dfrac{\left( 36-24 \right)m}{\left( 6-4 \right)\sec }=\dfrac{12m}{2\sec }\]

\[\therefore Slope\text{ }of\text{ }the\text{ }graph=6m/sec\] 

What is the speed of the motion? 

Ans:  Slope of the graph of a displacement-time graph = speed 

Hence speed \[=6m/sec\].

Important Questions of Chapter Motion Class 9 – PDF Download

Science being one of the broader and more stringent subjects requires much practice and preparation. Furthermore, Chapter motion also plays a key role in further academics as its concepts are involved in all the higher classes for science stream students. The students are thus advised to refer to the Class 9th Science Chapter 8 Important questions. 

The PDF version of Class 9th Science Ch 8 Important questions is available easily on Vedantu website. The students can access them for free online, and can also download it on any device for future reference. Important questions for Class 9 Science Chapter Motion helps the students get the best practice and preparation for the examinations, and pass the exams with flying colours. 

Vedantu has a team of experts with immense Science qualifications who design the PDF of Ch 8 Motion Class 9 important questions, keeping the past statistics and records in mind. The solutions of the problems are also available along with them, to provide the best assistance to the students. During the examination times, and when the students need the perfect resources for practice during revision, they must refer to these essential questions and get well-versed in the type of questions asked in the examinations, and the essential concepts related to them.

Chapter 8 Class 9 Science Important Questions – Related Essential Concepts

Important questions for Class 9 Science Chapter 8 are specifically designed using Chapter Motion's essential topics and sub-topics. Here is a brief on all the vital concepts and related topics involved in the chapter.

The motion refers to the movement of anybody from one position to another concerning the object's observer. When the object's motion goes in a straight line, it is known as a rectilinear motion or motion in a straight line. For example, when a car moves on a straight highway for some distance, it is known as a rectilinear motion. 

Motion is of Two Types

Uniform Motion – Uniform motion refers to the one in which the object covers equal distance in an equal time interval.

Non-Uniform Motion – Non-uniform motion refers to the one in which the object covers the unequal distance in an equal time interval.

Scalar and Vector Quantities

A scalar quantity is the one whose direction is not associated with it. For example, time, mass, etc. On the other hand, a vector quantity is the one that has both direction and magnitude associated with it—for example, position, force, etc.

Distance and Displacement

Distance is the length of the path that any moving object covers in the given period, and it is independent of its direction. Distance is a scalar quantity, and its SI unit is metre (m).

In contrast, Displacement is the shortest distance between the initial and final position of the object. The unit of Displacement is same as that of distance, but it is a vector quantity.

Speed and Velocity

Speed refers to the distance per unit time for any object. Speed is a scalar quantity, and its SI unit is metre/ second. For any given non-uniform motion under consideration, the average speed for any object is the total distance covered in the total time taken.

On the other hand, Velocity is a vector quantity, and it refers to the speed of any object travelling in a particular direction. The Velocity of any object is the Displacement per unit time. Average Velocity is the arithmetic mean of the final and initial Velocity for a certain period.

Acceleration is the rate of change of Velocity. Acceleration for an object is taken as the difference between the final Velocity and initial Velocity per unit time. SI unit of acceleration is metre / second 2 . Acceleration is a vector quantity, and it is positive when the acceleration is in the direction of Velocity, else vice versa. Retardation or deceleration is the term used to describe the negative acceleration.

Uniform and Non-Uniform Acceleration

When there is a change of Velocity for an object by an equal amount for an equal time period, the acceleration is uniform. For example, the motion of a freely falling object is in uniform acceleration.

On the other hand, non-uniform acceleration is when the object's velocity changes unequally in equal time intervals. For example, the motion of a vehicle on the road is in non-uniform acceleration.

More Topics Related to Important Questions for Class 9 Science Chapter 8

Class 9 Science Chapter 8 extra questions also carry some weightage from the other topics like the derivation of different motion laws and graphical representation of motion. And the students going through the entire set of questions get the best practice and score well in the examinations.

Motion Class 9th Important Questions

The expert team from Vedantu thoroughly researched the previous year question papers and statistics. It came up with the set of Class 9 Science Chapter 8 important questions to enhance the students' practice and preparation for school and competitive examinations. Here are the ten important questions of Chapter Motion Class 9:

Derive the three equations of motion using graphical representation and arithmetic expressions.

The Displacement of an object in the given time interval is 0, would the distance travelled also be zero? Justify the answer with examples.

Explain how there will be a change in the motion equations for an object moving with a uniform velocity.

A car starts from rest and moves towards the x-axis with a constant acceleration of 5 m/s for 8 seconds. If the car continues with a constant velocity, calculate the distance it will travel in 12 seconds.

A person moves from a given point A to another point B with a uniform speed of 30 km/h, and it then returns with the speed 20 km/h. Find the average speed of the person.

Draw the velocity-time graph of a piece of item thrown upwards and then it's way back downwards after reaching the maximum height.

Calculate the relation of distance that an object travelled while moving with uniform acceleration in the given time interval between the 4th and the 5th second.

A ball drops from rest at height 150 m and at the same time other drops from rest at height 100m. Calculate the difference in their heights after the end of 2 s if both of them drop with the same accelerations. How does the difference in heights vary with the given period?

A car started from rest travels 30 m in the first 2 seconds and then 170 m in the next 4 seconds. Calculate the Velocity after 7 seconds from the starting.

An electron moves with a velocity of 5 * 104 m/s, and it enters a uniform electric field and acquires uniform acceleration of 104 m/s 2 , in initial motion's direction. For this situation:

How much distance will be covered by the electron in that time?

Calculate the time by when the electron will acquire the velocity double of its initial Velocity.

Benefits of Important Questions for Class 9 Science Chapter 8

Motion Class 9 important questions help the students in their preparations, and they are very beneficial for the students, because:

Having the Class 9 Science Chapter 8 important questions handy helps the students get the instant practice material during the revision times before the examinations.

Precisely designed Class 9 Science Motion Important questions carry all the questions that either occurred in the previous year papers or have similar problem statements. Thus they help the students get an idea about the upcoming papers.

The students practising through the important questions, get a confidence boost for the examinations and prepare for solving all types of problems in the examinations.

Crucial Class 9 Science Chapter Motion questions are simply available on Vedantu websites and are vital and valuable for students' preparations. After extensive investigation, an expertly created set of critical questions provides students with a compact study resource that serves as the ideal revision material during test timings. It is required for both school and competitive examinations and provides with an understanding of the sorts of all-important and commonly asked questions.

Important Related Links for CBSE Class 9 

FAQs on Important Questions for CBSE Class 9 Science Chapter 8 - Motion

1. What are some important questions of Chapter 8 of Class 9 Physics?

All of the relevant questions for Chapter 8 of Class 9 Physics may be found on the Vedantu website and mobile app. Pupils may also obtain free PDFs of crucial problems for all chapters in Science for Class 9. This chapter contains a variety of significant test subjects, so students will benefit from reading and answering all of the important questions when studying for the exam and revising.

Some of the important questions are;

What is velocity?

Define Speed 

What is odometer 

What is uniform acceleration?

2. If an object has travelled a distance can its displacement be zero?

Displacement is determined by a change in the position of a body from point A to point B. In the case where point B is the same as Point A, the displacement would be zero. For example, if a boy moves from his home to go to the market and comes back home after covering a distance of 2 km, the displacement would be zero. This is because the initial starting point and the point after the movement are the same.

3. What is motion (long answer)?

Motion is defined as a change in a body's or object's location with relation to its surroundings and time. The two forms of motion are uniform and non-uniform motion. This category includes speed, velocity, distance, displacement, and acceleration. The revision notes and crucial questions for this chapter are available in offline PDF format from the Vedantu website and mobile app. The answers to the main questions are selected in a way that follows the intended reply strategy.

4. How do you solve numericals in Chapter 8 of Class 9 Science?

To solve the numerical problems in this chapter, you must first memorise all of the formulae and subjects presented in the chapter. A thorough knowledge of the ideas and formula derivations leads to a better understanding of the numerical difficulties. Students can use the PDF for Key Questions from Vedantu to obtain some assistance on how to apply each formula to different sorts of numericals. The PDF includes one of each type of question, ranked by significance.

5. How does solving the important questions help for Chapter 8 of Class 9 Physics?

The best strategy to revise the chapter is to answer the essential questions. Given the lengthy course, it is crucial to ensure that no important topic is overlooked for preparation. These questions provide you the opportunity to go over all of the chapter's themes in a short amount of time. It saves you time and energy because instead of going through all of the questions, you simply look at the ones that are important.

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Class 9th Science - Force and Laws of Motion Case Study Questions and Answers 2022 - 2023

Class 9th Science - Force and Laws of Motion Case Study Questions and Answers 2022 - 2023 Study Materials Sep-09 , 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 9th Science Subject - Force and Laws of Motion, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Force and laws of motion case study questions with answer key.

Final Semester - June 2015

(iii) If the above coin is replaced by a heavy five rupee coin, what will be your observation. Give reason. (a) Heavy coin will possess more inertia so it will not fall in tumbler. (b) Heavy coin will possess less inertia so it will fall in tumbler. (c) Heavy coin will possess more inertia so it will fall in tumbler. (d) Heavy coin will possess less inertia so it will not fall in tumbler. (iv) Name the law which provides the definition of force.

(v) State Newton's first law of motion. (a) Energy can neither be created nor be destroyed, it can be converted from one form to another, total amount of energy always remains constant. (b) A body at rest remains at rest or, if in motion, remains in motion at constant velocity unless it is acted upon by an external unbalanced force. (c) For every action in nature there is an equal and opposite reaction. (d) The acceleration in an object is directly related to the net force and inversely related to its mass.

(ii) What is the total momentum of car A and car B before collision?

(iii) What is the momentum of the car A after collision?

(iv) What is the total momentum of car A and car B after collision?

(v) What is the velocity of car B after the collision?

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Class 9 Science Case Study Questions Chapter 9 Force and Laws of Motion

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Case study Questions in Class 9 Science Chapter 9 are very important to solve for your exam. Class 9 Science Chapter 9 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving  Class 9 Science Case Study Questions  Chapter 9 Force and Laws of Motion

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In CBSE Class 9 Science Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Force and Laws of Motion Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 9 Science  Chapter 9 Force and Laws of Motion

Case Study/Passage-Based Questions

Case Study 1: The sum of the momentum of the two objects before the collision is equal to the sum of momentum after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum. This statement can alternatively be given as the total momentum of the two objects is unchanged or conserved by the collision. The Law of conservation of momentum is applicable to the system of particles. Answer the following questions.

(i)Law of conservation of momentum is applicable to

(a) A system of particles

(b) Only for 2 particles

(c) Only for 1 particle

(d) None of the above

Answer: (a) A system of particles

(ii) Law of conservation of momentum holds good provided that

(a) There should be external unbalanced force acting on particles

(b) There should not be any external unbalanced force acting on particles

(c) No internal forces acting on particles

Answer: (b) There should not be any external unbalanced force acting on particles

(iii)The total momentum of the two objects when collision occurs is

(a) Changed

(b) Remains conserved

(c) Become zero

Answer: (b) Remains conserved

(iv) State law of conservation of momentum.

Answer: The sum of momentum of the two objects before collision is equal to the sum of momentum after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum. This statement can alternatively be given as the total momentum of the two objects is unchanged or conserved by the collision.

(v) If action and Reaction are equal in magnitude and opposite in direction then why they do not cancel each other?

Answer: Action and Reaction are equal in magnitude and opposite in direction but they do not cancel each other because they are not action on sane object. As these forces are acting on different object hence produces different acceleration and does not cancel each other.

Case Study 2: The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and never on the same object. It is important to note that even though the action and reaction forces are always equal in magnitude; these forces may not produce accelerations of equal magnitudes, this is because each force acts on a different object that may have a different mass. The two opposing forces are also known as action and reaction forces. Answer the following questions.

(i) Action reaction forces are always

(a) Equal and in the same direction

(b) Equal and in the opposite direction

(c) Unequal and in the same direction

Answer: (b) Equal and in the opposite direction

(ii) Which of the following is correct about action reaction forces?

(a) They act on different objects

(b) They are equal in magnitude and opposite in direction

(c) Both forces acted on different object simultaneously

(d) All the above

Answer: (d) All the above

(iii) State third law of motion

Answer: The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and neveron the same object.

(iv) Give 5 examples of third law of motion

Answer: Examples of third law of motion are Swimming or rowing a boat. •Static friction while pushing an object. •Walking. •Standing on the ground or sitting on a chair. •The upward thrust of a rocket. •Resting against a wall or tree.

Case Study 3:

Force is a push or pull that can change the state of motion of an object. According to Newton’s first law of motion, an object at rest will remain at rest, and an object in motion will continue moving with a constant velocity in a straight line unless acted upon by an external force. This is known as the law of inertia. Newton’s second law of motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, this can be expressed as F = ma, where F is the force, m is the mass of the object, and a is the acceleration produced. Newton’s third law of motion states that for every action, there is an equal and opposite reaction. This means that whenever an object exerts a force on another object, the second object exerts an equal and opposite force on the first object. Understanding the concepts of force and the laws of motion helps us explain the behavior of objects and the factors that influence their motion.

What is force? a) A change in the state of motion of an object b) A push or pull that can change the state of motion of an object c) The mass of an object d) The velocity of an object Answer: b) A push or pull that can change the state of motion of an object

What does Newton’s first law of motion state? a) An object at rest will remain at rest, and an object in motion will continue moving with a constant velocity in a straight line unless acted upon by an external force. b) The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. c) For every action, there is an equal and opposite reaction. d) The force exerted by an object is equal to its mass multiplied by its acceleration. Answer: a) An object at rest will remain at rest, and an object in motion will continue moving with a constant velocity in a straight line unless acted upon by an external force.

What is Newton’s second law of motion? a) An object at rest will remain at rest, and an object in motion will continue moving with a constant velocity in a straight line unless acted upon by an external force. b) The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. c) For every action, there is an equal and opposite reaction. d) The force exerted by an object is equal to its mass multiplied by its acceleration. Answer: b) The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

What is Newton’s third law of motion? a) An object at rest will remain at rest, and an object in motion will continue moving with a constant velocity in a straight line unless acted upon by an external force. b) The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. c) For every action, there is an equal and opposite reaction. d) The force exerted by an object is equal to its mass multiplied by its acceleration. Answer: c) For every action, there is an equal and opposite reaction.

How do the concepts of force and the laws of motion help us? a) Explain the behavior of objects and the factors that influence their motion. b) Calculate the speed of objects. c) Classify objects into different categories. d) Determine the position of objects. Answer: a) Explain the behavior of objects and the factors that influence their motion.

Hope the information shed above regarding Case Study and Passage Based Questions for Class 9 Science Chapter 9 Force and Laws of Motion with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 9 Science Force and Laws of Motion Case Study and passage-based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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