NCERT Solutions Class 9 Science Chapter 8
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NCERT solutions Class 9 Science Chapter 8: Motion
Science is an exciting subject; it teaches many things about the environment. It also opens doors of opportunity for the students who wish to make a career in science. However, to understand the concepts in this subject, students require a deep understanding to grasp the concepts. Besides, class 9 is also a crucial stage which prepares them for the upcoming board exams.
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ToggleClass 9 Science chapter 8 is about “Motion”, which introduces the simplest type of motion along a straight line. The students will learn how to describe the motion of an object moving along a straight path. Furthermore, students will get an overview of velocity, speed, distance, and time. Finally, they will learn how to calculate the distance and time based on the movement of an object with its uniform speed.
NCERT solutions Class 9 Science Chapter 8 is available on our Extramarks website. It covers the entire syllabus and helps in revising the chapter properly. It is helpful for students to grasp basic concepts better and faster. Class 9 Science chapter 8 question answers will make students able to solve the problematic questions given in the exercise. Students can analyse the problems and answer them with precision and the right concepts.
Extramarks is one of the trusted online learning platforms in India. Lakhs of students refer to our NCERT solutions and study materials. Students who face difficulty answering the questions can refer to NCERT solutions. The solutions are essential in steering students toward their goals.
Students can regularly visit our Extramarks website for the latest updates. Further, students can also refer to NCERT solutions Class 10, NCERT solutions Class 11 and NCERT solutions Class 12.
Key Topics Covered in NCERT solutions Class 9 Science Chapter 8:
NCERT solutions Class 9 Science Chapter 8 introduces many new concepts to the students. For example, this chapter talks about the motion along a straight line and measuring the rate of motion. With unique concepts and theories, students will get to learn in the Class 9 Science Chapter 8.
Some of the key topics featured in NCERT solutions Class 9 Science Chapter 8 are:
Describing motion
We describe the location of an object by specifying a reference point. Thus, to describe the position of an object, we need to specify a reference point called the origin. For example, a passenger inside a bus sees the other passengers at rest, whereas an observer outside the bus sees the passengers in motion. Thus, a standard reference point frame is needed to make observations easy.
Motion along a straight line explained in NCERT solutions Class 9 Science Chapter 8:
The simplest type of motion is the motion along a straight line. Considering an object moving from one point to another, we will notice that the magnitude of the displacement for the course of motion is zero. However, the corresponding distance covered is not zero. Therefore, two different physical quantities the distance and the displacement are used to describe the overall motion of an object.
Uniform and Nonuniform Motion: A body is in uniform motion when it travels the equal distances in equal intervals. Uniform motion happens when a body travels unidirectional distances with equal intervals or equally long distances at different intervals.
Measuring the rate of motion
The distance travelled by the object over the period during which motion takes place is defined as the speed. The measurement unit for speed is the metres per second. So,
Speed=Distance travelled/time taken=s/t
Here’s the distance travelled by the object, and t is the amount of time required for the object’s journey to cover that distance. A body’s speed measures how slowly or faster it moves. The proportion of total distance measures the speed of a human body travelling for the duration of the body. A body’s instantaneous speed refers to its speed at any given moment if a body can travel the same distance at equal time intervals then it is considered to be having a constant or uniform speed.
To understand the relation between distance and time, students can refer to NCERT solutions Class 9 Science Chapter 8.
Rate of change in velocity
Velocity refers to the speed of an object moving in a particular direction. The velocity of an object can be either constant or fluctuating. It is influenced by altering the object’s movement’s direction, speed, or both aspects. Therefore, it is considered a vector value and the speed of a vehicle is a scalar value with only its magnitude without a specific direction.
If a body covers the same distance over the same amount of time and in the same direction, it is considered moving with uniform speed. Likewise, if a person covers different distances at equal intervals, and vice versa in a particular direction or changes direction, it’s classified as moving at a uniform speed.
Two methods can alter a body’s speed, one by altering its speed and the second is by changing the direction of movement while keeping the speed constant. To modify the speed of your body’s motion, velocity and the direction of the body may be altered. If an object’s velocity changes at a constant speed, the arithmetic average of the final and initial velocity over a specific time is used to calculate the average velocity. This is
Average velocity=(u+v)/2
In this equation, you are the initial speed of the body. V is the final velocity that the body will achieve. To understand the fundamental factors of average velocity, students can refer to our Extramarks NCERT solutions Class 9 Science Chapter 8.
Graphical Representation of Motion
Graphs are a great way to provide the essential details about various events. For example, to explain the movement or motion of an object, it is possible to employ line graphs. In this scenario, line graphs demonstrate the dependence of one physical measurement like velocity or distance on another, such as time.
Graph of Distance and Time
The shift in the position of an object as time passes can be represented by a distancetime graph using a suitable scale. The graph shows time calculated on the xaxis, while distance is calculated on the yaxis.
Graph velocitytime
The change in velocity as the duration of an object on a straight path may be represented in an inverse graph of velocity and time. This graph represents time by the vertical axis while velocity is shown on the yaxis. If the object is moving at a consistent speed, the elevation of its graph of velocity and time does not change as time passes.
Students can visit our Extramarks website to refer to NCERT solutions Class 9 Science Chapter 8. We have elaborated the principles of velocity and time with illustrative examples.
Equations of Motion by Graphical Method
Suppose an object moves along a straight line and has an equal acceleration. In that case, it’s possible to determine its speed, acceleration, and velocity during movement and the distance travelled by it over a specific period by a series of equations referred to as the formulas for motion. For ease of use, there is a set of three such equations is provided in NCERT solutions Class 9 Science Chapter 8 below:
 v = u + at
 s = ut + ½ at2
 2 a s = v2 – u2
Uniform Circular Motion
An object’s movement can be known as uniform circular motion when it travels in a circular direction at a constant rate. Numerous cases of circular motion can be seen in our everyday lives, including cars that drive on a circular route and innumerable others. Furthermore, the earth and various planets orbit the sun in circles.
Although the speed of particles moving in a circular motion is constant, it accelerates due to the constant change in speed. Therefore, for circular motions, we utilise an angular velocity instead as we do with linear motion. The force needed to propel a body through a circular path is known as a centripetal force.
Various types of motions are based on the motion’s path, which are elaborated in NCERT solutions class 9. Science chapter 8 includes:
 Translational Motion
 Rotational Motion
 Linear Motion
 Rectilinear Motion
 Curvilinear Motion
 Periodic Motion
 Simple Harmonic Motion
 Projectile Motion
 Oscillatory Motion
NCERT solutions Class 9 Science Chapter 8: Exercise & Answer Solutions
NCERT solutions Class 9 Science Chapter 8 offers detailed explanations which help students understand the concepts better. They will be able to develop higherorder thinking capabilities, which also helps prioritise their understanding. The solution guide will improve their awareness skills sharply, and they can quickly memorise more things precisely.
Class 9 NCERT solutions are available in understanding language and can make things a little easier for you. It helps students cope with the pressure of the extensive board exam syllabus. As a result, they can study in an organised manner and outperform their classmates. Students can start with a trial account by registering on our website.
Class 9 chapter 8 offers questions such as fill in the blanks, match the pair, and true or false. At the end of the solution, students will witness descriptive and objectivetype questions. Students can click on the links below to the exercisespecific questions and their answers:
 Chapter 8: Exercise 8.1 Solutions: 31 Questions
Along with this, students can also explore NCERT solutions for other classes on our website:
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NCERT Exemplar for Class 9 Science:
Science can be a difficult subject because it is a complex one with many ideas and concepts. Therefore, finding different problems with more difficulty levels can be beneficial. In addition, it improves students’ understanding of the most important topics and problemsolving ability.
NCERT’s Exemplar for Class 9 Science provides the most critical questions from the science topic of the 9th grade. The Exemplar was prepared by Extramarks subject matter experts who have carefully arranged the questions. Using an NCERT sample can help students score better if they are preparing for the test. It contains a variety of questions, such as multiplechoice and descriptive questions.
Our NCERT solutions for cClass 9 Science Chapter 8 have problems from the NCERT example. Students are provided with questions in a proper arrangement and organised manner. Further, each topic has a mix of questions and solutions with illustrations to help understanding.
Key Features of NCERT solutions Class 9 Science Chapter 8:
NCERT solutions play an essential part in preparation for entrance exams and board exams. Extramarks has a team of science specialists that offer appropriate solutions for each type of question. They will help students comprehend the topics covered in chapter 8 of class 9.
Some of the essential aspects of NCERT solutions Class 9 Science Chapter 8:
 It assists students in clearing all doubts regarding the various types of motion.
 NCERT solutions Class 9 Science Chapter 8 consists of past years’ questions and the current year’s answers.
 This NCERT solution is an important study material that provides answers to the questions by the text.
 The solutions provide a clear understanding of the chapter’s ideas so that students can comprehend these concepts before appearing on the test.
 NCERT solutions Class 9 Science Chapter 8 helps students understand the subject matter and build a solid foundation for conceptual understanding.
Q.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?
Ans
\begin{array}{l}\text{Given, diameter of circular track, d = 200 m}\\ \text{time required to cover the distance of circular}\\ \text{track = 40 s}\\ \text{Now, radius of the track, r =}\frac{d}{2}\text{=}\frac{\mathrm{200\; m}}{2}\text{= 100 m}\\ \text{Circumference = 2}\pi \text{r = 2}\pi \text{}(\mathrm{100\; m})=200\pi \text{m}\\ \because \text{distance covered by athlete in 40 s =}200\pi \text{m}\\ \therefore \text{in 1 s, the distance covered by the athlete}\\ \text{=}\frac{200\pi \text{m}}{40}\text{=5}\pi \text{m}\\ \text{So, the distance covered in 2 minutes and 20 seconds,}\\ \text{i}\text{.e}\text{., 140 s is given by = (140}\times \text{5}\pi )\text{m =700}\pi \text{m}\\ \text{= 700}\times \frac{22}{7}\text{m =2200 m}\end{array} \begin{array}{l}\text{The athlete completes one round in 4}0\text{s}\\ \text{So, the time taken in 3 complete rounds = 3}\mathrm{}\times \text{40 s}=\text{12}0\text{s}\\ \text{The displacement after 3 complete rounds would be zero}\text{.}\\ Now,\text{the time left = (140 s 120 s) = 20 s}\\ In\text{20 s},\text{he moves at the opposite end of the}\\ \text{initial position hence, the displacement will be equal to the}\\ \text{diameter of the circular track i}\text{.e}\text{., 200 m}\text{.}\end{array}Q.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?
Ans
(a)
\begin{array}{l}\text{Given, distance covered from A to B = 300 m}\\ \text{time taken = 2 min 30 seconds = 150 s}\\ \text{Average speed =}\frac{\text{Total distance covered}}{\text{Total time taken}}\\ \text{=}\frac{\text{300 m}}{\text{150 s}}\text{= 2}{\text{ms}}^{\text{1}}\text{}\\ \text{displacement between A to B = 300 m}\\ \text{time taken = 2 min 30 seconds = 150 s}\\ \text{Average velocity =}\frac{\text{Net displacement}}{\text{Total time taken}}\text{}\\ \text{=}\frac{\text{300 m}}{\text{150 s}}{\text{= 2 ms}}^{\text{1}}\text{}\end{array}
(b)
\begin{array}{l}\text{Total distance covered from A to C = AB +BC = 300 m + 100 m = 400}\\ \text{Total time taken = time taken from A to B + time taken from B to C =}\end{array} \begin{array}{l}\text{150 s + 60 s = 210 s}\\ \text{Average speed =}\frac{\text{Total distance covered}}{\text{Total time taken}}\\ \text{=}\frac{\text{400 m}}{\text{210 s}}\text{= 1}{\text{.90 ms}}^{\text{1}}\text{}\\ \text{Net displacement = AB \u2013 BC = 300 m \u2013 100 m = 200 m}\\ \text{time taken = 210 s}\\ \text{Average speed =}\frac{\text{Total distance covered}}{\text{Total time taken}}\text{}\\ \text{=}\frac{\text{200 m}}{\text{210 s}}{\text{= 0.952 ms}}^{\text{1}}\text{}\end{array}Q.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^{−1}. What is the average speed for Abdul’s trip?
Ans
\begin{array}{l}\text{While going to school:}\\ {\text{Given, Average speed = 20 kmh}}^{\text{1}}\\ \text{Let, the distance travelled to reach school = d}\\ {\text{time taken = t}}_{\text{1}}\\ \text{Average speed =}\frac{\text{total distance}}{\text{total time taken}}\\ \therefore \text{20 =}\frac{\text{d}}{{\text{t}}_{\text{1}}}\text{}\\ {\text{or, t}}_{\text{1}}\text{=}\frac{\text{d}}{\text{20}}\\ \text{While returning from school:}\\ {\text{Given, average speed = 40 kmh}}^{\text{1}}\text{}\\ \text{Let, the distance travelled while returning from school = d}\\ {\text{time taken = t}}_{\text{2}}\\ \therefore \text{40 =}\frac{\text{d}}{{\text{t}}_{\text{2}}}\\ {\text{or, t}}_{\text{2}}\text{=}\frac{\text{d}}{\text{40}}\\ \\ \text{Now, Average speed =}\frac{\text{Total distance travelled}}{\text{total time taken}}\\ \text{=}\frac{\text{2d}}{{\text{t}}_{\text{1}}{\text{+t}}_{\text{2}}}\\ \text{=}\frac{\text{2d}}{\frac{\text{d}}{\text{20}}\text{+}\frac{\text{d}}{\text{40}}}\text{=}\frac{\text{80}}{\text{3}}\text{=26}{\text{.67 ms}}^{\text{1}}\end{array}Q.4 A motorboat 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?
Ans
\begin{array}{l}{\text{Given, u= 0, a= 3 ms}}^{\text{\u2013}}{}^{\text{2}}\text{, t = 8 s}\\ \\ \text{On using second equation of motion,}\\ \text{s = ut +}\frac{\text{1}}{\text{2}}{\text{at}}^{\text{2}}\\ On\text{putting given values,}\\ \text{s = 0 +}\frac{\text{1}}{\text{2}}{\text{\xd73 ms}}^{2}\text{\xd7}{\left(\text{8 s}\right)}^{\text{2}}\text{= 96 m}\\ \text{Hence, the boat travels a distance of 96 m}\text{.}\end{array}Q.5 A driver of a car travelling at 52 kmh^{−1} applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 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 the two cars. Which of the two cars travelled farther after the brakes were applied?
Ans
\begin{array}{l}{1}^{st}\text{Case:}\\ {\text{Given, u = 52 kmh}}^{\text{1}}\text{= 14}{\text{.4 ms}}^{\text{1}}\\ Time\text{taken to stop the car, t = 5 s}\\ \text{v = 0}\\ Dis\mathrm{tan}ce{\text{covered, s}}_{\text{1}}\text{= area OAC}\\ \text{=}\frac{1}{2}\times \text{14}{\text{.4 ms}}^{1}\times \text{5 s = 36 m}\\ {2}^{nd}\text{Case:}\\ {\text{Given, u = 3 kmh}}^{\text{1}}\text{= 0}{\text{.833 ms}}^{\text{1}}\\ Time\text{taken to stop the car, t = 10 s}\\ \text{v = 0}\\ Dis\mathrm{tan}ce{\text{covered, s}}_{\text{2}}\text{= area ODB}\\ \text{=}\frac{1}{2}\times 0.833\text{\hspace{0.17em}}m{s}^{1}\times 10\text{\hspace{0.17em}}s\text{= 4}\text{.15 m}\\ Therefore,\text{the distance covered in first case is greater}\\ \text{than the distance covered in second case}\text{.}\end{array}
Q.6 The given figure shows the distancetime 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?
Ans
(a)
$\begin{array}{l}\text{speed =}\frac{\text{distance}}{\text{time}}\\ \mathrm{slope}\text{of graph =}\frac{y\text{axis}}{x\text{axis}}=\frac{\text{distance}}{\text{time}}\\ \therefore \text{Speed = Slope of the graph}\\ \text{Slop of B is greater than A and C}\text{. Hence,}\\ \text{it is travelling the fastest}\text{.}\end{array}$
(b) As, A, B and C will never meet at a single point at same time therefore, they will never be at the same point on the road.
(c)
\begin{array}{l}\text{From graph, 7 small boxes = 4 km}\\ \therefore \text{1 box =}\frac{4}{7}\text{km}\\ \text{Initially, C is 4 blocks away from the}\\ \text{origin}\text{.}\\ \therefore \text{Intial distance of object C from origin =}\frac{16}{7}\text{km}\end{array}
\begin{array}{l}\text{B passes A at point P. Object C is at point Q when B passes A.}\\ \text{Distance of point Q from origin}\\ \text{= 8 km}\\ \text{Distance covered = 8 km \u2013}\frac{\mathrm{16\; km}}{7}\text{=}\frac{\mathrm{56\; km}\mathrm{16\; km}}{7}\text{=}\frac{\mathrm{40\; km}}{7}\\ \text{= 5}\text{.714 km}\\ \text{Therefore, C has travelled a distance of 5}\text{.714 km}\\ \text{when B passes A}\text{.}\end{array}
\begin{array}{l}\text{(d)}\end{array}
\begin{array}{l}\text{B passes C at point R.}\end{array} \begin{array}{l}\text{Distance travelled by B at the time it passes C}\\ \text{= 9 boxes =}\frac{4}{7}\times 9.25=\frac{36}{7}=5.28\text{km}\\ \text{Therefore, B has travelled a distance of 5}\text{.28 km when}\\ \text{B passes A}\text{.}\end{array}
Q.7 A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms^{−2}, with what velocity will it strike the ground? After what time will it strike the ground?
Ans
\begin{array}{l}Given,{\text{s = 20 m, a = 10 ms}}^{2},\text{u = 0}\\ \text{v = ?}\\ \text{On using third equation of motion,}\\ {\text{v}}^{2}={\text{u}}^{2}+2as\\ \text{= 0 + 2}\times \text{10}\times 20=400\\ or\text{v =}\sqrt{400}=20{\text{ms}}^{1}\\ Now,\text{using first equation of motion}\\ \text{v = u + at}\\ \therefore \text{20 = 0 + 10}\times \text{t}\end{array}
⇒ t = 2 s.
Q.8 The speedtime graph for a car is shown here.
(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?
Ans
(a)
Distance travelled by the car the first 4s = shaded area = 12 × 4s × 6ms^{1} = 12cm
(b)
The part of the graph from 6 s to 10 s represents uniform motion of the car.
Q.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.
Ans
(a) It is possible.
Example: An apple has zero velocity at maximum height when it is thrown upward. Although, it will have constant acceleration due to gravity.
(b) It is possible.
Example: When a truck is moving in a circular track, it is moving with an acceleration but with uniform speed.
(c) It is possible.
Example: When a truck is moving in a circular track, its acceleration is perpendicular to its direction of motion.
Q.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.
Ans
\begin{array}{l}\text{Given, radius of circular path, r = 42250 km = 42250 km \xd7}\frac{1000\text{m}}{1\text{km}}\\ \text{= 4}{\text{.225\xd710}}^{\text{7}}\text{m}\\ \text{time period, T = 24 hours = 24 hr \xd7}\frac{\text{60 min}}{\text{1 hr}}\text{\xd7}\frac{\text{60 s}}{\text{1 min}}\text{= 86400 s}\\ \text{orbital velocity of satellite, v = ?}\\ \text{On using the relation,}\\ \text{v =}\frac{\text{2\pi r}}{\text{T}}\\ \text{On putting the given values,}\\ \text{=}\frac{\text{2\xd73}\text{.14\xd7}\left(\text{4}{\text{.225\xd710}}^{\text{7}}\text{m}\right)\text{}}{\text{86400 s}}\text{= 3070}{\text{.94 ms}}^{\text{1}}\text{}\end{array}
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FAQs (Frequently Asked Questions)
1. Which are the essential topics covered in NCERT solutions Class 9 Science Chapter 8?
The following are the essential topics covered in the NCERT solutions:
 Describing Motion
 Measuring the rate of motion
 Rate of change in velocity
 Graphical Representation of Motion
 Equations of Motion by Graphical Method
 Uniform Circular Motion
2. What is the importance of NCERT solutions?
The concepts covered in chapter 8 of class 9 are complicated to grasp. For better performance in the exam, students should look up NCERT solutions in Class 9 Science Chapter 8. The solutions cover all the questions, including fill in the blanks, matching the pairs, and true or false. It consists of a graphic presentation and a summary of the answers.
3. What are the different types of motion?
There are various types of motions based on the path of the motion, which include:
 Translational Motion
 Rotational Motion
 Linear Motion
 Rectilinear Motion
 Curvilinear Motion
 Periodic Motion
 Simple Harmonic Motion
 Projectile Motion
 Oscillatory Motion