NCERT Solutions for Class 9 Science Chapter 8 Motion

# NCERT Solutions for Class 9 Science Chapter 8 Motion

Edited By Vishal kumar | Updated on Sep 07, 2023 04:11 PM IST

## NCERT Solutions Class 9 Science Chapter 8 – Free PDF Download

NCERT Solutions for Class 9 Science Chapter 8 Motion: Welcome to the updated NCERT solutions for Class 10. On this Careers360 page, you'll find comprehensive NCERT solutions for class 9 science chapter 8 exercise questions crafted by subject experts, presented in detail and easy-to-understand language. Additionally, students can download the Class 9 PDF for free. Our expert team has developed class 9 science chapter 8 solutions that simplify complex concepts. This chapter includes ten exercise questions, along with important formulas and diagrams to aid students in exam preparation, as well as completing assignments and homework effectively."

Let's now read the comprehensive article on class 9 chapter 8 science solution, enabling us to solve the provided questions and gain insights into the essential formulas and key points covered in this chapter.

### NCERT Solutions for Class 9 Science: Important Formulas and Diagrams + eBook link

Understanding formulas is crucial for solving motion class 9 numericals and grasping the concepts of the chapter. Here are some important formulas from the Motion chapter that will aid you in solving questions:

• Speed(s) = Distance (d) / Time (t) [SI unit of speed= m/s]

• Average Speed = Total distance travelled / Total time taken [SI unit of average speed= m/s]

• Velocity(v) = Displacement/Time=Δx/Δt [SI unit of velocity= m/s]

• Average velocity = Total displacement / total time [SI unit of average velocity= m/s]

• Acceleration (a) = Change in Velocity (Δv) / Change in time (Δt) [SI unit of acceleration= m/s2]

#### Equation Of Motion

• v= u+at

• S = ut + ½ at2

• V2 = u2+2as

By comprehending the formulas mentioned above, students will be equipped to solve questions within the motion chapter class 9 and successfully complete assignments and tests. Formulas serve as essential tools for tackling numerical problems. Furthermore, Careers360 experts have compiled a comprehensive list of important chapter-wise formulas for NCERT Solutions for Class 9 Science, which students can access by clicking the link provided below:

This resource will further aid students in their understanding and application of formulas across various chapters, thereby enhancing their overall preparation and performance.

### NCERT Solutions for Class 9 Science Chapter 8 motion - Important Topics

Below is a list of important topics covered in motion class 9 solutions that will greatly assist you in exam preparation, completing assignments, and tackling tests:

• Distance and displacement: Understanding the difference between distance (the total length of the path travelled) and displacement (the change in position between the starting and ending positions).
• Speed and velocity: Differentiating between velocity (speed in a certain direction) and speed (rate of motion).
• Acceleration: Understanding acceleration as the speed at which a moving object changes speed.
• Position-time graphs: Analyzing graphs that depict the relationship between an object's position and time.
• Velocity-time graphs: Interpreting graphs that represent an object's velocity over time.
• Uniform circular motion: Understanding motion in a circular path at a constant speed.

Additionally, it's crucial to keep in mind the following motion-related key points:

• Motion is relative and depends on the observer's frame of reference.
• An object can have a constant speed even if its velocity (including direction) changes.
• An object can experience changing velocity even if its acceleration is zero, as long as its direction changes.
• The equations of motion can be used to solve problems involving motion by calculating an object's position, velocity, and acceleration at different time points.

By studying these important topics and concepts, students will develop a solid understanding of motion and be well-prepared to solve related problems.

This chapter has been renumbered as Chapter 7 in accordance with the CBSE Syllabus 2023–24.

## NCERT Solutions for Class 9 Science Chapter 8 Motion

Class 9 Science Chapter 8 Motion: Solved In-text Questions-

## Topic 8.1 Describing Motions

Yes, an object has moved through a distance can have zero displacements.

If an object moves and returns to the original position, the displacement will be zero. Consider the movement in a circular path. A man walks from point A in a circular path in a park and comes back to point A.

The distance traveled is equal to the circumference of the circular path, but displacement is zero.

Side of the square field = $10\ m$
$\therefore$ The perimeter of the square = $4\times10 = 40\ m$
According to question,
He completes 1 round in $40\ s$ .
$\therefore$ Speed of the farmer = $\frac{40}{40}= 1\ m/s$
$\therefore$ Distance covered in $2\ min\ 20\ s\ (=140\ s)$ = $140 \times 1 = 140\ m$
Now,
Number of round trips completed traveling = $\frac{140}{40} = 3.5$
We know, in 3 round trips the displacement will be zero.
In $0.5$ round, the farmer will reach diametrically opposite to his initial position.
$\therefore$ Displacement = $AC = \sqrt{(AB^2 + BC^2 )} = \sqrt{(100^2+100^2)} = 10\sqrt{2}\ m$

(a) It cannot be zero.

(b) Its magnitude is greater than the distance traveled by the object.

(a) The first statement is false. Because displacement can be zero when the initial point coincides with the final point.

(b) The second statement is false. The magnitude of displacement can never be greater than the distance travelled by the object. It can be either equal or less.

## Solutions of NCERT for Class 9 Science Chapter 8 Motion

Topic 8.2 Measuring the Rate of Motion

 Speed Velocity Speed is the distance travelled by an object in unit time Velocity is the speed of an object moving in a definite direction. Speed is a scalar quantity Velocity is a vector quantity Speed does not depend on the direction Velocity changes with change in direction Speed can never be negative Velocity can be positive, negative or zero.

When the total distance traveled by the object is equal to the displacement, the magnitude of the average velocity will be equal to the average speed. Average speed is the total distance upon the time taken, whereas average velocity is the total displacement upon time taken.

Odometer is a device that measures the total distance traveled by automobile.

An object is having a uniform motion if it covers equal distance in equal interval of time (which implies speed is constant!). So the path can be straight or curved.

For eg. Consider a circular path. For understanding purposes, divide the circumference of the circle in six equal parts each subtending $60^{\circ}$ at the centre. The object covers each equal part in equal amount of time. Hence, by definition, this object is in uniform motion.

Q 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 , $3 \times 10^{8}ms^{-1}$ .

Given, the signal travels at the speed of light, $v = 3 \times 10^{8}ms^{-1}$ .

Time taken by the signal = $5\ mins = 5\times60 s= 300\ s$

Let the distance of the spaceship from the ground station be $D\ m$

We know, $Speed = \frac{Distance\ travelled }{Time\ taken}$

$\\ \implies v = \frac{D}{t} \\ \\ \implies D = v\times t = 3\times10^8 \times 300$

$\\ \implies D = 9\times10^{10}\ m$

Therefore, the distance of spaceship from the ground station is $9\times10^{10}\ m$

## NCERT Textbook Solutions for Class 9 Science Chapter 8 Motion

Topic 8.3 Rate of Change of Velocity

(i) uniform acceleration?

(ii) nonuniform acceleration?

(i) If the velocity of an object traveling in a straight line increases or decreases by equal amounts in equal intervals of time, then the acceleration of the object is said to be uniform. For example, An apple having a free-fall motion.

(ii) On the other hand, if the velocity of the object increases or decreases by unequal amounts in equal intervals of time, then the acceleration of the object is said to be non-uniform. For example, A car travelling along a straight road increasing its speed by unequal amounts in equal intervals of time.

(We know, $1\ km = 1000\ m ; 1\ hr = 3600\ s$ )

Given, Initial speed of the bus, $u$ = $80\ kmh^{-1} = \frac{80\times1000}{3600} = \frac{200}{9} ms^{-1}$

The final speed of the bus, $v$ = $60\ kmh^{-1} = \frac{60\times1000}{3600} = \frac{100}{6} ms^{-1}$

Time is taken, $t= 5s$

We know, $v = u+at$

$\\ \implies \frac{100}{6} = \frac{200}{9}+a(5) \\ \implies 5a = \frac{900-1200}{6\times9} = \frac{-300}{6\times9} \\ \implies a = -\frac{100}{9} = -1.11\ ms^{-2}$

The negative sign implies retardation.

Therefore, the acceleration of the bus is $-1.11\ ms^{-2}$

Or, the retardation(de-acceleration) of the bus is $1.11 ms^{-2}$

(We know, $1\ km = 1000\ m ; 1\ hr = 3600\ s$ )

Given, The train starts from rest. Hence, the initial speed of the train = $0\ kmh^{-1} = 0\ ms^{-1}$

Final speed of the train = $40\ kmh^{-1} = \frac{40\times1000}{3600} = \frac{100}{9} ms^{-1}$

Time taken, $t = 10\ min = 10\times60\ s = 600\ s$

We know, $v = u+at$

$\\ \implies \frac{100}{9} = 0+a(600) \\ \implies a = \frac{100}{9\times600} = \frac{1}{6\times9} \\ \implies a = \frac{1}{54} = 0.0185\ ms^{-2}$

Therefore, the acceleration of the train is $0.0185\ ms^{-2}$

Motion Class 9 Topic 8.4 Graphical Representation of Motion

Distance-time graph is the plot of distance travelled by an object along x-axis against time along y-axis.

For the uniform motion of an object, the distance-time graph is a straight line with a constant slope.

For non-uniform motion of an object, the distance-time graph is a curved line with an increasing or decreasing slope.

If the distance-time graph of an object is a straight line parallel to the time axis, it means that the distance of the object is the same from its initial position at any point of time. This implies that the object is not moving and is at rest.

If the speed-time graph of an object is a straight line parallel to the time axis, it means that the speed of the object is not changing with time. Hence the speed of the object is constant. This also implies that the acceleration of the object is zero.

The area occupied below the velocity-time graph denotes the total distance travelled by an object in the given time frame.

We know,

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

Topic 8.5 Equations of motion by graphical method

• the speed acquired

Given, The bus starts from rest. Hence, the initial speed of the bus = $0\ ms^{-1}$

Acceleration of the bus, $a$ = $0.1\ ms^{-2}$

Time is taken, $t = 2\ min = 2\times60\ s = 120\ s$

(a) We know, $v = u+at$

$\\ \implies v = 0+(0.1)(120) = 12\ ms^{-1}$

Therefore, the speed acquired by the bus is $12\ ms^{-1}$

• The distance travelled.

Given, The bus starts from rest. Hence, the initial speed of the bus, u = $0\ ms^{-1}$

Acceleration of the bus, $a$ = $0.1\ ms^{-2}$

Time taken, $t = 2\ min = 2\times60\ s = 120\ s$

(b) We know, $s = ut+\frac{1}{2}at^2$

$\\ \implies s = 0(120) + \frac{1}{2}(0.1)(120)^2 \\ \implies s = \frac{1}{2}(0.1)(14400) \\ \\ \implies s = 720\ m$

Therefore, the distance travelled by bus is $720\ m$

(We know, $1\ km = 1000\ m ; 1\ hr = 3600\ s$ )

Given, Initial speed of the train, $u$ = $90\ kmh^{-1} = \frac{90\times1000}{3600} = 25\ ms^{-1}$

Acceleration of the train, $a = -0.5\ ms^{-2}$ (Negative sign implies retardation)

Since, the train has to be brought to rest, final speed of the train, $v$ = $0\ ms^{-1}$

We know, $v^2 = u^2 + 2as$

$\\ \implies 0^2 = 25^2 + 2(-0.5)s \\ \implies 0 = 625 -s \\ \implies s = 625\ m$

Therefore, the train travels a distance of $625\ m$ before coming to rest.

Given, The trolley starts from rest. Hence, the initial speed of the trolley, $u$ = $0\ ms^{-1}$

Acceleration of the trolley, $a$ = $2\ cms^{-2}$

Time is taken, $t =3\ s$

(a) We know, $v = u+at$

$\\ \implies v = 0+(2)(3) = 6\ cms^{-1}$

Therefore, the velocity of the trolley after 3 sec is $6\ cms^{-1}$

Given, Initial speed of the racing car, u = $0\ ms^{-1}$

Acceleration of the car, $a$ = $4\ ms^{-2}$

Time taken, $t = 10\ s$

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

$\\ \implies s = 0(10) + \frac{1}{2}(4)(10)^2 \\ \implies s = 2(100) \\ \implies s = 200\ m$

Therefore, the distance travelled by the racing car in $10\ s$ is $200\ m$

Taking upward direction as positive (+) direction:

Given,

$u = 5\ ms^{-1}$

$a = -10\ ms^{-2}$ (This is due to gravitational force!)

The stone will move up until its velocity becomes zero.

$\therefore v = 0\ ms^{-1}$

We know, $v^2 = u^2 + 2as$

$\\ \implies 0^2 = 5^2 + 2(-10)s \\ \implies 20 s = 25 \\ \implies s = 1.25\ m$

Therefore, the stone reaches to a height of $1.25\ m$

Now,

We know, $v = u + at$

$\\ \implies 0 = 5 + (-10)t \\ \implies t = 0.5\ s$

Therefore, the time taken by the stone to reach the maximum height is $0.5\ s$ .

## NCERT Solutions for Class 9 Science Chapter 8 Motion: Solved Exercise Questions-

Given, Diameter of the circular track = $200\ m$

$\therefore$ The circumference of the circular track, $d = 2\pi\frac{D}{2} = 200\pi\ m$

The athlete completes one round of a circular track in 40 s.

$\therefore$ Speed of the athlete = $u = \frac{200\pi\ m}{40\ s} = 5\pi\ ms^{-1}$

In $t = 2\ min\ 20\ s = 140\ s$ ,

Distance travelled by the athlete = $Speed\times time = (5\pi)\times(140)$

$= 5\times\frac{22}{7}\times140 = 2200\ m$

Also, number of rounds the athlete will complete in $140\ s$ = $\frac{140}{40} = 3.5$

Therefore, the final position of the athlete after $140\ s$ will be diametrically opposite to his initial point.

(3 complete rounds and one half round.)

Hence, displacement of the athlete = magnitude of diameter of the circle =200m

(a) from A to B and

(b) from A to C?

Given,

(a) Distance between A and B = $300\ m$

Time taken to reach from A to B = $2\ min\ 30\ s = 150\ s$

$\therefore$ Average speed from A to B = $\frac{Distance}{Time} = \frac{300}{150} = 2\ ms^{-1}$

And, Average velocity from A to B = $\frac{Displacement}{Time} = \frac{300}{150} = 2\ ms^{-1}$

(In this case, average speed is equal to the average velocity)

(b) Distance travelled from A to reach C = $AB + BC = 300 + 100 = 400\ m$

And, Displacement from A to C = $AC= 300-100 = 200\ m$

Also, time taken to reach C from A = $2\ min\ 30\ s + 1\ min= (150+60)\ s =210\ s$

$\therefore$ Average speed from A to C = $\frac{Distance}{Time} = \frac{400}{210} = 1.90\ ms^{-1}$

And, Average velocity from A to C = $\frac{Displacement}{Time} = \frac{200}{210} = 0.95\ ms^{-1}$

(In this case, average speed is not equal to the average velocity)

Given, Average speed while going to school, $v_1 = 20\ km h^{-1}$

And Average speed while returning back from school, $v_2 = 30\ km h^{-1}$

Let the distance between starting point and school be $x\ m$

And time taken by Abdul during the two trips be $t_1\ s\ and\ t_2\ s$

We know, $Speed = \frac{Distance}{Time}$

$\therefore v_1 = \frac{x}{t_1} = 20$

And, $\therefore v_2 = \frac{x}{t_2} = 30$ -(i)

Now, Total distance that Abdul covers = $x +x = 2x$

And total time Abdul takes to cover this distance = $t_1 + t_2$

$\\ \therefore v_{avg} = \frac{2x}{t_1+t_2} \\ = \frac{2x}{\frac{x}{20}+\frac{x}{30}} \\ \\ = \frac{60\times2x}{5x} \\ \\ = 24\ ms^{-1}$

Therefore, the average speed for Abdul's trip is $24\ ms^{-1}$

(Note: $\frac{20+30}{2} =25\ ms^{-1} \neq 24\ ms^{-1}$ )

Given, The motorboat starts from rest. Hence, initial speed of the motorboat, u = $0\ ms^{-1}$

Acceleration of the motorboat, $a$ = $3.0\ m s^{-2}$

Time taken, $t = 8\ s$

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

$\\ \implies s = 0(8) + \frac{1}{2}(3)(8)^2 \\ \implies s = \frac{1}{2}(3)(64) \\ \\ \implies s = 96\ m$

Therefore, the distance travelled by the motorboat is $96\ m$

The initial speed $=52\times \frac{5}{18}=14.4 \frac{m}{s}$

After 5 sec the car stops

The graph is represented by the blue line ( x-axis is time and the y-axis is speed)

For the car with 3Kmh -1 . Initial speed $=3\times\frac{5}{18}=0.833\frac{m}{s}$ . The graph which is represented by the golden line

The area covered by the blue graph is greater than the golden graph so the car with 15 m/s initial velocity travells large distance.

• Which of the three is travelling the fastest?

Given is a distance-time graph. The slope of this graph gives us speed. Hence, the graph with the highest slope will have the highest speed.

Since B has the highest slope(inclination), it travels the fastest.

• Are all three ever at the same point on the road?

Given is a distance-time graph. Any point on the curve will give the distance of object from O. Since there is no intersection point of all the three graphs, they never meet at the same point on the road.

(Although any two of them do meet at some point on the road!)

• How far has C travelled when B passes A?

Given is a distance-time graph. Any point on the curve will give the distance of object from O. To find how far C has travelled when B passes A, draw a perpendicular from the intersection point of A and B on time axis. The point where it intersects on the C graph, from C draw a perpandicular to y axis . Therefore, distance travelled by C will be (Final distance from O - Initial distance from O)

Therefore, C has traveled 6.5 km when B passes A.

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

Given is a distance-time graph. The graph of B and C intersect at a point whose y-coordinate is 5. Hence, B has travelled $5\ km$ by the time it passes C.

Considering downward direction as positive direction.

Given, Height from which ball is dropped, $s = 20\ m$

Acceleration of the ball, $a$ = $10\ ms^{-2}$

Initial velocity, $u = 0\ ms^{-1}$

(i) We know, $v^2 = u^2 + 2as$

$\\ \implies v^2 = 0^2 + 2(10)(20) \\ \implies v^2 = 400 \\ \implies v = 20\ ms^{-1}$ (In downward direction)

Therefore, the ball will strike the ground with a velocity of $20\ ms^{-1}$

(ii) Now, we know, $v = u + at$

$\\ \implies 20 = 0 + 10t \\ \implies t = 2\ s$

Therefore, the ball reaches the ground in $2\ s$ .

Note: $v = -20\ ms^{-1}$ was rejected because in this case, the negative sign implies the velocity in upward direction, which is opposite to the direction of the motion of the ball(before collision).

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.

Given is a speed-time graph. The area under the curve will give the distance travelled by the car.

In time $t= 4\ s$ , the distance travelled by the car will be equal to the area under the curve from $x = 0\ to\ x=4$

Considering this part of the graph as a quarter of a circle whose radius = 4 unit.

Therefore, required area = $\frac{1}{4}\pi r^2 = \frac{1}{4}\pi (4)^2 = 12.56\ m$

Therefore, distance the car travelled in the first 4 seconds is $12.56\ m$

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

In uniform motion, the speed of car will become constant which is represented by line parallel to the time axis. In the given figure, the straight line graph from $t = 6\ s\ to\ t= 10\ s$ represents the uniform motion of the car.

• an object with a constant acceleration but with zero velocity

(a) The given situation is possible.

When an object is thrown upwards (under gravity only), it reaches to a maximum height where its velocity becomes zero. However, it still has an acceleration acting in the downward direction (acceleration due to gravity).

Note: This is possible for a given point of time, however, it is not possible for a period of time.

• an object moving with an acceleration but with uniform speed.

(b) The given situation is possible.

An object moving in a circular path with uniform speed, i.e covering equal distance in equal amount of time is still under acceleration. Because, the velocity keeps on changing due to continuous change in the direction of motion. Therefore, circular motion is an example of an object moving with an acceleration but with uniform speed.

• an object moving in a certain direction with an acceleration in the perpendicular direction.

(c) The given situation is possible.

For an object moving in a circular path with constant speed, the direction of its velocity at any point will be tangential to that point. However, its acceleration will be directed radially inwards. (Constant speed but still having an acceleration - Due to continuous change in direction.)

Given, Radius of the circular orbit = $42250\ km$

$\therefore$ Circumference of the orbit = $2\pi r = 2.\frac{22}{7}. (42250) = 265571.4\ km$

The satellite takes 24 hours to revolve around the earth.

We know, $Speed = \frac{Distance}{Time}$

$\\ = \frac{265571.4}{24} \\ \\ = 11065.4\ kmh^{-1} \\ \\ = \frac{11065.4}{3600}\ kms^{-1} \\ \\ = 3.07\ kms^{-1}$

Therefore, the speed of the artificial satellite is $3.07\ kms^{-1}$

The 8th chapter of NCERT Solutions for Class 9 Science titled "Motion," provides a comprehensive understanding of the term motion. After studying this chapter, you will be able to answer the question. Motion is a relative concept, where the perception of movement depends on the observer's frame of reference. For instance, when we are travelling on a bus, a person sitting behind us appears to be at rest in relation to us. However, for someone outside the bus, that person is in motion. Moreover, while travelling, we may notice objects like trees seemingly moving, which can be attributed to the phenomenon of relative motion. This chapter, Motion Class 9 Science, presents numerous examples and explanations to deepen your comprehension of this fundamental physical concept.

By using some good examples, NCERT solutions for Class 9 Science chapter 8 Motion will also give you a better understanding of the concept. For example, if a person says that my home is 60 Km north of the airport. Here, the reference point is the airport. To specify the position of an object we need to choose a reference point and a direction. If in the previous case the person says that my home is 60 Km from the airport then we can go 60 Km in any direction. To know the exact point specifying direction along with reference is also important. Along with class 9 science chapter 8 question answer for Motion Class 9 Science Chapter, you will also get NCERT solutions for the additional questions mentioned in between the chapter.

## Motion Class 9 Chapter -Important Formulas

Formulae are essential for solving motion class 9 questions answers. These formulae allow you to calculate various parameters related to motion, such as speed, velocity, acceleration, and distance travelled.

• $Average \ speed = \frac{distance\ travelled}{total\ time\ taken}$
• If we specify the direction along with speed the term is known as velocity. Velocity is the speed of an object in a definite direction
• $Average \ velocity = \frac{displacement}{total\ time\ taken}$
• If the object is moving with varying velocity then the

$Average \ velocity = \frac{initial\ velocity+final\ velocity}{2}$

• Acceleration is the rate of change of velocity
• $acceleration=\frac{change\ in\ velocity }{time\ taken}$
• The SI unit of acceleration is

$ms^{-2}or\ \frac{m}{s^2}$

An important topic of the NCERT Solutions for Class 9 Science Chapter 8 is equations of motion for an object moving with uniform acceleration. If the initial velocity is u, the final velocity is v and t is the time taken and s is the distance travelled then the following are the three equations of motion.

$\\v=u+at\\s=ut+\frac{1}{2}at^2\\v^2-u^2=2as$

Motion Class 9 Numericals

i) what is the distance travelled when the ant is at B. Is the distance equal to displacement?

Solution:

$distance=ACB=\frac{1}{2}circumferance=\frac{1}{2}\times2\pi r=\frac{1}{2}\times2\pi \times 7=22cm$

Displacement =AB=diameter =14cm. So distance not equal to the displacement

ii) what is the total distance travelled and the final displacement?

Total distance travelled = circumference of the circle = $2\pi r=2\pi \times 7=44cm$

Displacement =0 as the initial and final position are the same

iii) What is the time taken by the ant to travel from A to B

The time = distance / speed = 22 / 1 = 22 sec

Class 9 Chapter 8 Science Topics

Topics for class 9 physics motion are given below:

8.1 Describing Motion

8.1.1 Motion along a straight line

8.1.2 Uniform Motion and Nonuniform Motion

8.2 Measuring the rate of motion

8.2.1 Speed with Direction

8.3 Rate of Change of Velocity

8.4 Graphical Representation of Motion

8.4.1 Distance- Time Graphs

8.4.2 Velocity-Time Graphs

8.5 Equations of Motion by Graphical Method

8.5.1 Equation for Velocity-Time Relation

8.5.3 Equation for Position-Velocity Relation

8.6 Uniform Circular Motion

## Key features of NCERT Solutions for Class 9 Science Chapter 8 Motion

• Completion of Homework and Assignments: These ncert solutions for class 9 science chapter 8 exercise questions provide comprehensive answers and explanations, enabling you to complete your homework and assignments accurately.
• Coverage of NCERT Book Questions: The motion chapter class 9 questions and answers cover all the questions given in the NCERT book, ensuring that you have a complete understanding of the chapter.
• Easy-to-Understand Language: The class 9 science chapter 8 solutions are written in a language that is easy to comprehend, making complex concepts more accessible to you. They have been crafted by highly experienced subject matter experts.
• Detailed Step-by-Step Solutions: The solution provides a step-by-step breakdown of the problems, helping you to understand the logic and methodology behind each solution.
• Time-Saving and Quick Revision: By having access to these class 9 science chapter 8 exercise question answer, you can save time in searching for answers and can quickly revise the chapter by referring to the detailed solutions.

## Highlight Points

• In order to determine the position of an object, a reference point or origin is necessary. The motion of an object can appear different to different observers, depending on their perspective. For example, a person sitting inside a moving bus may see other passengers as being stationary, while an observer standing outside the bus sees the passengers as moving.
• To make observations easier, a standard reference point or frame of reference is required. This ensures that all objects have the same reference frame for consistency. It's essential to use the same reference point to accurately describe the motion of an object, regardless of the observer's perspective.

The study of motion in physics is a part of classical mechanics and is taught in class 9 science ch 8. However, if students did not understand the chapter or missed out on theclass 9 science chapter 8 question answer, they need not worry as they have come to the right place. In this article, we provide a comprehensive guide on NCERT solutions for class 9 science chapter 8.

Additionally, students can also access NCERT Solutions for Class 9 Maths, which will enable them to revise the complete syllabus and score better marks in their examinations. These resources will help students understand and master the concepts of motion and related numerical problems for a better understanding and performance in their exams.

If you stuck anywhere or want to complete your homework on time, refer to the solutions mentioned below:

NCERT Books and NCERT Syllabus:

NCERT Solutions for Class 9 Science- Chapter Wise

 Chapter No. Chapter Name 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 8 Motion 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

NCERT Solutions for Class 9 - Subject Wise

 NCERT Solutions for Class 9 Maths NCERT Solutions for Class 9 Science

NCERT Science Exemplar Solutions Class 9 - Chapter Wise

 Chapter No. Chapter Name Chapter 1 Chapter 1 Matter in our Surroundings Chapter 2 Chapter 2 Is Matter Around Us Pure? Chapter 3 Chapter 3 Atoms and Molecules Chapter 4 Chapter 4 Structure of the Atom Chapter 5 Chapter 5 The Fundamental Unit of Life Chapter 6 Chapter 6 Tissues Chapter 7 Chapter 7 Diversity in Living Organisms Chapter 8 Chapter 8 Motion Chapter 9 Chapter 9 Forces and Laws of Motion Chapter 10 Chapter 10 Gravitation Chapter 11 Chapter 11 Work and Energy Chapter 12 Chapter 12 Sound Chapter 13 Chapter 13 Why do We Fall ill? Chapter 14 Chapter 14 Natural Resources Chapter 15 Chapter 15 Improvement in Food Resources

1. List out the topics covered in the Class 9 Science NCERT book chapter 8
• The topics covered in the NCERT syllabus of Cass 9 Science chapter 8 are listed below-
• Describing Motion
• Motion along a straight line
• Uniform Motion and Nonuniform Motion
• Measuring the rate of motion
• Speed with Direction
• Rate of Change of Velocity
• Graphical Representation of Motion
• Distance- Time Graphs
• Velocity-Time Graphs
• Equations of Motion by Graphical Method
• Equation for Velocity-Time Relation
• Equation for Position-Velocity Relation
• Uniform Circular Motion
2. Where I can get more practice problems of Motion Class 9 NCERT chapter.

The questions on the chapter Motion can be practiced using NCERT exercise and NCERT exemplar questions.

3. what is motion class 9

In class 9, motion refers to the movement of an object from one location to another and includes measuring how fast the thing moves, in which direction it moves, and how much it moves. It assists students in understanding the fundamental physics principles relating to moving things.

4. What is the main topic of motion Class 9?

In class 9 of chapter motion, Students learn about motion and different types of motion, such as rectilinear, circular, and periodic motion, and the concepts related to motion, including speed, velocity, acceleration, and displacement. They also learn about the graphical representation of motion, such as distance-time graphs and velocity-time graphs, and how to solve numerical problems related to motion. Understanding the fundamentals of motion is crucial for students studying science and physics, as it forms the basis for many concepts and theories in higher classes.

5. What is the SI unit of motion?

The motion itself is not a physical quantity, so it does not have an SI unit. However, there are several physical quantities related to motion that have their respective SI units, including:

Length or distance: meter (m)

Time: second (s)

Speed: meter per second (m/s)

Velocity: meter per second (m/s)

Acceleration: meter per second squared (m/s²)

Force: Newton (N)

Momentum: kilogram meter per second (kg·m/s)

Energy: Joule (J)

Power: Watt (W)

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Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9