NCERT Solutions for Class 9 Science Chapter 7 Motion - Free PDF

NCERT Solutions for Class 9 Science Chapter 7 Motion - Free PDF

Vishal kumarUpdated on 15 Sep 2025, 01:07 AM IST

Have you ever questioned yourself about how sportsmen can measure their speed or why a vehicle becomes slower when the brakes are pressed? Motions form a key component of our everyday existence, be it in a roller coaster ride or in a bus or on a street walk. Class 9 Science Chapter 7 - Motion is a topic that delves into the scientific concepts that support these common processes to enable the student to relate what he has been learning in the classroom with what he experiences in life.

This Story also Contains

  1. Class 9 Science Chapter 7 - Motion Question Answers: Download Solution PDF
  2. Class 9 Science Chapter 7 - Motion Question Answers: In-text Questions
  3. NCERT Solutions for Class 9 Science Chapter 7 Motion: Solved Exercise Questions
  4. Class 10 Science NCERT Chapter 7: Higher Order Thinking Skills (HOTS) Questions
  5. NCERT Solutions for Class 9 Science Chapter 7 Motion - Key Topics
  6. NCERT Solutions for Class 9 Science Chapter 7 Motion - Important Formulas
  7. Approach to Solve Questions of Class 9 NCERT Chapter 7:Motion
  8. Benefits of NCERT Solutions for Class 9 Science Chapter 7 Motion
  9. NCERT Solutions for Class 9 Science Chapter-wise
NCERT Solutions for Class 9 Science Chapter 7 Motion - Free PDF
NCERT Solutions for Class 9 Science Chapter 7 Motion

NCERT Solutions for Class 9 Science Chapter 7 - Motion are perfectly designed to be interesting and effective in learning, with the subject experts working smoothly on easy-to-understand language. All important notions, such as speed, velocity, acceleration, displacement, and distance, are included in these NCERT Solutions for Class 9 Science Chapter 7 - Motion, and a variety of solved problems and practice problems are provided to help the students develop a better understanding of the concepts and practice using numerical problems. Step-by-step NCERT solutions of in-text questions, exercise questions, and HOTS (Higher Order Thinking Skills) problems are also beneficial to students, and it is easier to learn such topics as types of motion, motion graphs, and equations of motion. These Motion NCERT Solutions are ideal for examination preparation, self-study, and revision, as the students will learn the theory and numbers effectively.

Class 9 Science Chapter 7 - Motion Question Answers: Download Solution PDF

Class 9 Science Chapter 7 - Motion question answers are provided to students in a well-structured and easy-to-follow format, giving clear answers to all in-text and exercise questions in the NCERT textbook. These solutions embrace significant concepts such as speed, velocity, acceleration, displacement, distance, motion graphs, and equations of motion, which make studying exam preparation less difficult and effective. These NCERT solutions can help the student to reinforce his or her knowledge, practice numbers, and revise in a hurry before the examination.

Download PDF

Class 9 Science Chapter 7 - Motion Question Answers: In-text Questions

Class 9 Science Chapter 7 Motion In-text Questions also offer step-by-step answers to the questions in the NCERT textbook in a general, easy-to-understand format. Motion NCERT Solutions assist in the clarification of concepts like speed, velocity, acceleration, distance and displacement and also simplify numerical and conceptual problems. They can be used to prepare, revise and do homework with ease.

Topic 7.1 Describing Motions

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

Answer:

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

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 travelled is equal to the circumference of the circular path, but the displacement is zero.

Q 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?

Answer:

Side of the square field = 10 m
The perimeter of the square = 4×10=40 m
According to the question,
He completes 1 round in 40 s.
Speed of the farmer = 4040=1 m/s
Distance covered in 2 min 20 s (=140 s) = 140×1=140 m
Now,
Number of round-trip completed travelling = 14040=3.5
We know, in 3 round-trips, the displacement will be zero.
In 0.5 rounds, the farmer will reach diametrically opposite to his initial position.
Displacement = AC=(AB2+BC2)=(1002+1002)=102 m

Q 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) 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.

Topic 7.2 Measuring the Rate of Motion

Q 1. Distinguish between speed and velocity.

Answer:

SpeedVelocity

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 quantityVelocity is a vector quantity
Speed does not depend on the directionVelocity changes with change in direction
Speed can never be negativeVelocity can be positive, negative or zero.

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

Answer:

When the total distance travelled 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 uponthe time taken.

Q 3. What does the odometer of an automobile measure?

Answer:

An odometer is a device that measures the total distance travelled by an automobile.

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

Answer:

An object has a uniform motion if it covers an equal distance in an 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 into six equal parts, each subtending 60 at the centre. The object covers each equal part in an 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 , 3×108ms1.

Answer:

Given, the signal travels at the speed of light, v=3×108ms1 .

Time taken by the signal = 5 mins=5×60s=300 s

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

We know, Speed=Distance travelledTime taken

v=Dt

D=v×t=3×108×300

D=9×1010 m

Therefore, the distance of the spaceship from the ground station is 9×1010 m

Topic 7.3 Rate of Change of Velocity

Q 1. When will you say a body is in

(i) uniform acceleration?

(ii) nonuniform acceleration?

Answer:

(i) If the velocity of an object travelling 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 has 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 increases its speed by unequal amounts in equal intervals of time.

Q.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.

Answer:

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

Given, Initial speed of the bus, u = 80 kmh1=80×10003600=2009ms1

The final speed of the bus, v = 60 kmh1=60×10003600=1006ms1

Time is taken, t=5s

We know, v=u+at

1006=2009+a(5)

5a=90012006×9=3006×9

a=1009=1.11 ms2

The negative sign implies retardation.

Therefore, the acceleration of the bus is 1.11 ms2

Or, the retardation(de-acceleration) of the bus is 1.11ms2

Q 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.

Answer:

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

Given that the train starts from rest. Hence, the initial speed of the train = 0 kmh1=0 ms1

Final speed of the train = 40 kmh1=40×10003600=1009ms1

Time taken, t=10 min=10×60 s=600 s

We know, v=u+at

1009=0+a(600)

a=1009×600=16×9

a=154=0.0185 ms2

Therefore, the acceleration of the train is 0.0185 ms2

Topic 7.4 Graphical Representation of Motion

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

Answer:

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

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

1687160020858

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

1687160041034

Q2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

Answer:

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 in time. This implies that the object is not moving and is at rest.

Q3. What can you say about the motion of an object if its speedtime graph is a straight line parallel to the time axis?

Answer:

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.

Q4. What is the quantity which is measured by the area occupied below the velocity-time graph?

Answer:

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

We know,

Speed=DistanceTimeDistance=Speed×Time

Topic 7.5 Equations of motion by graphical method

Q1.(a) A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find the speed acquired

Answer:

Given that the bus starts from rest. Hence, the initial speed of the bus = 0 ms1

Acceleration of the bus, a = 0.1 ms2

Time is taken, t=2 min=2×60 s=120 s

(a) We know, v=u+at

v=0+(0.1)(120)=12 ms1

Therefore, the speed acquired by the bus is 12 ms1

Q1.(b) A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find the distance travelled.

Answer:

Given that the bus starts from rest. Hence, the initial speed of the bus, u = 0 ms1

Acceleration of the bus, a = 0.1 ms2

Time taken, t=2 min=2×60 s=120 s

(b) We know, s=ut+12at2

s=0(120)+12(0.1)(120)2

s=12(0.1)(14400)

s=720 m

Therefore, the distance travelled by bus is 720 m

Q2. 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.

Answer:

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

Given, Initial speed of the train, u = 90 kmh1=90×10003600=25 ms1

Acceleration of the train, a=0.5 ms2 (Negative sign implies retardation)

Since the train has to be brought to rest, final speed of the train, v = 0 ms1

We know, v2=u2+2as

02=252+2(0.5)s

0=625s

s=625 m

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

Q3. 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:

Given that the trolley starts from rest. Hence, the initial speed of the trolley, u = 0 cms1

Acceleration of the trolley, a = 2 cms2

Time is taken, t=3 s

We know, v=u+at

v=0+(2)(3)=6 cms1

Therefore, the velocity of the trolley after 3 sec is 6 cms1

Q4. A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 seconds after start?

Answer:

Given, Initial speed of the racing car, u = 0 ms1

Acceleration of the car, a = 4 ms2

Time taken, t=10 s

We know, s=ut+12at2

s=0(10)+12(4)(10)2

s=2(100)s=200 m

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

Q 5. A stone is thrown in a vertically upward direction with a velocity of 5 ms1 . If the acceleration of the stone during its motion is 10 ms2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Answer:

Taking upward direction as positive (+) direction:

Given,

u=5 ms1

a=10 ms2 (This is due to gravitational force!)

The stone will move up until its velocity becomes zero.

v=0 ms1

We know, v2=u2+2as

02=52+2(10)s

20s=25s=1.25 m

Therefore, the stone reaches a height of 1.25 m

Now,

We know, v=u+at

0=5+(10)tt=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 7 Motion: Solved Exercise Questions

The Class 9 Science Chapter 7 - Motion question answers provide well-articulated answers to all the end-of-chapter questions, expressing the key ideas of distance, speed, displacement, velocity, and acceleration. The solutions serve as the means of effective practice and foundation in solving problems on the basis of physics.

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?

Answer:

Given, Diameter of the circular track = 200 m

The circumference of the circular track, d=2πD2=200π m

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

Speed of the athlete = u=200π m40 s=5π ms1

In t=2 min 20 s=140 s ,

Distance travelled by the athlete = Speed×time=(5π)×(140)

=5×227×140=2200 m

Also, the number of rounds the athlete will complete in 140 s = 14040=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 the diameter of the circle =200m

Q2. 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?

Answer:

Given,

(a) Distance between A and B = 300 m

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

Average speed from A to B = DistanceTime=300150=2 ms1

And, Average velocity from A to B = DisplacementTime=300150=2 ms1

(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=300100=200 m

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

Average speed from A to C = DistanceTime=400210=1.90 ms1

And, Average velocity from A to C = DisplacementTime=200210=0.95 ms1

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

Q 3. Abdul, while driving to school, computes the average speed for his trip to be 20 kmh1 . On his return trip along the same route, there is less traffic and the average speed is 30 kmh1. What is the average speed for Abdul’s trip?

Answer:

Given, Average speed while going to school, v1=20 kmh1

And Average speed while returning back from school, v2=30 kmh1

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

And the time taken by Abdul during the two trips be t1 s and t2 s

We know, Speed=DistanceTime

v1=xt1=20

And, v2=xt2=30 -(i)

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

And the total time Abdul takes to cover this distance = t1+t2

vavg=2xt1+t2=2xx20+x30=60×2x5x=24 ms1

Therefore, the average speed for Abdul's trip is 24 ms1

(Note: 20+302=25 ms124 ms1 )

Q 4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms2 for 8.0 s. How far does the boat travel during this time?

Answer:

Given that the motorboat starts from rest. Hence, the initial speed of the motorboat, u = 0 ms1

Acceleration of the motorboat, a = 3.0 ms2

Time taken, t=8 s

We know, s=ut+12at2

s=0(8)+12(3)(8)2

s=12(3)(64)

s=96 m

Therefore, the distance travelled by the motorboat is 96 m

Q 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?

Answer:

The initial speed =52×518=14.4ms

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 3 kmh -1. Initial speed =3×518=0.833ms . The graph which is represented by the golden line

1651223289557

The area covered by the blue graph is greater than the golden graph, so the car with 15 m/s initial velocity travels la arge distance.

Q 6. (a) Figure shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions:

1651223337605

Which of the three is travelling the fastest?

Answer:

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.

Q 6. (b) Figure shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions:

1651223370102

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

Answer:

Given is a distance-time graph. Any point on the curve will give the distance of the object from O. Since there is no intersection point of all 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!)

Q 6.(c) Figure shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions:

1651223382702

How far has C travelled when B passes A?

Answer:

Given is a distance-time graph. Any point on the curve will give the distance of the 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 the time axis. The point where it intersects the C graph, from C, draw a perpendicular to the y-axis. Therefore, the distance travelled by C will be (Final distance from O - Initial distance from O)

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

Q 6. (d) Figure shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions:

1651223407329

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

Answer:

Given is a distance-time graph. The graphs 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.

Q 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?

Answer:

Considering the downward direction as the positive direction.

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

Acceleration of the ball, a = 10 ms2

Initial velocity, u=0 ms1

(i) We know, v2=u2+2as

v2=02+2(10)(20)

v2=400

v=20 ms1 (In downward direction)

Therefore, the ball will strike the ground with a velocity of 20 ms1

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

20=0+10tt=2 s

Therefore, the ball reaches the ground in 2 s.

Note: v=20 ms1 was rejected because in this case, the negative sign implies the velocity in the upward direction, which is opposite to the direction of the motion of the ball(before collision).

Q 8.(a) The speed-time graph for a car is shown is Figure:

1651223440314

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.

Answer:

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 units.

Therefore, required area = 14πr2=14π(4)2=12.56 m

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

Q 8. (b) The speed-time graph for a car is shown is Figure:

1651223470213

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

Answer:

In uniform motion, the speed of the car will become constant, which is represented by a 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.

Q 9. (a) State which of the following situations are possible and give an example for each of these: an object with a constant acceleration but with zero velocity

Answer:

(a) The given situation is possible.

When an object is thrown upwards (under gravity only), it reaches 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 in time; however, it is not possible for a period of time.

Q 9.(b) State which of the following situations are possible and give an example for each of these: an object moving with an acceleration but with uniform speed.

Answer:

(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.

Q 9. (c) State which of the following situations are possible and give an example for each of these: an object moving in a certain direction with an acceleration in the perpendicular direction.

Answer:

(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.)

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.

Answer:

Given, Radius of the circular orbit = 42250 km

Circumference of the orbit = 2πr=2.227.(42250)=265571.4 km

The satellite takes 24 hours to revolve around the Earth.

We know, Speed=DistanceTime

Speed=265571.424=11065.4 kmh1=11065.43600 kms1=3.07 kms1

Therefore, the speed of the artificial satellite is 3.07 kms1

Class 10 Science NCERT Chapter 7: Higher Order Thinking Skills (HOTS) Questions

The HOTS questions in Chapter 7, Motion, provide challenges to students to use such concepts as displacement, velocity, acceleration, and representation of motion in graphs to solve complex problems in situations. Such questions develop analytical abilities and stimulate a better comprehension of real-life motion phenomena.

Q1:

The velocity-time graph of an object is shown in the figure. Identify the correct statement(s) regarding this graph.

(i)This is a non-uniform velocity-time graph of the object.
(ii) The velocity of the object is increasing at the same rate during OP and QR.
(iii) The velocity of the object is decreasing at the same rate during PQ and RT.

Answer:

From the graph, the velocity of the object is increasing at a constant rate for OP and QR and decreasing at a constant rate for PQ and RT. However, the rate of increasing or decreasing the velocity is not the same for all intervals. It is a non-uniform velocity-time graph.

Hence, the answer is option (1).


Q2:

A train is moving at a speed of 40 km/hr at 10:00 a.m., and at 50 km/hr at 10:02 a.m.. Assuming that the train moves along a straight track and the acceleration is constant, find the value of the acceleration.

Answer:

a=vut=50 km/h40 km/h2 min=10 km/h(3/20)h=10×602 km/h2=300 km/h2a=300×10003600×3600 m/s2=0.02 m/s2


Q3:

While driving through Napa, you observe a hot air balloon in the sky with tourists on board. One of the passengers accidentally drops a wine bottle, and you note that it takes 2.3 seconds for it to reach the ground. Find how high the balloon is.

Answer:

Given wine bottle is dropped, i.e.,u=0

Δx=ut+12at2 or h=ut+12gt2 h=0+12(10 m/s2)(2.3 s)2=26.5 m


Q4:

A person starts jogging on a circular track and completes 5 full rounds. The circumference of the circular track is 88 m. Then find the net velocity of that person.

Answer:

Since the net displacement is zero, the net velocity will be zero.


Q5:

A bus covers a distance of 250 km from Delhi to Jaipur towards the West in 5 hours in the morning and returns to Delhi in the evening covering the same distance of 250 km in the same time of 5 hours. Find (a) the average speed, and ( b ) the average velocity, of the bus for the whole journey.

Answer:

(a)

Average speed = Total distance travelled Total time taken =250 km+250 km5 h+5 h=500 km10 h=50 km/h

(b) In this case, the bus travels 250 km from Della to Jaipur towards the West and then comes back to the starting point Delhi in the reverse direction. So, the total displacement (or total distance travelled in a specified direction) will be 250 km-250 km=0

Average velocity = Total displacement Total time taken =250 km250 km5 h+5 h=0 km10 h=0 km/h


NCERT Solutions for Class 9 Science Chapter 7 Motion - Key Topics

Motion Class 9 question answers cover all the key topics that explain how objects move, including concepts like speed, velocity, acceleration, and graphical representation of motion. These topics help students understand the fundamental laws that govern motion in everyday life.

7.1 Describing Motion

7.1.1 Motion along a straight line

7.1.2 Uniform Motion and Nonuniform Motion

7.2 Measuring the rate of motion

7.2.1 Speed with Direction

7.3 Rate of Change of Velocity

7.4 Graphical Representation of Motion

7.4.1 Distance- Time Graphs

7.4.2 Velocity-Time Graphs

7.5 Equations of Motion by Graphical Method

7.5.1 Equation for Velocity-Time Relation

7.5.3 Equation for Position-Velocity Relation

7.6 Uniform Circular Motion


NCERT Solutions for Class 9 Science Chapter 7 Motion - Important Formulas

Motion important formulas in Class 9 Science are a rapid guide to the solutions of numerical problems concerning distance, displacement, speed, velocity, acceleration, and equations of motion. The formulas guide students to enhance their problem-solving capabilities as well as effectively prepare them to sit exams.

1. Speed

Speed = Distance Time

2. Velocity

Velocity = Displacement Time

3. Acceleration

Acceleration = Change in velocity Time =vut
Where:
v= final velocity
u= initial velocity
t= time taken
Unit: m/s2

4. Uniform Acceleration Equations of Motion

1. v=u+at
2. s=ut+12at2
3. v2=u2+2as

Where:
s= displacement
a= acceleration

5. Average Speed

Average Speed = Total Distance Total Time

Approach to Solve Questions of Class 9 NCERT Chapter 7:Motion

Approach to Solve Questions of Class 9 Science Chapter 7 -Motion is used to guide students to have a clear system and structure to solve the theoretical and numerical problems in the chapter. Students will be able to reinforce their knowledge by paying attention to the main terms such as speed, velocity, acceleration, distance, and displacement to answer questions correctly. This strategy will help learners relate real life observations with what they learn in the textbook, which makes it easier and more effective to prepare exams.

  • Read the question in order to know the quantities provided (such as distance, time, or the type of motion).

  • Determine the nature of the motion involved; is it non-uniform or uniform (with a steady speed).

  • Understand what is asked, i.e. which speed, in which direction, which distance is covered or which comparison of two motions.

  • Apply conceptual knowledge: apply the purpose of the question to definitions of motion, such as the distinction between speed and velocity.

  • It is also important to draw diagrams or motion graphs (as appropriate) to make the problem easier to understand.

  • Write down and organise the information in a step-by-step kind of representation before one tries the solution.

  • Break down units and directions. Units and directions are important to analyse, especially when it comes to displacement and velocity.

Benefits of NCERT Solutions for Class 9 Science Chapter 7 Motion

Advantages of NCERT Solutions for Class 9 Science Chapter 7 - Motion are that it assists the student to acquire the basic concepts of motion easily. These resources are interactive and less stressful since they provide detailed explanation, solved examples and step by step solutions. They also facilitate frequent revision, develop knowledge of numerical problems, and confidence in both school examinations and competitive examinations.

  • Clear Concept Understanding- The solutions are written in a clear and well-organised way that can make the learner understand motions, distance, displacement, speed, velocity and acceleration.
  • Step-by-Step Solutions- The numerical problems are solved step-by-step, enabling the students to easily learn formulas and approaches.
  • Exam-Oriented Preparation: Relevant exercises, QMCs and solved examples assist the students to be well prepared to face the board examination and competitive exams.
  • Enhances Problem-Solving Skills -Different kinds of problems practised enhance such skills of analysis and application.
  • Time-Saving Material- Ready-to-use solutions enable effective time management and quick revision of the study material during examination preparation.
  • Supports Self-Learning – Students can use these solutions for self-study without much external guidance.

Frequently Asked Questions (FAQs)

Q: How to solve motion questions?
A:

Read the question carefully to identify the type of motion (rectilinear, circular, etc.). For numerical questions, use the formula Speed = Distance ÷ Time, and ensure units are correct. For descriptive questions, name the motion and give a real-life example.

Q: Why do we need to study motion?
A:

Understanding motion helps us describe how objects move, predict their positions, and apply these concepts in transportation, sports, machines, and more.

Q: Can displacement be zero even if distance is not?
A:

Distance is the total path covered (scalar), while displacement is the shortest path from start to end (vector).

Q: What is the difference between distance and displacement?
A:

Distance is the total path covered (scalar), while displacement is the shortest path from start to end (vector).

Q: What is uniform motion?
A:

When an object covers equal distances in equal time intervals.

Articles
Upcoming School Exams
Ongoing Dates
UP Board 12th Others

11 Aug'25 - 30 Sep'25 (Online)

Ongoing Dates
UP Board 10th Others

11 Aug'25 - 30 Sep'25 (Online)