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 18 Aug 2025, 05:10 PM IST

Have you ever wondered how sportsmen determine how fast they are racing or why cars reduce in speed when a driver presses a brake? Motion is the very essence of our day-to-day lives through a ride on the roller coaster or going down the street or traveling in a bus. Chapter 7 Motion of Class 9 Science studies the scientific laws that govern such everyday phenomena, and in this manner, the students are able to relate what is learnt in the classroom with the observations in life. These NCERT Solutions are thoroughly delivered by subject experts and in a simple-to-understand pattern, in a way that the students will study more and with interest.

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  1. NCERT Solutions for Class 9 Science Chapter 7 Motion: Download Solution PDF
  2. Class 9 Science Chapter 7 Motion: Solved 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

The chapter also gives enough chances to the students to exercise the important concepts, that include speed, velocity, acceleration, displacement and distance by completing numerous solved examples, and problems with numbers. These NCERT solutions for class 9 Science aid in the mastery of the logic and application of concepts of motion. The NCERT Solutions have solved In-text Questions to get an in-depth perception of concepts in the textbook, problem-solving skills are expanded with some solved exercise questions, Higher Order Thinking Skills (HOTS) questions for advanced knowledge and thinking, all Important topics such as types of motion, graphs of motion and equations of motion are covered and approach to solve questions which offers step-by-step plan to adequately address both the theory and the numericals.

NCERT Solutions for Class 9 Science Chapter 7 Motion: Download Solution PDF

The chapter Motion emphasises essential concepts such as speed, velocity, acceleration, and types of motion, supported by a variety of questions and answers aimed at strengthening these fundamentals. These solutions are especially useful for self-study, enabling students to clearly understand the core ideas and improve their performance in exams.

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Class 9 Science Chapter 7 Motion: Solved In-text Questions

The in-text solved questions, NCERT Class 9 Science Chapter 7, Motion, will present step-wise solutions to allow students to gain knowledge to grasp concepts such as displacement, velocity, acceleration, and graphs of motion. Such NCERT solutions for class 9 make learning easy, enhance problem-solving abilities as well and guarantee proper exam preparation.

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 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.

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 = $10\ m$
$\therefore$ 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 = $AC = \sqrt{(AB^2 + BC^2 )} = \sqrt{(100^2+100^2)} = 10\sqrt{2}\ m$

Q 3. Which of the following is true for displacement?

(a) It cannot be zero.

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

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

Answer:

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

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

Answer:

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

Answer:

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 $9\times10^{10}\ m$

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 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.

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, $u$ = $80\ kmh^{-1} = \frac{80\times1000}{3600} = \frac{200}{9} ms^{-1}$

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 $1.11 ms^{-2}$

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, 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 $0.0185\ ms^{-2}$

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 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.

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

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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 of 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,

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

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, 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 $12\ ms^{-1}$

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, 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 $720\ m$

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, $u$ = $90\ kmh^{-1} = \frac{90\times1000}{3600} = 25\ ms^{-1}$

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.

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

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

Time is taken, $t =3\ s$

We know, $v = u+at$

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

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

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

Answer:

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 $200\ m$

Q 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?

Answer:

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 7 Motion: Solved Exercise Questions

The Chapter 7, Motion, solved exercise questions 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 = $200\ m$

$\therefore$ 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

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 = $2\ min\ 30\ s = 150\ s$

$\therefore$ 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)

Q 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?

Answer:

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

Q 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?

Answer:

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 $96\ m$

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

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

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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.

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 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!)

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 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.

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 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.

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 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).

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 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 $12.56\ m$

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 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.

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 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.

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 = $42250\ km$

$\therefore$ 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, $Speed = \frac{Distance}{Time}$

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

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

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 the 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:

$\begin{aligned} a & =\frac{v-u}{t}=\frac{50 \mathrm{~km} / \mathrm{h}-40 \mathrm{~km} / \mathrm{h}}{2 \mathrm{~min}} \\ & =\frac{10 \mathrm{~km} / \mathrm{h}}{(3 / 20) \mathrm{h}}=\frac{10 \times 60}{2} \mathrm{~km} / \mathrm{h}^2=300 \mathrm{~km} / \mathrm{h}^2 \\ a & =\frac{300 \times 1000}{3600 \times 3600} \mathrm{~m} / \mathrm{s}^2=0.02 \mathrm{~m} / \mathrm{s}^2\end{aligned}$


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

$\begin{aligned} & \Delta x=u t+\frac{1}{2} a t^2 \text { or } h=u t+\frac{1}{2} g t^2 \\ & \mathrm{~h}=0+\frac{1}{2}\left(10 \mathrm{~m} / \mathrm{s}^2\right)(2.3 \mathrm{~s})^2=26.5 \mathrm{~m}\end{aligned}$


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, then the net velocity will be zero.


Q5:

A bus covers a distance of 250 km from Delhi to Jaipur towards 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)

$\begin{aligned} \text { Average speed } & =\frac{\text { Total distance travelled }}{\text { Total time taken }} \\ & =\frac{250 \mathrm{~km}+250 \mathrm{~km}}{5 \mathrm{~h}+5 \mathrm{~h}} \\ & =\frac{500 \mathrm{~km}}{10 \mathrm{~h}}=50 \mathrm{~km} / \mathrm{h}\end{aligned}$

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

$\begin{aligned} \text { Average velocity } & =\frac{\text { Total displacement }}{\text { Total time taken }} \\ & =\frac{250 \mathrm{~km}-250 \mathrm{~km}}{5 \mathrm{~h}+5 \mathrm{~h}} \\ & =\frac{0 \mathrm{~km}}{10 \mathrm{~h}} \\ & =0 \mathrm{~km} / \mathrm{h}\end{aligned}$


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

NCERT Solutions for Class 9 Science Chapter 7 – Motion 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 $=\frac{\text { Distance }}{\text { Time }}$

2. Velocity

Velocity $=\frac{\text { Displacement }}{\text { Time }}$

3. Acceleration

$
\text { Acceleration }=\frac{\text { Change in velocity }}{\text { Time }}=\frac{v-u}{t}
$
Where:
$v=$ final velocity
$u=$ initial velocity
$t=$ time taken
Unit: $\mathrm{m} / \mathrm{s}^2$

4. Uniform Acceleration Equations of Motion

1. $v=u+a t$
2. $s=u t+\frac{1}{2} a t^2$
3. $v^2=u^2+2 a s$

Where:
$s=$ displacement
$a=$ acceleration

5. Average Speed

Average Speed $=\frac{\text { Total Distance }}{\text { Total Time }}$

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

  • 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

  • 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.

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