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NCERT Solutions for Class 9 Science Chapter 7 Motion

NCERT Solutions for Class 9 Science Chapter 7 Motion

Edited By Vishal kumar | Updated on Jun 26, 2025 08:39 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.

This Story also Contains
  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 10: 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 Formulaes
  7. Approach to Solve Questions of Class 9 NCERT Chapter 7:Motion
  8. NCERT Solutions for Class 9 Science Chapter-wise
NCERT Solutions for Class 9 Science Chapter 7 Motion
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 mastering 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 emphasizes 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.

Download PDF

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

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
The perimeter of the square = 4×10=40 m
According to 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 trips completed traveling = 14040=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.
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 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 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 , 3×108ms1.

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 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 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 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, 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 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.

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 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=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, 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, 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, The trolley starts from rest. Hence, the initial speed of the trolley, u = 0 ms1

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 s 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 to 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-

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, 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 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 starting point and school be x m

And 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, Total distance that Abdul covers = x+x=2x

And 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, The motorboat starts from rest. Hence, 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 3Kmh -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 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 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 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 = 14πr2=14π(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

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 10: Higher Order Thinking Skills (HOTS) Questions

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:

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 is the balloon.

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, then the net velocity will be zero.


Q5:

A bus covers a distance of 250 km from Delli to Jaipur towards West in 5 hours in the morning and returns to Dellii 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 Dellhi 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

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 Formulaes

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

  • 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 purpose of the question to definitions of motion such as 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 organize 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 analyze especially when it comes to displacement and velocity.

NCERT Solutions for Class 9 Science Chapter-wise

NCERT Solutions for Class 9 - Subject Wise

Also Read,

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Questions (FAQs)

1. Can displacement be zero even if distance is not?

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

2. What is the difference between distance and displacement?

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

3. What is uniform motion?

When an object covers equal distances in equal time intervals.

4. What does the slope of a distance-time graph represent?

It gives the speed of the object.

5. Why do we need to study motion?

To understand how things move and to solve real-life problems like travel, transport, etc.

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Option 1)

0.34\; J

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0.67\; J

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)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

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

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} 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

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