NCERT Solutions for Class 7 Science Chapter 13 Motion and Time

Edited By Vishal kumar | Updated on Apr 11, 2025 07:09 PM IST

Hey there! If you're studying Class 7 Science Chapter 9 – Motion and Time, you're in the right place. On this page from Careers360, you’ll find easy-to-understand solutions to all the textbook questions, prepared by subject experts just for you. These are simple answers written in plain English, and the best news is that you can download them as a PDF file and learn them anywhere, even offline.

In this chapter, you’ve already learned about different types of motion, like straight-line motion, circular motion, and periodic motion. The first few questions in the book are based on these. You also know that some vehicles are faster than others. By comparing how much distance two buses cover in one hour, you can easily say which one is quicker—that’s what speed is all about! This chapter mainly helps you understand speed and time and how to measure them. The solutions will help you clear your doubts and, most importantly, help you score well in exams. NCERT Solutions will be your main guide to prepare for your exams

Download the Class 7 Science Chapter 9 – Motion and Time NCERT Solutions PDF for free and become exam-ready with easy, correct answers. Ideal for CBSE students, this PDF allows you to revise anytime, anywhere—no internet required!

This Story also Contains

Motion and Time Class 7 Questions and Answers: Exercise
Chapter-wise NCERT Solutions of Class 7 Science
Class 7 Motion and Time NCERT Solutions: Important Formulas and Points
Motion and Time Class 7 Science Chapter 9-Topics
Benefits of NCERT Motion and Time Class 7 Solutions

NCERT Solutions for Class 7 Science Chapter 13 Motion and Time

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Motion and Time Class 7 Questions and Answers: Exercise

1. Classify the following as motion along a straight line, circular, or oscillatory motion:

(i) Motion of your hands while running.
(ii) Motion of a horse pulling a cart on a straight road.
(iii) Motion of a child in a merry-go-round.
(iv) Motion of a child on a see-saw.
(v) Motion of the hammer of an electric bell.
(vi) Motion of a train on a straight bridge.

Answer:

(i) Oscillatory motion
While running, our hands move in back and forth direction and repeat its movement after some time. Hence it is an oscillatory motion.

(ii) Motion along the straight line-
A horse is pulling a cart on a straight road. Therefore it follows the motion along a straight line.

(iii) Circular motion-
Since the merry-go-round moves in a circular motion

(iv) Oscillatory motion-
The see-saw moves up and down continuously. It oscillates up-down.

(v) Oscillatory motion
(vi) Motion along a straight line-
The train is moving on a straight bridge. Therefore it follows the motion along the straight line.

2. Which of the following are not correct?

(i) The basic unit of time is second.
(ii) Every object moves with a constant speed.
(iii) Distances between two cities are measured in kilometres.
(iv) The time period of a given pendulum is constant.
(v) The speed of a train is expressed in m/h.

Answer:

The option (ii), (iv) and (v) are not correct.

(ii) The object can move with variable or constant speed.

(iv) The time period of a pendulum depends on the length of the thread. So, it is constant for a particular length.

(v) Speed of train is measured either in km/h or m/s not in m/h

3 . A simple pendulum takes 32 s to complete 20 oscillations. What is the time period of the pendulum?

Answer:

Given that,
A simple pendulum takes 32 s to complete 20 oscillations. It means,
No. of oscillation = 20, and
Total time is taken = 32 s
According to question,
Time period = (T)
$T = \frac{t o t a l t i m e t a k e n}{N o . o f o s c i l l a t i o n}$
$T = \frac{32}{20} = 1.6 s$

4 . The distance between two stations is 240 km. A train takes 4 hours to cover this distance. Calculate the speed of the train.

Answer:

Given that,
Distance between the two stations = 240 km
and, time taken = 4 hour

Therefore, Speed = Distance / Time
= 240/4
= 60 km/h

5. The odometer of a car reads 57321.0 km when the clock shows the time 08:30 AM. What is the distance moved by the car, if at 08:50 AM, the odometer reading has changed to 57336.0 km? Calculate the speed of the car in km/min during this time. Express the speed in km/h also.

Answer:

We have,
The initial reading of odometer = 57321 km
The final reading of odometer = 57336 km

So, distance covered by the Car = Final reading - Initial reading = 15 km

The total time taken by the car to travel 15 km = 8:50 AM - 8:30 AM = 20 min

According to question,
Speed of the car = Distance / time taken = 15/20 = 0.75 km/min

Again, 60 minute = 1 hr
so, 1 min = 1/60 hr

Speed in terms of km/hr = $0.75 \times 60 = 45 k m / h r$

6 . Salma takes 15 minutes from her house to reach her school on a bicycle. If the bicycle has a speed of 2 m/s, calculate the distance between her house and the school.

Answer:

Given that,
time taken by Salma to reach her school from house = 15 min = $15 \times 60 = 300 s$
and, speed of the bicycle = 2 m/s

It is known that,
Distance covered = Speed $\times$ time taken
$= 2 \times 900 = 1800 m$
= 1.8 km

7 . Show the shape of the distance-time graph for the motion in the following cases:

(i) A car moving with a constant speed.
(ii) A car parked on a side road.

Answer:

(i) A car is moving with a constant speed covers equal distance in equal interval of time. Therefore,
1653597264891 (ii) Distance-time graph of a car parked on a roadside is such that an increase in time, there is no change in distance.
1653597310214

8 . Which of the following relations is correct?

$(i) S p e e d = D i s t a n c e \times T i m e$

$(i i) S p e e d = \frac{D i s t a n c e}{T i m e}$

$(i i i) S p e e d = \frac{T i m e}{D i s t a n c e}$

$(i v) S p e e d = \frac{1}{D i s t a n c e \times T i m e}$

Answer:

The option (ii) is the correct answer.
Speed = distance /time , this relation holds correctly

9. The basic unit of speed is:

(i) km/min (ii) m/min (iii) km/h (iv) m/s

Answer:

The option (iv) is the correct answer.
The SI unit of speed is m/s (meter per second)

10. A car moves with a speed of 40 km/h for 15 minutes and then with a speed of 60 km/h for the next 15 minutes. The total distance covered by the car is:

(i) 100 km (ii) 25 km (iii) 15 km (iv) 10 km

Answer:

The correct option is (ii)
Answer = 25 km

1st case,
Speed = 40 km/hr
and time taken = 15 min $= 15 / 60 = 0.25 h$
So, distance = $0.25 \times 40 = 10 k m$

Case 2nd,
Speed = 60 km/h
Time taken = 15 min = 0.25 h
So, Distance travelled = $0.25 \times 60 = 15 k m$

So, total distance = 15 + 10 = 25 km

11. Suppose the two photographs, shown in Fig. 13.1 and Fig. 13.2, had been taken at an interval of 10 seconds. If a distance of 100 metres is shown by 1 cm in these photographs, calculate the speed of the fastest car.

car

Answer:

Given, Scale: $1 c m = 100 m$

The blue car is the fastest.
Distance travelled by the blue car in the picture= $2 c m$
$∴ 2 c m = 200 m$

Therefore, the actual distance travelled by blue car = $200 m$

Time taken = $10 s e c$

Speed of the blue car = $\frac{d i s t a n c e}{t i m e t a k e n}$

$= \frac{200}{10} = 20$

Therefore, the Speed of the blue car is $20 m / s$

12 . Fig. 9.15 shows the distance-time graph for the motion of two vehicles A and B. Which one of them is moving faster?

1596114827076

Answer:

It is known that,
(speed = distance/time)
It means greater the speed, more distance is covered by the car in a given interval of time.
From the graph, we can see that The distance covered by the vehicle A is more than the vehicle B. Hence A is faster than B.

13. Which of the following distance-time graphs shows a truck moving with a speed which is not constant?

132

1321

Answer:

The correct option is (iii)

In $d - t$ graphs, the constant speed is shown by a straight line.
A straight line parallels to the time axis, it indicates that the body is at rest.
A curved line on $d - t$ graphs means the speed is not constant.

The Class 7 Science Chapter 9 – Motion and Time solution includes 13 questions, with both multiple-choice and short-answer types. Experts at Careers360 have explained each answer step by step in simple and easy language. You can download the PDF for free and use it anytime, even without internet—super helpful for quick revision!

Chapter-wise NCERT Solutions of Class 7 Science

Get chapter-wise NCERT Solutions for Class 7 Science in simple language to help you understand concepts and ace your exams with ease! Science Chapter 9 class 7 solution contains chapter-wise solutions for class 7 science, which are given below:

Chapter 1	Nutrition in Plants
Chapter 2	Nutrition in Animals
Chapter 3	Fibre to Fabric
Chapter 4	Heat
Chapter 5	Acids, Bases and Salts
Chapter 6	Physical and Chemical Changes
Chapter 7	Weather, Climate and Adaptations of Animals to Climate
Chapter 8	Winds, storms and cyclones
Chapter 9	Soil
Chapter 10	Respiration in Organisms
Chapter 11	Transportation in Animals and Plants
Chapter 12	Reproduction in Plants
Chapter 13	Motion and Time
Chapter 14	Electric Current and its Effects
Chapter 15	Light
Chapter 16	Water: A Precious Resource
Chapter 17	Forests: Our Lifeline
Chapter 18	Wastewater Story

Class 7 Motion and Time NCERT Solutions: Important Formulas and Points

The numerical problems discussed in the NCERT solutions for Class 7 Science chapter 9 Motion and Time is based on the following formulas which are given below:

Motion - Motion is the change in position of an object with respect to its surroundings over time.

A stationary object is said to be at rest.

Types of Motion

Rectilinear motion: Motion along a straight line.
Circular motion: Motion in a circular path.
Periodic motion: Motion that repeats at regular intervals.

Uniform and Non-Uniform Motion

Uniform Motion: An object in uniform motion moves at a constant speed along a straight path, covering equal distances in equal intervals of time. For example, a car is driving steadily on a straight road.

Non-Uniform Motion: Non-uniform motion occurs when an object's speed changes as it moves along a straight path, covering unequal distances in equal intervals of time. For instance, a bicycle's speed fluctuates when riding through hilly terrain.

Distance and Displacement

Distance is the total path covered by an object.
Displacement is the change in position of an object from its initial position to the final position in a straight line.

Speed

Speed is a measure of how fast an object is moving. It represents the rate of change of distance with respect to time.

Average Speed

Average speed is calculated by dividing the total distance travelled by the object by the total time taken to cover that distance.
The formula for average speed is: Speed = Distance travelled / Time taken.
Units of Speed: SI (International System of Units) unit for speed is meters per second (m/s).

Measurement of Time

Time is typically measured in units such as seconds, minutes, and hours.
The time period refers to the total time taken by an object to complete one oscillation or cycle.

Units of Time

The primary units for measuring time are seconds, minutes, and hours.

Time period

$T = \frac{t o t a l t i m e t a k e n}{N o . o f o s c i l l a t i o n}$

Speedometer and Odometer

A speedometer records the speed of a vehicle directly in kilometres per hour (km/h). It provides real-time information about the vehicle's speed.
An odometer measures the total distance moved by the vehicle directly in kilometres (km). It keeps a cumulative record of the distance travelled over time.

Questions based on the distance-time graph are also discussed in the NCERT solutions for Class 7 Science chapter 13 Motion and Time.

Motion and Time Class 7 Science Chapter 9-Topics

Here are the topics covered in Class 7 Science Chapter 9 – Motion and Time:

Types of Motion (rectilinear, circular, and periodic motion)
Measurement of Time
Units of Time and Speed
Speed and its Calculation
Distance-Time Graphs
Standard Units and Instruments (like clocks and stopwatches)
Story of Timekeeping Devices
Practical Applications of Speed and Motion

Benefits of NCERT Motion and Time Class 7 Solutions

Here are the benefits of using NCERT Solutions for Class 7 Science Chapter 9 – Motion and Time:

Clear Understanding: Concepts like speed, time, and types of motion are explained in simple language.
Perfect for Homework: Helps you solve textbook questions quickly and correctly.
Exam Ready: Prepares you well for tests with step-by-step answers.
Doubt Clearing: It removes confusion with easy-to-follow explanations.
Free PDF Access: It can be downloaded and used offline anytime.
Time-Saving: It saves time during revision with ready-made solutions.
Concept Building: Builds a strong foundation for higher classes and competitive exams.

NCERT Solutions for Class 7 Subject Wise

Access subject-wise NCERT Solutions for Class 7 to get clear, step-by-step answers and boost your exam preparation for every subject!

NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7 Science

Also, check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

1. Whether the class 7 science chapter 13 motion and time is useful in higher studies ?

Yes, this chapter is very important for higher studies in the field of science and technology. The chapter motion is studied in Class 9 NCERT book and also in Class 11 NCERT Physics book

2. How many questions in NCERT Solutions for Class 7 Science chapter 13?

There are 13 questions in NCERT Solution Class 7 Science chapter 13

3. What are the topics covered in NCERT Class 7 Science chapter 13?

Here are the topics covered in NCERT Class 7 Science chapter 13

Slow or Fast
Speed
Measurement of Time
Units of Time and Speed
Measuring Speed
Distance-time Graph

4. Is class 7 science ch 13 question answer scoring ?

Yes, scoring well in Class 7 Science, Chapter 13, is possible with proper understanding and practice.

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NCERT Solutions for Class 7 Science Chapter 13 Motion and Time

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