NCERT Solutions for Class 7 Science Chapter 13 Motion and Time

Edited By Vishal kumar | Updated on Feb 16, 2024 06:19 PM IST

NCERT Solutions for Class 7 Science Chapter 13 Motion and Time: Welcome to the updated NCERT Solutions for Class 7! On this page from Careers360, you will find comprehensive class 7 science chapter 13 question answer for exercises crafted by subject experts in simple language. Additionally, these motion and time class 7 ncert solutions are accessible in PDF format, allowing students to use them anytime, anywhere, without any internet constraints.

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 13-Topics
Benefits of NCERT Motion and Time Class 7 Solutions

NCERT solutions for class 7 Science chapter 13 Motion and Time helps students in solving the homework questions. Earlier you have learnt about different types of motions. You learned the motion along a straight line, circular path and periodic motion. The first question explained in the NCERT solutions for Class 7 Science chapter 13 Motion and Time is based on these classifications. You know that some vehicles move faster than others. If you know the distance covered by two buses in one hour, it is easy to tell which one is faster. The distance covered by an object per unit time as the speed. All you will study in this chapter is the speed and time measurements. NCERT Solutions for Class 7 Science chapter 13 Motion and Time is helpful in scoring good marks in the examination.

Motion and time class 7 solutions is an essential tool for the preparation of the exam. Students are recommended to go through NCERT Solutions for Class 7 to score high marks in their examinations. If you have any problem in answering the exercise questions or you are not getting the correct answers then don't worry, NCERT solutions for Class 7 Science Chapter 13 Motion and Time is there for you to help.

**As per the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 9 in Class 7 Science.

Free download class 7 science chapter 13 ncert solutions PDF for CBSE exam.

Download PDF of NCERT Solutions for Class 7 Science Chapter 13 Motion and Time

Download PDF

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)
$T=\frac{total\ time\ taken}{No.\ of\ oscillation}$
$T=\frac{32}{20} = 1.6 s$

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 = $0.75 \times 60 = 45\ km/hr$

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
$=2\times900 =1800\ m$
= 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?

$(i)Speed=Distance\times Time$

$(ii)Speed=\frac{Distance}{Time}$

$(iii)Speed=\frac{Time}{Distance}$

$(iv)Speed=\frac{1}{Distance\times Time}$

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 = $0.25\times 40 =10 km$

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

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: $1\ cm = 100\ m$

The blue car is the fastest.
Distance travelled by the blue car in the picture= $2\ cm$
$\therefore 2\ cm = 200\ m$

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

Time taken = $10\ sec$

Speed of the blue car = $\frac{distance}{time\ taken}$

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

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

12 . Fig. 13.15 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 position.
A curved line on $d-t$ graphs means the speed is not constant.

Solution for motion and time class 7 consists of a total of thirteen questions that are a mix of multiple-choice and short-answer types. Experts at Careers360 have created a step-by-step class 7 science chapter 13 question answer in easy-to-understand language. Students can access and download the PDF version of the solutions for free of cost and use them offline anytime and anywhere according to their convenience.

Chapter-wise NCERT Solutions of Class 7 Science

Science Chapter 13 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 13 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 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

$T=\frac{total\ time\ taken}{No.\ of\ oscillation}$

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 13-Topics

Topics for motion and time class 7 questions and answers are given below:

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

Benefits of NCERT Motion and Time Class 7 Solutions

Motion and time class 7 ncert solutions simplify homework and revision tasks, making them more manageable.
These class 7 science chapter 13 question answer help students prepare for exams by addressing similar types of questions that are likely to appear in their class assessments.
Utilizing class 7 science chapter 13 ncert solutions is instrumental in achieving good scores in examinations, thanks to the comprehensive and clear explanations they offer.
Class 7 motion and time ncert solutions offer students the benefit of free access, making them an affordable and accessible resource for all learners.
These motion and time class 7 questions and answers serve as an excellent self-study resource, allowing students to independently practice and reinforce their knowledge of the chapter's content.

NCERT Solutions for Class 7 Subject Wise

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

Also Read

NCERT Solutions for Class 7 Science Chapter 18 Wastewater Story

NCERT Solutions for Class 7 Science Chapter 17 Forests Our Lifeline

NCERT Solutions for Class 7 Science Chapter 16 Water A Precious Resource

NCERT Solutions for Class 7 Science Chapter 14 Electric Current and Its Effects

NCERT Solutions for Class 7 Science Chapter 13 Motion and Time

NCERT Solutions for Class 7 Science Chapter 12 Reproduction in Plants

Articles

How to Track NSP Payment Status – Step-by-Step Guide for Scholarship Updates

Nov 14, 2024

What is OTR in NSP? – Comprehensive Guide to One Time Registration

Nov 14, 2024

Best Hobbies to Mention in Resume - [Updated for 2024]

Nov 13, 2024

Silverzone IOEL Answer Key 2024 – Download International Olympiad of English Language Solutions

Nov 12, 2024

Allen Tallentex Answer Key 2025, Download TALLENTEX 2025 Answer Key Online/Offline Exam

Nov 12, 2024

Banking Courses After 12th - Top Colleges, Fees, Salary, Scope

Nov 11, 2024

TOSS Time Table 2025 for SSC & Inter: Telangana Open School Society Date Sheet for April Exam

Nov 11, 2024

Aakash ANTHE Exam 2024-25, Check ANTHE Scholarship Dates, Admit Card, Exam Timings Here

Nov 09, 2024

Upcoming School Exams

National Institute of Open Schooling 12th Examination

Admit Card Date:04 October,2024 - 29 November,2024

Application Process

Exam Pattern

Admit Card

National Institute of Open Schooling 10th examination

Admit Card Date:04 October,2024 - 29 November,2024

Application Process

Exam Pattern

Mock Test

Maharashtra School Leaving Certificate Examination

Application Date:07 October,2024 - 19 November,2024

Exam Pattern

Admit Card

Result

Mizoram Board of School Education 12th Examination

Application Date:07 October,2024 - 22 November,2024

Exam Pattern

Result

Mock Test

Mizoram Board 10th Examination

Application Date:07 October,2024 - 22 November,2024

Exam Pattern

Admit Card

Result

View All School Exams

Get answers from students and experts

Popular Questions

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

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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 3.8×10⁷J 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