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NCERT Solutions for Class 7 Science Chapter 8 Measurement of Time and Motion

NCERT Solutions for Class 7 Science Chapter 8 Measurement of Time and Motion

Edited By Vishal kumar | Updated on Jul 23, 2025 01:23 AM IST

Ever wondered how we can tell how far something moves or how quickly it happens? That is exactly what Class 7 Science Chapter 8 Motion and Time helps you learn. This chapter explains how different objects move, how to measure time accurately and how to calculate speed in easy ways. The NCERT Solutions for class 7 science chapter 8 are prepared to help you to understand these basic concepts easily.

These NCERT solutions for Class 7 Science provide complete answers to all textbook exercises and explain important topics like motion, speed, measurement of time and the Simple Pendulum. With step-by-step explanations, simple language and helpful examples, these NCERT solutions make studying easier and quicker. Along with this a clear method to approach and solve questions is included so you can practice better and perform very well in your school exams.

This Story also Contains
  1. NCERT Solutions for Class 7 Science Chapter 8: Exercise Solutions
  2. Measurement of Time and Motion Class 7 Science Chapter 8: Topics
  3. Approach to Solve the NCERT Class 7 Science Chapter 8 Questions
  4. NCERT Solutions for Class 7 Science – Chapter-Wise
NCERT Solutions for Class 7 Science Chapter 8 Measurement of Time and Motion
NCERT Solutions for Class 7 Science Chapter 8 Measurement of Time and Motion

NCERT Solutions for Class 7 Science Chapter 8: Exercise Solutions

Below are the detailed exercise solutions for Class 7 Science Chapter 8 is given. These answers are written in a simple and clear manner to help you understand the concepts better and prepare well for your exams.

Q1. Calculate the speed of a car that travels 150 metres in 10 seconds. Express your answer in km/h.

Answer:

Distance =150 m
Time taken =10 s

 Speed = Distance covered  Time taken =150 m10 s Speed in km/hr=15×185=54 km/h

Q2. A runner completes 400 metres in 50 seconds. Another runner completes the same distance in 45 seconds. Who has a greater speed and by how much?

Answer:

Runner 1:
Distance =400 m,
Time =50 s

 Speed = Distance covered  Time taken =400 m50 s=8 m/s


Runner 2:
Distance =400 m,
Time =50 s

 Speed = Distance covered  Time taken =400 m45 s=8.89 m/s


Difference =8.898=0.89 m/s
Hence, speed of runner 2 is greater by approximately 0.89 m/s.

Q3. A train travels at a speed of 25 m/s and covers a distance of 360 km. How much time does it take?

Answer:

 Speed =25 m/s Distance =360 km=3,60,000 m Time taken = Distance  Speed =3,60,0000 m25 m/s=14,400 s=240mins=4 hours 

Q4. A train travels 180 km in 3 h. Find its speed in:
(i) km/h
(ii) m/s
(iii) What distance will it travel in 4 h if it maintains the same speed throughout the journey?

Answer:

Distance =180 km, time =3 h
(i) Speed = Distance  Time =180 km3 h=60 m/h
(ii) Speed in m/s=60×518=16.677 m/s
(iii) Time =4 h, Speed =60 km/h

Distance = Speed × Time

=60×4=240 km

Q5. The fastest galloping horse can reach the speed of approximately 18 m/s. How does this compare to the speed of a train moving at 72 km/h?

Answer:

Speed of horse =18 m/s
Speed of train =72 km/h=72×518=20 m/s
The train is faster by 2m/s than the fastest galloping horse.

Q6. Distinguish between uniform and non-uniform motion using the example of a car moving on a straight highway with no traffic and a car moving in city traffic.
Answer:

Uniform motion: If an objects covers equal distances in equal distance in equal intervals of time, its motion is said to be uniform. A car moving on a straight highway with no traffic is an example of uniform motion.

Non-uniform motion: If an object covers unequal distances in equal interval of time, its motion is said to be non-uniform. A car moving in a traffic is an example of nonuniform motion.

Q7. Data for an object covering distances in different intervals of time are given in the following table. If the object is in uniform motion, fill in the gaps in the table.

Answer:

In uniform motion, object covers equal distances in equal intervals of time. Hence, the speed of an object remains constant throughout the motion.

The object covers 8 m in every 10 seconds. Hence, the speed of an object remains constant at 0.8 m/s throughout the motion.

Q8. A car covers 60 km in the fi rst hour, 70 km in the second hour, and 50 km in the third hour. Is the motion uniform? Justify your answer. Find the average speed of the car.

Answer:

The car covers different distances in each hour. Hence, the motion of the car is nonuniform.
Total distance =60 km+70 km+50 km=180 km
Total time =3 hours

 Average speed = Total distance travelled  Total time taken =180 km3 h=60 km/h


Hence, the average speed of the car is 60 km/h.

Q9. Which type of motion is more common in daily life — uniform or non-uniform? Provide three examples from your experience to support your answer.

Answer:

In our daily life, most motions are nonuniform because object do not move at the same speed all the time. Their speed changes due to factors like traffic, rough or uneven roads and other obstacles.

Examples:
• Travelling in a bus on an uneven road
• Playing cricket
• Walking through a crowded market

Q10. Data for the motion of an object are given in the following table. State whether the speed of the object is uniform or non-uniform. Find the average speed.

Answer:

The object exhibits non-uniform motion, because it covers

 Total distance travelled =60 m Total time taken =100 s Average speed = Total distance  Total time taken =60100=0.6 m/s

Q11. A vehicle moves along a straight line and covers a distance of 2 km. In the fi rst 500 m, it moves with a speed of 10 m/s and in the next 500 m, it moves with a speed of 5 m/s. With what speed should it move the remaining distance so that the journey is complete in 200 s? What is the average speed of the vehicle for the entire journey?

Answer:

Given
Total distance =2 km=2000 m,
Total time =200
Step 1: Time taken to cover the first 500 m

 Time = Distance  Speed =50010=50 s


Step 2: Time taken to cover the next 500 m

 Time =50010=100 s


Step 3: Remaining distance =20001000=1000 m

 Remaining time =200150=50 s


Step 4: Speed required to cover the remaining 1000 m

 Speed =100050=20 m/s


Step 5: Average speed = Totaldistance  Totaltimetaken =2000200=10 m/s

Measurement of Time and Motion Class 7 Science Chapter 8: Topics

Below are the important topics covered in Class 7 Science Chapter 8 Motion and Time. These topics will help you understand different types of motion, how to measure time, and calculate speed accurately

8.1 Measurement of Time
8.1.1 A simple pendulum
8.1.2 SI unit of time
8.2 Slow or Fast
8.3 Speed
8.4 Uniform and Non-uniform Linear Motion

Approach to Solve the NCERT Class 7 Science Chapter 8 Questions

To solve questions from Class 7 Science Chapter 8 start by identifying what the question is asking whether it is about measuring time, calculating speed, or identifying types of motion. For most of the numerical problems, use the formula:
Speed = Distance  Time 
Make sure all values are in the correct SI units (distance in meters, time in seconds, and speed in m/s ). For concept based questions, recall important terms like uniform motion non-uniform motion, and a simple pendulum.

NCERT Solutions for Class 7 Science – Chapter-Wise

In addition to the Class 7 Science chapter 8 question answer, students can access comprehensive chapter-wise solutions for Class 7 Science by clicking the link provided below:

NCERT Solutions for Class 7- Subject Wise

Also Check NCERT Books and NCERT Syllabus here

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

1. What topics are included in the NCERT curriculum Class 7 Science?

The NCERT curriculum Class 7 Science covers various chapters such as nutrition, heat transfer, plant reproduction, motion, and electric current, with exercises and activities to make learning interactive.

2. Where can I study from the Science textbook for Class 7 NCERT?

You can follow the Science textbook for Class 7 NCERT which is designed according to the CBSE syllabus and covers all important scientific concepts in simple language with examples.

3. What are some devices that measure time?

Some devices we can use to measure time are clocks, watches, stopwatches, sundials, water clocks, and sand timers.

4. Is the swing in the playground an example of periodic movement?

Yes, a swing moves back and forth on a regular basis and is a type of periodic motion, just like a simple pendulum

5. Does the weight of an object play a role in how fast it moves?

Not really. Speed is mainly dependent upon the force applied to the object and the surface it is travelling on. A heavy object might travel slower if not enough force is applied.

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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 3.8×107 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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