NCERT Solutions for Class 7 Science Curiosity Chapter 8 Time and Motion

NCERT Solutions for Class 7 Science Curiosity Chapter 8 Measurement of Time and Motion help students understand important topics like how time was measured in the past, types of motion, speed, and how to calculate it.

These solutions are based on the latest NCERT Class 7 Science Syllabus and cover all the in-text and exercise questions in a simple and clear way. With the help of these solutions, students can easily follow the concepts and strengthen their understanding of motion and time-related topics.

NCERT Solutions for Class 7 Science Curiosity Chapter 8 Introduction

Chapter 8 of Class 7 Science Curiosity, Measurement of Time and Motion, introduces students to the basic concepts of how time and motion are measured. The chapter begins by exploring how people measured time before modern clocks existed, using devices like sundials and water clocks.

It then explains different types of motion, uniform and non-uniform and how to measure speed using distance and time. Students learn to calculate average speed and compare speeds for different moving objects. Through real-life examples, activities, and simple calculations, the chapter helps build a strong foundation in understanding how motion is observed and measured scientifically. This chapter is essential for developing logical thinking and problem-solving skills related to time and movement.

Class 7 Science Chapter 8 Measurement of Time and Motion Question Answer

Below are the solutions for Class 7 Science Chapter 8  Measurement of Time and Motion: 

Activity 8.1: Let us construct  Take a used transparent plastic bottle (1/2 litre or larger) with its cap.  Cut it into two, roughly in the middle as shown in Fig. 8.5a.  

Using a drawing pin, make a small hole in the cap of the bottle (Fig. 8.5b). Fig. 8.5: Making a simple water clock (a) (b) (c) (d)  Place the upper part of the bottle in an inverted position over the lower half (Fig. 8.5c).  

Fill the upper part of the bottle with water. You may add a few drops of ink or colour to make the water level easily visible (Fig. 8.5d).

The water will start dripping into the lower part of the bottle. Using a watch, mark the level of water after every one minute till all the water drips down.

 Answer:

Objective:
To make a basic time-measuring device using a plastic bottle, demonstrating how time was measured in ancient times.

What You Did:
You created a simple water clock by:

• Cutting a plastic bottle in half.

• Making a small hole in the cap.

• Inverting the top half into the bottom half.

• Filling the upper portion with colored water.
  
  

Observation: The water started dripping through the hole at a steady rate. By using a watch or clock to mark water levels every minute, you noticed that the water level rose in equal steps over equal intervals of time.

Conclusion:This activity helped you understand that in ancient times, people used the steady flow of water (like in this water clock) to measure short intervals of time. It also showed how time can be measured using natural, consistent processes.

Activity 8.2: Let us experiment  Collect a piece of string around 150 cm long, a heavy metal ball with hook/ a stone (bob), a stopwatch/ watch, and a ruler.  

Tie the bob at one end of the string. 

Fix the other end of the string to a rigid support such that the length of the string in between support and bob is around 100 cm.

 Wait for the bob to come to rest. Your pendulum is now ready. 

Gently hold the bob, move it slightly to one side and release it. Take care not to push the bob while releasing it and that the string is taut. Is your pendulum now oscillating?  Using a watch, measure the time it takes for the pendulum to complete 10 oscillations. Record the time in Table 8.1.  Repeat this activity 3–4 times. 

Divide the time taken for 10 oscillations by 10 to calculate the time period of your pendulum. Note it down in Table 8.1. Table 8.1 Time period of a simple pendulum Length of the string = 100 cm S.No. Time taken for 10 oscillations (seconds) 1. 2. 3. Time period (seconds)

Answer:

Objective:
To observe the oscillatory motion of a simple pendulum and calculate its time period.

Materials Required:

• 150 cm string

• Metal ball or stone (bob)

• Ruler

• Stopwatch or watch
  
  

Procedure:

1. Tie the bob to one end of the string.

2. Fix the other end of the string to a rigid support, ensuring the length between the support and the bob is 100 cm.

3. Let the bob come to rest.

4. Gently pull the bob to one side (without pushing it) and release it.

5. Start your stopwatch and record the time taken for 10 complete oscillations.

6. Repeat the activity 3–4 times for accuracy.

7. Divide each recorded time by 10 to get the time period (time for one oscillation).

Conclusion: The average time period of the simple pendulum is approximately 2 seconds. This activity helps us understand oscillatory motion and how a pendulum moves back and forth in equal intervals of time. This principle is used in pendulum clocks to measure time.

Activity 8.3: Let us identify  Look at the wall clock shown in Fig. 8.9 carefully. What is the smallest interval of time you can measure with it? One second is the smallest interval of time that we can measure using this clock. 

Answer:

 The smallest interval of time that we can measure using this clock is one second, as the second hand completes one full tick every second. This allows us to track the passage of time in one-second intervals.

Conclusion: This activity helps us understand how to read a standard wall clock and recognize the smallest unit of time it can measure accurately, which is one second.

 Activity 8.4: Let us calculate  Look up at the railway timetable on the internet.  Identify a train stopping at the railway station nearest to your place of stay.  Find out the name of the next station where this train stops. Also, fi nd the distance to that station as given in the timetable.  Note the time at which the train departs from your station and arrives at the next station. Find the diff erence to calculate the time taken by the train to cover the distance till the next station.  Calculate the speed of the train between the two stations and record it in Table 8.2.  Repeat for 4–5 diff erent types of trains (Passenger/ Express/ Superfast).

 Answer:

Conclusion: This activity gives a practical approach to understanding the relationship between distance, time, and speed. It also encourages students to use online tools and timetables to gather and interpret real data effectively.

 (In-Text Questions):

Question 1. How was time measured when there were no clocks and watches?
Answer:
Before clocks and watches were invented, time was measured using natural and simple devices such as sundials, water clocks, hourglasses, and candle clocks. These instruments relied on consistent natural events like the movement of the sun or the flow of water.

Question 2. For races covering the same distance, we can tell who was faster by measuring time. But how can we tell that when comparing races for different distances?
Answer:
When races cover different distances, we cannot compare speed by time alone. Instead, we use the formula for speed (Speed = Distance ÷ Time) to determine who was faster by calculating how much distance each person covered per unit of time.

Question 3. I once watched a part of a marathon on a straight road. I noticed that some people ran at the same speed, while others sped up or slowed down. How were their motions different?
Answer:
Runners moving at a constant speed showed uniform motion, where equal distances are covered in equal time intervals. Those who sped up or slowed down had non-uniform motion, where the distance covered in equal time intervals varies.

 Let Us Enhance Our Learning 

Question 1.
Calculate the speed of a car that travels 150 metres in 10 seconds. Express your answer in km/h.
Answer:

• Distance = 150 m

• Time = 10 s

• Speed in m/s = 150 ÷ 10 = 15 m/s

• Speed in km/h = 15 × (18 ÷ 5) = 54 km/h

Question 2. 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:

• Speed = 400 ÷ 50 = 8 m/s

Runner 2:

• Speed = 400 ÷ 45 ≈ 8.89 m/s

Difference in speed = 8.89 − 8 = 0.89 m/s
So, Runner 2 is faster by 0.89 m/s.

Question 3. A train travels at a speed of 25 m/s and covers a distance of360 km. How much time does it take?
Answer:
Distance = 360 km = 360,000 m

• Speed = 25 m/s

• Time = 360,000 ÷ 25 = 14,400 seconds
  Convert to minutes: 14,400 ÷ 60 = 240 minutes = 4 hours

Question 4. 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 4h if it maintains the same speed throughout the journey?
Answer:

• Distance = 180 km
  
  

• Time = 3 

(i) Speed = 180 ÷ 3 = 60 km/h

(ii) Speed in m/s = 60 × (1000 ÷ 3600) = 16.67 m/s

(iii) Distance in 4 h = 60 × 4 = 240 km

Question 5. The fastest galloping horse can reach a speed of approximately 18 m/s. How does this compare to a train moving at 72 km/h?
Answer:

• Speed of train in m/s = 72 × (5 ÷ 18) = 20 m/s

• Difference = 20 − 18 = 2 m/s

The train is faster by 2 m/s.

Question 6. 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: When an object covers equal distances in equal time intervals, it is said to be in uniform motion.
  Example: A car moving steadily on a straight highway.

• Non-uniform Motion: When an object covers unequal distances in equal time intervals, its motion is non-uniform.
  Example: A car moving in city traffic where speed varies due to signals and congestion.

Question 7. 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.

Question 8. A car covers 60 km in the first hour, 70 km in the second hour, and 50 km in the third hour. Is the motion uniform? Justify your Answer:

Answer:

• The car covers different distances in equal time intervals (1 hour each).

• Therefore, the motion is non-uniform.

To find the average speed:

• Total Distance = 60 km + 70 km + 50 km = 180 km

• Total Time = 3 hours

Average Speed = Total Distance ÷ Total Time = 180 ÷ 3 = 60 km/h

Question 9. 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 daily life, non-uniform motion is more common because objects rarely move at a constant speed. Their speed changes due to various factors like traffic, obstacles, road conditions, or interactions with people.

Examples:

• Traveling in a bus on a bumpy or uneven road.

• Playing cricket, where players speed up or slow down frequently.

• Walking through a crowded market where pace constantly changes.

Question 10. 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:

• Since the object covers unequal distances in equal intervals of time, the motion is non-uniform.

Total Distance Travelled = 6 + 4 + 6 + 5 + 8 + 6 + 7 + 3 + 10 + 5 = 60 m
Total Time Taken = 100 s

Average Speed = Total Distance ÷ Total Time =
= 60 ÷ 100 = 0.6 m/s

Question 11. A vehicle moves along a straight line and covers a distance of 2 km. In the first 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:

• Total distance = 2 km = 2000 m

• Total time = 200 s

Step 1: Time to cover first 500 m at 10 m/s

Time = Distance ÷ Speed = 500 ÷ 10 = 50 s

Step 2: Time to cover next 500 m at 5 m/s

Time = 500 ÷ 5 = 100 s

Step 3: Remaining distance

= 2000 – 1000 = 1000 m
Remaining time = 200 – (50 + 100) = 50 s

Step 4: Required speed for remaining 1000 m

Speed = Distance ÷ Time = 1000 ÷ 50 = 20 m/s

Step 5: Average speed for entire journey

Average speed = Total Distance ÷ Total Time = 2000 ÷ 200 = 10 m/s

NCERT Solutions for Class 7 Science Curiosity Chapter 8 PDF Download

Students looking to strengthen their understanding of the topic Measurement of Time and Motion can easily download the NCERT Solutions for Class 7 Science Curiosity Chapter 8 in PDF format from the link below. 

Benefits of Using NCERT Solutions for Class 7 Science Curiosity Chapter 8

1. The solutions explain complex topics like speed, time, uniform and non-uniform motion, and average speed in a simple and easy-to-understand manner.

2. Numerical questions are solved with proper steps and units, helping students learn the correct approach to problem-solving.

3. All textbook activities and questions are covered, including in-text and end-of-chapter questions.

4. These solutions are aligned with the CBSE syllabus and follow the latest NCERT guidelines, making them highly useful for exam revision.

5. Students can quickly refer to these answers for homework or revision without spending extra time searching for correct responses.

6. Practicing with accurate solutions helps build confidence in answering application-based and conceptual questions.

7. Ideal for students who are preparing independently or revising at home without tutoring support.