NCERT Solutions for Class 7 Science (Curiosity) Chapter 8: Measurement of Time and Motion (NCERT 2026–27)

These Class 7 Science Curiosity Chapter 8 solutions cover Measurement of Time and Motion from the new NCF-2023 textbook (2026–27). The chapter explains how time has been measured from ancient sundials and water clocks to modern atomic clocks, introduces the simple pendulum and its time period, defines speed and its units, and explores uniform and non-uniform linear motion. Below you will find every “Let Us Enhance Our Learning” exercise question solved step by step, with all numeric answers verified.

Class: 7 Subject: Science Book: Curiosity Chapter: 8 Topic: Time, Pendulum, Speed & Motion Session: 2026–27
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Class 7 Science Curiosity Chapter 8 Solutions – Overview

Chapter 8 of Curiosity, Measurement of Time and Motion, traces how humans learned to keep track of time. Early people noticed repeating natural events — sunrise and sunset, the phases of the Moon and the seasons — and used them to build calendars. To measure shorter intervals within a day they invented sundials, water clocks, hourglasses and candle clocks. The discovery that a simple pendulum of a given length always takes the same time for one oscillation (its time period) led to the pendulum clock, and modern quartz and atomic clocks use far more rapid, regular vibrations. The chapter then defines speed as the distance covered in unit time, gives its SI unit (m/s) and the relation between speed, distance and time, and finally distinguishes uniform linear motion (constant speed, equal distances in equal times) from non-uniform linear motion (changing speed).

Key Concepts & Formulae

Time period of a pendulum: the time taken by the pendulum to complete one oscillation. It depends only on the length of the pendulum, not on the mass of the bob, and is constant at a place for a given length.

One oscillation: the bob moving from its mean position O to extreme position A, then to extreme position B, and back to O (or from A to B and back to A).

SI unit of time: the second (symbol s). 60 s = 1 min; 60 min = 1 h.

Speed: the distance covered by an object in unit time.
Speed = Total distance covered ÷ Total time taken. SI unit = metre/second (m/s); also km/h.

Rearranged forms: Distance = Speed × Time, and Time = Distance ÷ Speed.

Unit conversions: 1 km = 1000 m, 1 h = 3600 s. To change m/s → km/h, multiply by 3.6; to change km/h → m/s, divide by 3.6.

Uniform linear motion: motion along a straight line with constant speed (equal distances in equal time intervals). Non-uniform linear motion: motion along a straight line with changing speed (unequal distances in equal time intervals).

“Let Us Enhance Our Learning” — NCERT Solutions

All questions below are reproduced verbatim from the NCERT Curiosity (Grade 7) textbook, Chapter 8. Answers are original and the calculations have been independently verified.

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

ANSWER Speed = Total distance ÷ Total time = 150 m ÷ 10 s = 15 m/s. Converting to km/h: 15 m/s × 3.6 = 54 km/h. (Check: 15 m/s = (15 × 3600) m per hour = 54000 m/h = 54 km/h.)

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 Speed of first runner = 400 m ÷ 50 s = 8 m/s. Speed of second runner = 400 m ÷ 45 s = 8.89 m/s (approximately). The second runner has the greater speed, by about 8.89 − 8 = 0.89 m/s (roughly 0.9 m/s). The second runner is faster because she covers the same distance in less time.

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

ANSWER Time = Distance ÷ Speed. First make the units match: 360 km = 360 × 1000 m = 360000 m. Time = 360000 m ÷ 25 m/s = 14400 s. Converting to hours: 14400 s ÷ 3600 = 4 h (i.e. 14400 seconds, or 4 hours).

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 4 h if it maintains the same speed throughout the journey?

ANSWER (i) Speed = Distance ÷ Time = 180 km ÷ 3 h = 60 km/h. (ii) 60 km/h ÷ 3.6 = 16.67 m/s (approximately). (Check: 60 km = 60000 m, 3 h = 10800 s; 60000 ÷ 10800 = 16.67 m/s.) (iii) Distance = Speed × Time = 60 km/h × 4 h = 240 km.

5. 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 To compare them fairly, convert both to the same unit. Train speed = 72 km/h ÷ 3.6 = 20 m/s. Horse speed = 18 m/s; train speed = 20 m/s. So the train (20 m/s) is faster than the horse (18 m/s) by 2 m/s.

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 A car moving on a clear, straight highway with no traffic can keep a constant speed, covering equal distances in equal intervals of time. This is an example of uniform linear motion. A car moving in city traffic keeps speeding up and slowing down (and stopping at signals), so its speed keeps changing and it covers unequal distances in equal intervals of time. This is an example of non-uniform linear motion.

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 From the given data, at 10 s the object covers 8 m, so its constant speed = 8 m ÷ 10 s = 0.8 m/s. For uniform motion, distance = 0.8 × time. At 20 s: 0.8 × 20 = 16 m.   At 70 s: 0.8 × 70 = 56 m. (The values 24 m at 30 s, 32 m at 40 s, 40 m at 50 s are already consistent with 0.8 m/s.)
Time (s)0102030405070
Distance (m)081624324056
Note: the textbook table prints Time as 0, 10, 20, 30, 50, 70; reading the data row (8, —, 24, 32, 40, 56) with a 40 s column makes the uniform pattern of 0.8 m/s clear — the two gaps to fill are 16 m and 56 m.

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. Find the average speed of the car.

ANSWER Is it uniform? No. In equal time intervals (each of 1 hour) the car covers unequal distances — 60 km, 70 km and 50 km. Since the distances are not equal, the speed keeps changing, so the motion is non-uniform. Average speed = Total distance ÷ Total time = (60 + 70 + 50) km ÷ 3 h = 180 km ÷ 3 h = 60 km/h.

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 Non-uniform motion is far more common in daily life. Real objects seldom move at exactly the same speed for long, because they speed up, slow down and stop. Three examples: (i) a bus on a city road that keeps stopping at signals and bus stops; (ii) a person walking who slows at crossings and speeds up on a clear path; (iii) a bicycle going down a slope (speeding up) and then up a slope (slowing down). (Your own everyday examples are accepted.)

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
Time (s)0102030405060708090100
Distance (m)06101621293542455560
Uniform or non-uniform? In equal 10 s intervals the object covers different distances (6 m, then 4 m, then 6 m, then 5 m, and so on). Because the distances covered in equal time intervals are unequal, the speed is non-uniform. Average speed = Total distance ÷ Total time = 60 m ÷ 100 s = 0.6 m/s.

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. Distance already given = 500 m + 500 m = 1000 m, so the remaining distance = 2000 − 1000 = 1000 m. Time for first 500 m = 500 ÷ 10 = 50 s. Time for next 500 m = 500 ÷ 5 = 100 s. Time used so far = 50 + 100 = 150 s. Time available for the remaining 1000 m = 200 − 150 = 50 s. Required speed = 1000 m ÷ 50 s = 20 m/s. Average speed for the whole journey = Total distance ÷ Total time = 2000 m ÷ 200 s = 10 m/s.

Extra Practice Questions

Short Answer Type Questions

Q1. Name any four time-measuring devices used in ancient times.

ANSWERSundials, water clocks, hourglasses and candle clocks.

Q2. What is meant by the time period of a simple pendulum?

ANSWERIt is the time taken by the pendulum to complete one oscillation. For a pendulum of a given length, the time period is constant at a place.

Q3. State the SI unit of time and write the relations between the second, minute and hour.

ANSWERThe SI unit of time is the second (symbol s). 60 s = 1 min and 60 min = 1 h.

Q4. Define speed and give its SI unit.

ANSWERSpeed is the total distance covered by an object divided by the total time taken to cover it. Its SI unit is metre per second (m/s).

Q5. On what factors does the time period of a simple pendulum depend, and on what does it not depend?

ANSWERThe time period depends on the length of the pendulum (and the place), but it does not depend on the mass of the bob. All pendulums of the same length have the same time period at a given location.

Long Answer Type Questions

Q1. Describe how you would set up a simple pendulum and measure its time period.

ANSWERTie a metal bob (or a stone) to one end of a string and fix the other end to a rigid support so that the length between the support and the bob is about 100 cm. Wait for the bob to come to rest at its mean position. Gently move the bob a little to one side and release it without pushing, keeping the string taut, so it begins oscillating. Using a stopwatch, measure the time taken for 10 oscillations and repeat the measurement 3–4 times. Divide the time for 10 oscillations by 10 to get the time period. The readings are almost the same each time, which shows that the time period of a pendulum of a given length is constant at a place.

Q2. Explain the difference between uniform and non-uniform linear motion with the help of the data given for two trains X and Y travelling between 10:00 AM and 11:00 AM in equal 10-minute intervals.

ANSWER When an object moves along a straight line, its motion is linear. If it covers equal distances in equal intervals of time, it moves with a constant speed and is in uniform linear motion. If it covers unequal distances in equal intervals of time, its speed keeps changing and it is in non-uniform linear motion.
Time (AM)Train X distance in 10 min (km)Train Y distance in 10 min (km)
10:00–10:102020
10:10–10:202015
10:20–10:302015
10:30–10:402025
10:40–10:502020
10:50–11:002025
Train X covers 20 km in every 10-minute interval, so it is in uniform linear motion. Train Y covers unequal distances (20, 15, 15, 25, 20, 25 km) in equal intervals, so it is in non-uniform linear motion, even though both trains cover 120 km in one hour.

Q3. Trace the development of timekeeping devices from ancient to modern times.

ANSWERAncient people first used repeating natural events — sunrise and sunset, the Moon’s phases and the seasons — to make calendars and mark a day. To measure smaller intervals they built sundials (using the Sun’s moving shadow), water clocks (using water flowing out of or into a vessel), hourglasses (using flowing sand) and candle clocks (using a marked burning candle). From the fourteenth century, mechanical clocks driven by weights, gears and springs appeared. The invention of the pendulum clock by Christiaan Huygens in the seventeenth century, based on Galileo’s study of pendulums, was a major breakthrough. Today, quartz clocks use the rapid vibrations of a quartz crystal and atomic clocks use the vibrations of specific atoms; atomic clocks are so precise that they lose only one second in millions of years.

MCQs & Assertion–Reason

1. Which of the following is NOT an ancient time-measuring device?

(a) Sundial    (b) Water clock    (c) Atomic clock    (d) Candle clock

2. The SI unit of time is the:

(a) minute    (b) hour    (c) second    (d) day

3. The time taken by a pendulum to complete one oscillation is called its:

(a) frequency    (b) time period    (c) amplitude    (d) speed

4. The time period of a simple pendulum depends on its:

(a) mass of the bob    (b) colour    (c) length    (d) material of the thread

5. A speed of 36 km/h is equal to:

(a) 6 m/s    (b) 10 m/s    (c) 12 m/s    (d) 36 m/s

6. Speed is calculated as:

(a) time ÷ distance    (b) distance × time    (c) distance ÷ time    (d) distance + time

7. An object covering equal distances in equal intervals of time along a straight line is in:

(a) non-uniform motion    (b) uniform linear motion    (c) circular motion    (d) oscillatory motion

8. The instrument in a vehicle that displays its speed in km/h is the:

(a) odometer    (b) thermometer    (c) speedometer    (d) barometer

9. The pendulum clock was invented by:

(a) Galileo Galilei    (b) Christiaan Huygens    (c) Aryabhata    (d) Varahamihira

10. A car travels 100 m in 5 s. Its speed is:

(a) 5 m/s    (b) 20 m/s    (c) 50 m/s    (d) 500 m/s

Answer key: 1-(c), 2-(c), 3-(b), 4-(c), 5-(b), 6-(c), 7-(b), 8-(c), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: The time period of a simple pendulum of a given length is constant at a place.

Reason: A pendulum’s oscillatory motion is periodic, repeating its path after a fixed interval of time.

A-R 2. Assertion: The time period of a pendulum changes when the mass of the bob is changed.

Reason: The time period of a simple pendulum depends only on its length, not on the bob’s mass.

A-R 3. Assertion: A car moving in heavy city traffic is usually in non-uniform motion.

Reason: In city traffic the car keeps speeding up and slowing down, so it covers unequal distances in equal time intervals.

A-R 4. Assertion: To compare a speed in m/s with a speed in km/h, the two must first be converted to the same unit.

Reason: A speed of 72 km/h is equal to 20 m/s.

A-R 5. Assertion: A sundial can be used to measure time at night.

Reason: A sundial determines time from the changing position of a shadow cast by the Sun during the day.

Answer key: 1-(B), 2-(D), 3-(A), 4-(B), 5-(C).

Common Mistakes to Avoid

Watch out for these

  • Forgetting to convert units before calculating — always make distance and time match (e.g. change km to m and h to s) before dividing.
  • Confusing the m/s ↔ km/h conversion — multiply by 3.6 for m/s → km/h, and divide by 3.6 for km/h → m/s.
  • Thinking the pendulum’s time period depends on the bob’s mass — it depends only on the length.
  • Writing ‘sec’ for second or ‘hrs’ for hour — the correct symbols are s, min and h, written in lowercase and singular, with no full stop.
  • Mixing up the speedometer (measures speed) with the odometer (measures distance travelled).
  • Assuming average speed is the simple average of two speeds — always use total distance divided by total time.

Quick Revision Summary

  • Ancient clocks — sundials, water clocks, hourglasses and candle clocks — used repeating processes; modern quartz and atomic clocks use rapid, regular vibrations.
  • The time period of a pendulum is the time for one oscillation; it is constant for a given length and does not depend on the bob’s mass.
  • The SI unit of time is the second (s); 60 s = 1 min, 60 min = 1 h.
  • Speed = total distance ÷ total time; SI unit m/s, also km/h (multiply m/s by 3.6 to get km/h).
  • Uniform linear motion: constant speed, equal distances in equal times. Non-uniform linear motion: changing speed, unequal distances in equal times.
  • A speedometer shows a vehicle’s speed; an odometer shows the distance it has travelled.

How to score full marks in this chapter

In every numerical, write the formula first, convert all quantities to consistent units, then substitute and show each step with the correct unit on the answer. Remember the handy fact that 1 m/s = 3.6 km/h. For “uniform or non-uniform?” table questions, compare the distance covered in each equal time interval — equal distances mean uniform motion. Always find average speed as total distance ÷ total time, never by averaging the individual speeds.

Frequently Asked Questions

What is Class 7 Science Curiosity Chapter 8 about?

Chapter 8, Measurement of Time and Motion, explains how time was measured from ancient sundials and water clocks to modern atomic clocks, introduces the simple pendulum and its time period, defines speed and its units, and distinguishes uniform from non-uniform linear motion.

What is the time period of a simple pendulum?

The time period is the time taken by a pendulum to complete one oscillation. It is constant for a pendulum of a given length at a place and does not depend on the mass of the bob.

How do you convert m/s to km/h?

Multiply the speed in m/s by 3.6 to get km/h (because 1 km = 1000 m and 1 h = 3600 s, so 1 m/s = 3.6 km/h). To go the other way, divide the km/h value by 3.6.

Are these Class 7 Science Curiosity Chapter 8 solutions free?

Yes. All solutions are free and follow the official NCERT Curiosity textbook for session 2026–27, with every exercise question solved and the calculations verified.

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