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NCERT Class 7 Science Chapter 8 Measurement of Time and Motion
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Measurement of Time and Motion
Chapter: 8
Page No – 110
Table 8.1 Time period of a simple pendulum
(Length of the string = 100 cm)
| S.No. | Time taken for 10 oscillations (seconds) | Time period (seconds) |
| 1 | ||
| 2 | ||
| 3 |
Ans: Time Period of a Simple Pendulum, you need to record the time taken for 10 oscillations and then calculate the time period using the formula:
Time period = Time taken for 10 oscillations/10
| S.No. | Time taken for 10 oscillations (seconds) | Time period (seconds) |
| 1 | 20 | 2.0 |
| 2 | 18 | 1.8 |
| 3 | 22 | 2.2 |
Page No- 114
Table 8.2: Finding the speed of trains.
Name of the railway station nearest to your place of stay __________.
| S.No. | Name of the train | Name of the next station | Distance till the next station (km) | Time taken till the next station (h) | Speed of the train between these two stations (km/h) |
Ans:
| S.No. | Name of the train | Name of the next station | Distance till the next station (km) | Time taken till the next station (h) | Speed of the train between these two stations (km/h) |
| 1 | Guwahati–Rajdhani Express | Rangiya Junction | 65 | 1 | 65 |
| 2 | Kamakhya Intercity Express | Goalpara Town | 122 | 2 | 60 |
| 3 | Saraighat Express | New Bongaigaon | 155 | 2.5 | 62 |
Page No – 117
Table 8.3: Distances travelled by two trains in equal time intervals of 10 minutes.
| Time (AM) | Train X | Train Y | ||
| 10:00 | 0 | 0 | 0 | 0 |
| 10:10 | 20 | 20 | 20 | 20 |
| 10:20 | 40 | 20 | 35 | 15 |
| 10:30 | 60 | 20 | 50 | 15 |
| 10:40 | 80 | 20 | 75 | 25 |
| 10:50 | 100 | 20 | 95 | 20 |
| 11:00 | 120 | 20 | 120 | 25 |
Ans: Train X covers equal distances in equal intervals of time, so it is in uniform linear motion while Train Y is in non-uniform linear motion.
| Let Us Enhance Our Learning |
1. Calculate the speed of a car that travels 150 metres in 10 seconds. Express your answer in km/h.
Ans: Distance = 150m
Time = 10 sec
Speed = Distance / Time
= 150 / 10
= 15 m/sec
Now, convert this speed from meters per second (m/s) to kilometers per hour (km/h).
we multiply the speed in m/s by 3600/1000 or 3.6:
Speed in km/h = Speed in m/s×3.6
Speed in km/h = 15 m/s×3.6
= 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?
Ans: First Runner:
Distance = 400 metres
Time = 50 seconds
Speed = Distance / Time
= 400 / 50
= 8 m/s
Second Runner:
Distance = 400 metres
Time = 45 seconds
Speed = Distance / Time
= 400 / 45
= 8.89m/sec
Difference in speed = 8.89 − 8 = 0.89 m/s
∴ The second runner is faster by approximately 0.89 m/s.
3. A train travels at a speed of 25 m/s and covers a distance of 360 km. How much time does it take?
Ans: Speed = 25 m/s
Distance = 360 km.
= 360 km = 360 × 1000 = 360,000 m
Time = Distance / Speed
= 360,000 / 25 = 14,400 seconds
= 14,400 s = 3600 /14,400 = 4 hours
The train takes 4 hours to cover 360 km at a speed of 25 m/s.
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?
Ans: Given:
Distance = 180 km
Time = 3 hours
(i) Speed in km/h
Speed = Distance/Time = 180/3 = 60 km/h
(ii) Speed in m/s
To convert km/h to m/s:
Speed in m/s = 60 ×1000/3600=60×5/18=16.67 m/s
(iii) Distance in 4 hours (if speed is same)
Distance = Speed × Time = 60 × 4 = 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?
Ans: To compare the speeds, we need to have them in the same units. Let’s convert the train’s speed from kilometers per hour (km/h) to meters per second (m/s).
We know that:
1 km = 1000 meters
1 hour = 3600 seconds
So, to convert 72 km/h to m/s,
72km/h = 72 × 1000/3600 = 20m/s
The speed of the train is 20 m/s.
compare the two speeds:
Speed of the fastest galloping horse ≈ 18 m/s
Speed of the train = 20 m/s
∴ the train moving at 72 km/h is slightly faster than the fastest galloping horse.
The train is 20 −18 = 2 m/s faster than the horse.
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.
Ans: Uniform motion – Motion where an object covers equal distances in equal intervals of time.
For example, a car moving on a straight highway with no traffic at a constant speed is said to be in uniform motion because it travels the same distance every second.
Non-uniform motion – Motion where an object covers unequal distances in equal intervals of time.
For Example – A car moving through city traffic is an example of non-uniform motion, as it frequently slows down, stops, and speeds up due to signals, turns, and congestion. Thus, while the car on the highway maintains a steady speed, the car in the city experiences variable speeds, clearly distinguishing uniform motion from non-uniform 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.
| Time (s) | 0 | 10 | 20 | 30 | 50 | 70 | ||
| Distance (m) | 0 | 8 | 24 | 32 | 40 | 56 |
Ans: Calculate Speed (Assuming Uniform Motion)
From time = 0 s to 70 s,
Distance covered = 56 m
Time taken = 70 s
Speed=Distance / Time = 56/70 = 0.8m/s
Speed to Fill in Missing Distances
Distance=Speed×Time
At 10 s: 0.8 × 10 = 8 m
At 20 s: 0.8 × 20 = 16 m (instead of 24 m)
At 30 s: 0.8 × 30 = 24 m
At 50 s: 0.8 × 50 = 40 m
At 70 s: 0.8 × 70 = 56 m
| Time(s) | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 |
| Distance (m) | 0 | 8 | 16 | 24 | 32 | 40 | 48 | 56 |
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.
Ans: A car covers in the first hour = 60km.
A car covers in the second hour = 70 km.
A car covers in the third hour = 50km.
Average speed of the car = Total Distance / total time.
Total distance = 60 km + 70 km + 50 km = 180 km.
Total time = 1 hr + 1 hr + 1 hr = 3 hours.
Average speed = 180/3 = 60km/hr.
∴ The average speed of the car is 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.
Ans: In daily life, non-uniform motion is more common than uniform motion. Most objects around us do not move at a constant speed or in a straight line for a long time. They speed up, slow down, stop, or change direction due to various factors like traffic, obstacles, or human control. Therefore, non-uniform motion is observed more frequently in our daily activities.
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.
| Time (s) | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 100 |
| Distances(s) | 0 | 6 | 10 | 16 | 21 | 29 | 35 | 42 | 45 | 55 | 60 |
Ans: The motion of the object is non-uniform because it covers unequal distances in equal intervals of time.
Average speed= Total distance/Total time
Total distance = 60 m
Total time = 100 s
Average speed= 60/ 100 = 0.6m/s
The average speed of the object is 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?
Ans: Given:
Total distance = 2 km = 2000 m
First 500 m at 10 m/s
Next 500 m at 5 m/s
Remaining distance = 2000 – (500 + 500) = 1000 m
Total time for journey = 200 s
Time taken for first 500 m:
t1= 500/10 = 50s
Time taken for next 500 m:
t2= 500/5 = 100s
Time left for remaining 1000 m:
Time left = 200 − (50+100)=50 s
Speed needed for remaining 1000 m in 50 s:
V = 1000 /50 = 20m/s
Average speed for entire journey
Average speed = Total distance/Total time = 2000/200 = 10m/s
The average speed of the vehicle for the entire journey is 10 m/s.
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