Go through BSE Odisha Class 7 Science Solutions and Class 7 Science Chapter 8 Measurement of Time and Motion Question Answer to understand textbook questions more clearly.
Class 7 Science Curiosity Chapter 8 Question Answer
Science Curiosity Class 7 Chapter 8 Question Answer
Measurement of Time and Motion Class 7 Questions and Answers
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Question 1.
Calculate the speed of a car that travels 150 metres in 10 seconds. Express your answer in km/h.
Answer:
The speed of a car that travels 150 m in 10 s is
Speed = \(\frac{\text { Distance }}{\text { Time }}\) = \(\frac{150}{10}\) = 15 m/s
Convert metre per second to kilometre per hour, as follows:
= \(\frac{15 \mathrm{~m} \times \frac{\mathrm{km}}{1000 \mathrm{~m}}}{1 \mathrm{~s} \times \frac{\mathrm{h}}{3600 \mathrm{~s}}}\) = 54 km/h
[Trick: To convert m/s to km/h, multiply by \(\frac{18}{5}\)]
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:
Distance covered by each runner =400 m
Time taken by first runner = 50s
Time taken by second runner = 45s
Speed of first runner = \(\frac{400}{50}\) = 8m/s
Speed of second runner = \(\frac{400}{45}\) = 8.89m/s
Difference in speed = 8.89-8=0.89 m/s
Therefore, the second runner is faster, having a greater speed by 0.89 m/s
Question 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:
To convert m/s to km/h, multiply by \(\frac{18}{5}\):
So, speed = 25 × \(\frac{18}{5}\) = 90 km/h
Time taken = \(\frac{\text { Distance }}{\text { Speed }}\)=\(\frac{360}{90}\) = 4 hours
Hence, the train takes 4 hours to cover a distance of 360 km.
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 4 h if it maintains the same speed throughout the journey?
Answer:
(i) Speed in km/h:
Speed = \(\frac{\text { Distance }}{\text { Time }}\)= \(\frac{180}{3}\)=60km/h
(ii) Speed in m/s:
To convert km/h to m/s, we multiply by \(\frac{5}{18}\). Thus, 60 × \(\frac{5}{18}\)=16.67 m/s
(iii) Distance travelled in 4 h:
Distance = Speed × Time = 60 × 4 = 240km
Hence, the train’s speed is 60 km/h or 16.7 m/s, and it will travel 240 km in 4 hours.
Question 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 convert km/h into m/s, multiply by \(\frac{5}{18}\)
Speed of train in m/s=72 km /h × \(\frac{5}{18}\) =20m/ s
Thus, the train moves faster than the horse.
Difference in their speeds =20-18=2m/s.
Hence, the horse’s speed is 2 m/s less than the train’s speed.
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:
A car moving on a straight highway with no traffic covers equal distances in equal intervals of time. Its speed remains constant throughout the journey. Such motion, where the object moves at a constant speed along a straight line, is called uniform linear motion. On the other hand, a car moving in city traffic slows down and speeds up frequently because of traffic signals and other vehicles on the road. So it covers unequal distances in equal intervals of time. Such motion, where the speed keeps changing, is called non-uniform motion.
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:
The motion is uniform, which means the object covers equal distances in equal intervals of time. From the table, in every 10 -second interval, the object covers 8 metres. Hence, the distances increase by 8 m for each 10s.
| Time (S) | 0 | 10 | 20 | 30 | 40 | 50 | 70 |
| Distance(m) | 0 | 8 | 16 | 24 | 32 | 40 | 56 |
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. Find the average speed of the car.
Answer:
Given: Total time = 3h
The car covers unequal distances (60km, 70km, and 50 km ) in equal time intervals of one hour each. Hence, the motion is non-uniform because equal intervals of time do not correspond to equal distances.
Average speed = \(\frac{\text { Total distance travelled }}{\text { Total time taken }}\)
= \(\frac{60+70+50}{3}\) = 60km/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 most objects do not move at a constant speed for long durations. Their speeds keep changing due to various factors such as road conditions, obstacles, or the need to stop and start. Examples:
- A car moving through traffic, which slows down and speeds up repeatedly.
- A bicycle on a road slows down at turns and speeds up on straight paths.
- A vehicle moving on an uneven or bumpy road experiences irregular changes in speed.
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:
From the table, the distances covered in equal intervals of 10 seconds are not the same. For example, between 0-10 s, the object covers 6 m , and between 10-20s, it covers 4 m. Thus, the object’s motion is non-uniform.
Average speed = \(\frac{\text { Total distance travelled }}{\text { Total time taken }}\)
= \(\frac{60}{100}\) =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 10m/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 = 2km=2000m
Total time for completing the journey =200 s
Time taken to cover the first part:
t1= \(\frac{\text { Distance of first part }}{\text { Speed in first part }}\)=\(\frac{500}{10}\)=50s
Time taken to cover the second part:
t1= \(\frac{\text { Distance of first part }}{\text { Speed in first part }}\)=\(\frac{500}{5}\)=100s
Total time taken so far =50+100=150s
Time left =200-150=50s
Remaining distance =2000-(500+500)=1000m
Speed required for the remaining part:
Speed = \(\frac{\text { Remaining distance }}{\text { Time left }}\)=\(\frac{1000}{50}\) = 20m/s
Now, the average speed of the vehicle for the entire journey is given by:
Average speed = \(\frac{\text { Total distance travelled }}{\text { Total time taken }} \)
= \(\frac{2000}{200}\) = 10m/s
Thus, the vehicle must move at 20 m/s} for the remaining 1000 m; its average speed for the entire journey is 10 m/s.
Class 7 Curiosity Chapter 8 Question Answer
InText Questions
Question 1.
How was time measured when there were no clocks and watches? [Page 106]
Answer:
People used repeating natural events such as sunrise, sunset, moon phases and seasons. Later, simple devices like sundials, water clocks, hourglasses and candle clocks were used to measure shorter intervals of time.
Question 2.
Why does the Samrat Yantra sundial require correction to match Indian Standard Time? [Page 107]
Answer:
The Samrat Yantra at Jantar Mantar, Jaipur, is a huge stone sundial whose shadow moves at a steady rate and can measure very small time intervals as short as 2 seconds, though it shows local solar time rather than standard clock time. Local solar time depends on the Sun’s actual position in the sky at a specific place, whereas IST is a uniform time standard adopted for the entire country.
Question 3.
Why were early water clocks not very accurate? [Page 108]
Answer:
Early water clocks were not very accurate because their time measurement depended on the rate of water flow, which was difficult to keep constant. The flow changed as the water level in the container fell. Since there was no method to keep the flow of water constant, the clocks could not measure equal intervals of time accurately.
Question 4.
Does the mass of the bob affect the time period of a pendulum? [Page 110]
Answer:
No, the time period is independent of the bob’s mass. For small oscillations, it depends only on the length of the pendulum.
Question 5.
How can we compare the speeds of the runners in different races? [Page 112]
Answer:
The speeds of runners in different races can be compared by calculating their average speeds in the respective races.
Question 6.
How do we decide who is running faster in an ongoing race? [Page 113]
Answer:
The runner who has covered more distance in the same time is running faster.
Question 7.
When speed varies during a journey, what type of speed is calculated using total distance and total time? [Page 115]
Answer:
The speed calculated using the total distance covered and the total time taken is called the average speed.