A. Fill in the blanks

1. An object is said to be at rest if it does not change its position with time.
2. The SI unit of time is second (s).
3. A child sitting in a revolving giant wheel is an example of circular motion.
4. A car moving on a busy straight road is an example of non-uniform motion.
5. The speedometer of a motorbike measures its speed in kilometres per hour (km/h).

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B. Write True or False

1. The speed of a fast-moving train is usually measured in metre per hour. – False
2. The average speed remains constant for an object having a uniform motion. – True
3. A man walks for 1 minute at a speed of 1 m/s along a straight track. The total distance covered by him is 1 m. – False
4. An object moving along a straight line is said to be in uniform motion if it covers regularly increasing distances in equal intervals of time. – False
5. The time period of a simple pendulum that takes 42 seconds to complete 20 oscillations is 2.1 seconds. – True
6. The distance-time graph for a car kept parked on a side road is a straight line parallel to the time axis. – True

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C. Tick (✓) the correct option.

1. Out of the following, the only correct formula is—

• ☐ distance × time = speed
• ☑ distance = speed × time
• ☐ time = speed/distance
• ☐ speed = time/distance

Solution:
We know that:

Distance = Speed × Time

Therefore, the correct option is:

☑ distance = speed × time

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2. A man walks on a straight road from his home to market 3 km away with a speed of 6 km/h. The time taken by the man to go from his home to market equals—

• ☐ 18 minutes
• ☐ 20 minutes
• ☑ 30 minutes
• ☐ 40 minutes

Solution:
Distance = 3 km
Speed = 6 km/h$$\text{Time}=\frac{\text{Distance}}{\text{Speed}} =\frac{3}{6}\text{ hour} =\frac{1}{2}\text{ hour}$$Time=SpeedDistance​=63​ hour=21​ hour $$\frac{1}{2}\text{ hour}=30\text{ minutes}$$21​ hour=30 minutes

Therefore, the correct option is:

☑ 30 minutes

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3. The odometer of a car reads 57,321.0 km when the clock shows the time as 8:30 a.m. The odometer reading changes to 57,330.0 km at time 8:50 a.m. The distance moved by the car, in these 20 minutes, equals—

• ☐ 15 km
• ☐ 21 km
• ☑ 9 km
• ☐ 36 km

Solution:
Initial odometer reading = 57,321 km
Final odometer reading = 57,330 km$$\text{Distance travelled} =57,330-57,321 =9\text{ km}$$Distance travelled=57,330−57,321=9 km

Therefore, the correct option is:

☑ 9 km

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4. Out of the following distance-time graphs, the graph that represents a truck at rest is—

• ☑ A
• ☐ B
• ☐ C
• ☐ D

Solution:
A truck at rest does not change its position with time. Hence, its distance remains constant and the distance-time graph is a horizontal straight line, which is shown in Graph A.

Therefore, the correct option is:

☑ A

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5. In the given diagram of a simple pendulum, the time taken by the bob to move from X to Z is $t_2-t_1$t2​−t1​, then the time period of this simple pendulum is—

• ☐ $(t_2+t_1)$(t2​+t1​)
• ☑ 2(t2−t1)2(t_2-t_1)2(t2​−t1​)
• ☐ $\dfrac{(t_2+t_1)}{2}$2(t2​+t1​)​
• ☐ $4(t_2+t_1)$4(t2​+t1​)

Solution:
The motion from X to Z represents half an oscillation of the pendulum.

Time taken for half an oscillation = $t_2-t_1$t2​−t1​

Therefore, the time period (time for one complete oscillation) is:$$T=2(t_2-t_1)$$T=2(t2​−t1​)

Hence, the correct option is:

☑ 2(t2−t1)2(t_2-t_1)2(t2​−t1​)

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6. The SI unit of speed is—

• ☐ km/minute
• ☐ m/minute
• ☐ km/hour
• ☑ m/second

Solution:
In the SI system, distance is measured in metres (m) and time is measured in seconds (s). Therefore, the SI unit of speed is:$$\text{Speed}=\frac{\text{metre}}{\text{second}}=m/s$$Speed=secondmetre​=m/s

Hence, the correct option is:

☑ m/second (m/s)

D. Answer the following questions in brief.

1. A boy walks to his school with a constant speed of 4 km/h and reaches there in 30 minutes. Find the distance of the school from his house.

Solution:

Given:

• Speed of the boy = 4 km/h
• Time taken = 30 minutes = 1/2 hour

We know that,

Distance = Speed × Time$$\text{Distance} = 4 \times \frac{1}{2} = 2 \text{ km}$$Distance=4×21​=2 km

Answer: The distance of the school from the boy’s house is 2 km.

Explanation:
According to the Tata DAV Class 7 Science book, when an object moves with a constant speed, the distance covered can be calculated by multiplying speed by time. Since the boy walks at the same speed throughout his journey, he covers 2 km in 30 minutes.

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2. The distance between two stations is 216 km. A bus takes 4 hours to cover this distance. Calculate the average speed of the bus in km/h.

Solution:

Given:

• Distance = 216 km
• Time taken = 4 hours

We know that,

Average Speed = Distance ÷ Time$$\text{Average Speed} = \frac{216}{4} = 54 \text{ km/h}$$Average Speed=4216​=54 km/h

Answer: The average speed of the bus is 54 km/h.

Explanation:
The Tata DAV book explains that average speed is obtained by dividing the total distance travelled by the total time taken. Here, the bus covers 216 km in 4 hours, so its average speed is 54 km/h.

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3. Two cars, A and B, starting at the same time from the same point, are moving with average speeds of 40 km/h and 50 km/h respectively. At the end of one hour, find the distances covered by the cars.

Solution:

For Car A:

Speed = 40 km/h
Time = 1 hour$$\text{Distance} = 40 \times 1 = 40 \text{ km}$$Distance=40×1=40 km

Distance covered by Car A = 40 km

For Car B:

Speed = 50 km/h
Time = 1 hour$$\text{Distance} = 50 \times 1 = 50 \text{ km}$$Distance=50×1=50 km

Distance covered by Car B = 50 km

Answer:

• Distance covered by Car A = 40 km
• Distance covered by Car B = 50 km

Explanation:
The Tata DAV Class 7 Science chapter states that if two objects move for the same amount of time, the object with the greater speed covers a greater distance. Since Car B moves faster than Car A, it travels 10 km more in one hour.

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4. A car moves with a speed of 60 km/h for 15 minutes and then with a speed of 60 km/h for the next 15 minutes. Find the total distance covered by the car in these 30 minutes.

Solution:

Total time = 30 minutes = 1/2 hour

Speed = 60 km/h

Using the formula,$$\text{Distance} = \text{Speed} \times \text{Time}$$Distance=Speed×Time$$\text{Distance} = 60 \times \frac{1}{2} = 30 \text{ km}$$Distance=60×21​=30 km

Answer: The total distance covered by the car in 30 minutes is 30 km.

Explanation:
The speed remains the same during both intervals, so the car is in uniform motion. As explained in the Tata DAV book, an object moving with uniform speed covers equal distances in equal intervals of time. Therefore, the car covers a total distance of 30 km in half an hour.

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5. Define the term ‘Periodic motion’. Give two examples of periodic motions that can be used to measure time.

Answer:
Periodic motion is the motion that repeats itself after equal intervals of time.

Examples:

1. The oscillations of a simple pendulum.
2. The rotation of the Earth about its axis.

Explanation:
According to the Tata DAV Class 7 Science book, many natural and man-made clocks depend on periodic motions because they occur at regular intervals. The swinging of a pendulum and the Earth’s rotation are examples of periodic motions that help us measure time accurately.

E. Answer the following questions.

1. A farmer moves along the boundary of a rectangular field ABCD as shown in the figure. He takes 4 minutes to travel across each side. Is the motion of the farmer uniform or non-uniform? Find his average speed over one complete round of the field.

Solution:

From the figure,

• Length of the field = 80 m
• Breadth of the field = 40 m
• Time taken to travel each side = 4 minutes

Step 1: Find the total distance covered in one round.

The farmer covers the boundary of the field, which is its perimeter.$$\text{Perimeter} = 2(l+b)$$Perimeter=2(l+b)$$= 2(80+40)$$=2(80+40)$$= 2 \times 120 = 240 \text{ m}$$=2×120=240 m

Therefore, the total distance covered in one round is 240 m.

Step 2: Find the total time taken.

There are four sides and the farmer takes 4 minutes to cross each side.$$\text{Total time} = 4 \times 4 = 16 \text{ minutes}$$Total time=4×4=16 minutes

Step 3: Find the average speed.

$$\text{Average Speed}=\frac{\text{Total Distance}}{\text{Total Time}}$$Average Speed=Total TimeTotal Distance​$$=\frac{240}{16}=15 \text{ m/min}$$=16240​=15 m/min

Nature of Motion

The farmer covers 80 m in 4 minutes on the longer sides and 40 m in 4 minutes on the shorter sides. Thus, he covers unequal distances in equal intervals of time.

Hence, his motion is non-uniform.

Answer:

• Motion of the farmer: Non-uniform motion
• Average speed: 15 m/min

Explanation:
According to the Tata DAV Class 7 Science book, an object is in uniform motion only when it covers equal distances in equal intervals of time. Since the farmer covers different distances in the same time, his motion is non-uniform.

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2. During the rainy season, show that the thundering sound of clouds was heard 6 seconds after the lightning was seen by him. If the speed of sound is 340 m/s, find the distance of the point where the thundering sound was produced.

Solution:

Given:

• Speed of sound = 340 m/s
• Time taken = 6 s

We know that,

Distance = Speed × Time$$\text{Distance}=340 \times 6$$Distance=340×6$$=2040 \text{ m}$$=2040 m

Answer: The thunder was produced at a distance of 2040 m or 2.04 km from the observer.

Explanation:
As explained in the Tata DAV book, light travels much faster than sound. Therefore, we see lightning almost immediately, whereas the sound of thunder reaches us after a few seconds. Here, the delay of 6 seconds indicates that the thunder occurred 2040 m away.

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3. (a) How can we make a simple pendulum?

Answer:

A simple pendulum can be made by tying a small heavy object, such as a metal bob or a stone, to one end of a light, inextensible thread and suspending the other end from a fixed support so that the bob can swing freely.

Explanation:
According to the Tata DAV Class 7 Science book, a simple pendulum consists of a bob, a light string, and a fixed support. When the bob is displaced and released, it starts oscillating to and fro.

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3. (b) A simple pendulum takes 10 seconds to complete 5 oscillations. Find the time period of this pendulum.

Solution:

Given:

• Time taken for 5 oscillations = 10 s
• Number of oscillations = 5

We know that,$$\text{Time Period}=\frac{\text{Total Time}}{\text{Number of Oscillations}}$$Time Period=Number of OscillationsTotal Time​$$T=\frac{10}{5}=2 \text{ s}$$T=510​=2 s

Answer: The time period of the pendulum is 2 seconds.

Explanation:
The Tata DAV book defines the time period of a pendulum as the time taken to complete one oscillation. Since the pendulum takes 10 seconds for 5 oscillations, it takes 2 seconds for one oscillation.

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4. Carefully examine the data given below for motion of two different objects A and B. State whether the motion of these objects is uniform.

Solution:

Object A

Time	Distance (km)
7:00 a.m.	0
7:30 a.m.	20
7:45 a.m.	30
8:00 a.m.	40
8:15 a.m.	50
8:30 a.m.	60

After 7:30 a.m., the object covers 10 km every 15 minutes, and its speed remains constant.

Therefore, Object A shows uniform motion.

Object B

Time	Distance (km)
7:00 a.m.	0
7:30 a.m.	10
7:45 a.m.	24
8:00 a.m.	35
8:15 a.m.	38
8:30 a.m.	43

The distances covered in equal intervals of time are different.

Therefore, Object B shows non-uniform motion.

Answer:

• Object A – Uniform motion
• Object B – Non-uniform motion

Explanation:
According to the Tata DAV Class 7 Science book, an object is said to be in uniform motion if it covers equal distances in equal intervals of time. Object A satisfies this condition, whereas Object B does not.

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5. Observe the graph given here. From the graph, find (i) the values of distances ‘P’ and ‘Q’ in m and (ii) the value of time ‘T’ in seconds.

Solution:

From the graph:

• At 4 s, the distance covered is 3 m.

Therefore,$$P=3 \text{ m}$$P=3 m

• At 12 s, the distance covered is 9 m.

Therefore,$$T=12 \text{ s}$$T=12 s

• At 16 s, the distance covered is 12 m.

Therefore,$$Q=12 \text{ m}$$Q=12 m

Answer:

• $P = 3$P=3 m
• $Q = 12$Q=12 m
• $T = 12$T=12 s

Explanation:
The distance-time graph is a straight line passing through the origin, showing uniform motion. The values of $P$P, $Q$Q, and $T$T are obtained by reading the corresponding points directly from the graph.

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