# DAV Class 7 Science Book Solutions Chapter 6

DAV Class 7 Science Book Solutions Chapter 6: Here you will get the complete question answer for DAV Class 7 Science Book Chapter 6 Motion and Time. With the help of these solutions for DAV Class 7 Science the Living World book, you can easily grasp basic concepts better and faster.

## DAV Class 7 Science Book Solutions Chapter 6

Highlights

A. Fill in the blanks.

1. An object is said to be at rest if it ___________ change its position with time.

2. The SI unit of time is ___________.

3. A child, sitting in a revolving giant wheel, is an example of a ___________ motion.

4. A car, moving on a busy straight road, is an example of ___________ motion.

5. The speedometer of a motorbike measures its speed in ___________.

Answer: (1) does not (2) second (3) circular (4) non uniform or linear (5) km/h

B. Write True or False for the following statements.

1. The speed of a fast moving train is usually measured in metre per hour.

2. The average speed remains constant for an object having a uniform motion.

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.

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.

5. The time period of a simple pendulum, that takes 42 seconds to complete 20 oscillations, equals 2.1 seconds.

6. The distance-time graph, for a car kept parked on a side road, is a straight line parallel to the time axis.

Answer: (1) False (2) True (3) False (4) False (5) True (6) True

C. Tick the correct option.

1. distance = speed x time

2. 30 minutes

3. 15 km

4. option (a)

5. 4 (t1+t2)

6. m/second

D. Answer the following questions in brief.

Q. 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.

Speed = 4 km/h

Time taken = 30 minutes = ½ h = 0.5 h

Therefore,

Distance = speed x time

= 4 km/h x ½ h

= 2 km

Hence, distance of school from his house is 2 km.

Q. 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/hour.

Q. 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, in the same direction. Find how far will Car B be from Car A after 3 hours.

Q. 4. A car moves with a speed of 40 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.

Distance covered in 1st 15 minutes = 40 km/h x 15 minutes

= 40 km/h x ¼ h

= 10 km

Distance covered in 2nd 15 minutes = 60 km/h x 15 minutes

= 60 km/h x ¼ h

= 15 km

Therefore, Total distance covered = 10 km + 15 km

= 25 km

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

Periodic motion – motion, which repeats itself after a regular interval of time, is called periodic motion.

Examples: (1) motion of the hands of a clock.

(2) motion of the moon around the earth.

Q. 6. Distinguish between uniform and non-uniform motion. Give one example of each.

Uniform motion

An object is said to be in uniform motion when it moves along a straight path, and covers equal distances in equal intervals of time, howsoever small these intervals of time may be.

Example: Motion of a body along a straight path covering equal distances in equal intervals of time.

Non-uniform motion

When an object, (1) moving along a straight path, covers unequal distances in equal intervals of time or (2) does not move along a straight path, its motion, is called a non-uniform motion.

Example: Motion of a free-falling object.

Q. 7. Draw the shape of distance-time graph for:

(a) a man, waiting for a bus, standing at one point, on a bus-stand.

(b) a man, walking on a level, straight and narrow road, with a constant speed.

Motion of the farmer – non-uniform, as he covers unequal distances in equal intervals of time.

Average speed of the farmer = total distance covered/time taken

= 240 m/16 min.

= 240 m / 960 seconds {change min. into second}

= ¼ m/s

= 0.25 m/s

Speed of sound = 340 m/s

Time taken = 6 seconds

Therefore, distance = 340 m/s x 6 seconds

= 2040 m

= 2.04 km

Hence, thundering sound was produced at a distance of 2.04 km from Shivam.

(a) A simple pendulum can be made by a metal ball attached to taut light string, or thread, that is fixed rigidly at one end.

(b) Number of complete oscillations = 5

Total time taken = 10 seconds

Time period of the pendulum = 10/5 = 2 seconds

Object A is in uniform motion as it covers equal distances in equal intervals of time.

Object B is in non-uniform motion as it does not cover equal distances in equal intervals of time

(1)     Value of ‘P’ = 3 m, Value of ‘Q’ = 12 m

(2)     Value of ‘R’ = 12 seconds

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