Question 1.
Classify the following as motion along a straight line, circular or oscillatory motion:

1. Motion of your hands while running.
2. Motion of a horse pulling a cart on a straight road.
3. Motion of a child in a merry-go-round.
4. Motion of a child on a see-saw.
5. Motion of the hammer of an electric bell.
6. Motion of a train on a straight bridge.

Solution:

1. oscillatory
2. straight line
3. circular
4. oscillatory
5. oscillatory
6. straight line

Question 2.
Which of the following are not correct?
(i) The basic unit of time is second.
(ii) Every object moves at a constant speed.
(iii) Distances between two cities are measured in kilometers.
(iv) The time period of a given pendulum is not constant.
(v) The speed of a train is expressed in m/h.
Solution:
(ii), (v)

Question 3.
A simple pendulum takes 32 s to complete 20 oscillations. What is the time period of the pendulum?
Solution:
Given here,
No. of oscillations = 20
Total time taken = 32 sec.
We know that the time period of a pendulum is the time taken by it to complete one oscillation. Thus,
Time period = TotaltimetakenNo.ofoscillations=32sec/20=1.6seconds
Therefore, the time period of this pendulum will be 1.6 s.

Question 4.
The distance between the two stations is 240 km. A train takes 4 hours to cover this distance. Calculate the speed of the train.
Solution:
Here, it is given that,
The distance between two stations = 240 km
Time is taken to cover this distance = 4 hr.
Now, Speed = DistanceTime=240Km/4hr=60Km/h
Therefore, the speed of the train will be 60 km/h.

Question 6.
Salma takes 15 minutes from her house to reach her school on a bicycle. If the bicycle has a speed of 2 m/s, calculate the distance between her house and the school.
Solution:
According to the question,
Speed of the bicycle = 2 m/s
Total time taken = 15 min = 900 sec.
We know that,
The distance covered = Speed x Time
= 2 m/s x 900 sec.
= 1800 m
Therefore, the distance between her house and the school will be 1800 m or 1.8 km.

Question 7.
Show the shape of the distance-time graph for the motion in the following cases :
(i) A car moving at a constant speed.
(ii) A car parked on a side road.
Solution:
(i) A car moving with a constant speed covers equal distance in equal intervals of time

Question 10.
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. The total distance covered by the car is:
(i) 100 km
(ii) 25 km
(iii) 15 km
(iv) 10 km
Solution:
Distance travelled in first 15 min
= speed x time
= 40 km/h x 15 min
= 40 km/h x 15/60 h = 10 km
Distance travelled in last 15 min
= speed x time
= 60 km/h x 15 min
= 60 km/h x 15/60 h = 15 km
Total distance = (10 +15) km = 25 km
Hence, option (ii) is correct.

Question 11.
Suppose the two photographs, shown in Fig. 13.1 and Fig. 13.2, had been taken at an interval of 10 seconds. If a distance of 100 meters is shown by 1 cm. in these photographs, calculate the speed of the blue car.
Solution:
Speed = 100 m/10 s = 10 m/s

Question 12.
shows the distance-time graph for the motion of two vehicles A and B. Which one of them is moving faster?