**Question 1.**

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

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

**Solution:**

- oscillatory
- straight line
- circular
- oscillatory
- oscillatory
- 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?