**1) Classify the following as motion along a straight line,**

circular or oscillatory motion:

circular or oscillatory motion:

**(i) Motion of your hands while running.**

**(ii) Motion of a horse pulling a cart on a straight road.**

**(iii) Motion of a child in a merry-go-round.**

**(iv) Motion of a child on a see-saw.**

**(v) Motion of the hammer of an electric bell.**

**(vi) Motion of a train on a straight bridge.**

ANSWER: –

(i) Oscillatory

(ii) Straight Line

(iii) Circular

(iv) Oscillatory

(v) Oscillatory

(vi) Straight Line

**2)Which of the following are not correct?**

**(i) The basic unit of time is second.**

**(ii) Every object moves with a constant speed.**

**(iii) Distances between two cities are measured in kilometres.**

**(iv) The time period of a given pendulum is not constant.**

**(v) The speed of a train is expressed in m/h.**

ANSWER: –

(i) True

(ii) False. Different objects have different

speeds.

speeds.

(iii) True

(iv) False. The time period of a given pendulum is fixed.

(v) False. The speed of train is usually expressed in km/hr or miles/hr.

**3) A simple pendulum takes 32 s to complete 20 oscillations.**

What is the time period of the pendulum?

What is the time period of the pendulum?

ANSWER: – Time Period = Total Time Taken ÷ No. of Oscillations

No. of Oscillations = 20

Total Duration = 32s

Time Period = 32 ÷ 20 = 1.6s

**4) The distance between two stations is 240 km. A train takes 4**

hours to cover this distance. Calculate the speed of the train.

hours to cover this distance. Calculate the speed of the train.

ANSWER: – Distance between two stations = 240 km

Time taken by train to cover the distance = 4 hours

Speed = Distance ÷ Time

Speed of Train = 240 / 4 = 60km/hour.

**5) The odometer of a car reads 57321.0 km when the clock shows**

the time 08:30 AM. What is the distance moved by the car, if at 08:50 AM, the

odometer reading has changed to 57336.0 km? Calculate the speed of the car in

km/min during this time. Express the speed in km/h also.

the time 08:30 AM. What is the distance moved by the car, if at 08:50 AM, the

odometer reading has changed to 57336.0 km? Calculate the speed of the car in

km/min during this time. Express the speed in km/h also.

ANSWER:- Odometer reading at 8.30 AM (O1) =57321.0 km

Odometer reading at 8.50 AM (O2) = 57336.0 km

Distance covered by car = O 2 – O1 = 57336.0 –

57321.0 = 15 Km

Time interval between 8.30 AM to 8.50 AM = 20 min.

Speed of car = Distance ÷ Time = 15km ÷ 20 min

= 0.75 km/min

1 Hr = 60 min.

Speed of Car in km/hr = 0.75 × 60 = 45 km/hr.

**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.

a bicycle. If the bicycle has a speed of 2 m/s, calculate the distance between

her house and the school.

ANSWER: – Distance = Speed ×Time

Speed of the bicycle = 2 m/s

Time taken to reach school = 15 min = 15 × 60seconds = 900s

Distance = 2 × 900 = 1800 m = 1800/1000Km = 1.8 Km

**7) Show the shape of the distance- time graph for the motion in**

the following cases:

the following cases:

**(i) A car moving with a constant speed.**

**(ii) A car parked on a side road.**

ANSWER: i-

ii-

**8) Which of the following relations is correct?**

**(i) Speed = Distance × Time**

**(ii) Speed=Distance/time**

**iii) Speed=time/distance**

**iv) Speed=1/distance*time**

ANSWER: – speed=distance/time

**9) The basic unit of speed is:**

**(a) km/min**

**(b) m/min**

**(c) km/h**

**(d) m/s**

ANSWER:- (d) m/s

**10) A car moves with a speed of 40 km/h for 15minutes and then**

with a speed of 60 km/h for the next 15 minutes. The total distance covered by

the car is:

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**

ANSWER: – Case I

Speed of the car = 40 km/h

Time taken = 15 min =

Distance covered, d1 = Speed × Time taken = 40 × 0.25 = 10 km

Case II

Speed of the car = 60 km/h

Time taken = 15 min =

Distance covered, d2 = Speed × Time taken = 60× 0.25 = 15 km

Total distance covered by the car, d = d1 + d2 =10 + 15 = 25 km

Therefore, the total distance covered by the car is 25 km.

**11) Suppose the two photographs, shown in Figure 1 and Figure 2,**

had been taken at an interval of 10 seconds. If a distance of 100 metres is

shown by 1 cm in these photographs, calculate the speed of the blue car.

had been taken at an interval of 10 seconds. If a distance of 100 metres is

shown by 1 cm in these photographs, calculate the speed of the blue car.

ANSWER: – 1 cm = 100 m

Distance covered by blue car = 2.0 cm = 2.0 ×100 = 200 m

Time taken to cover 200m = 10s

Speed of Car = Distance ÷ Time = 200 / 10 =20m/s

Speed of Car (in km/hr) = 20 × 3600s/1000m =72 km/hr

**13)Which of the following distance- time graphs shows a truck**

moving with speed which is not constant?

moving with speed which is not constant?

ANSWER: – Graph (iii)