1) Classify the following as motion along a straight line, 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.
(ii) Straight Line
(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.
(ii) False. Different objects have different speeds.
(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?
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.
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.
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.
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:
(i) A car moving with a constant speed.
(ii) A car parked on a side road.
8) Which of the following relations is correct?
(i) Speed = Distance × Time
ANSWER: - speed=distance/time
9) The basic unit of speed is:
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:
(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
Speed of the car = 60 km/h
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.
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?
ANSWER: - Graph (iii)