THESE NOTES ARE PREPARED ACCORDING TO JKBOSE SYLLABI IN SIMPLIFIED AND CONCISE PATTERN
Exercises
1. Classify the following as motion along a straight line, circular or oscillatory motion:
I. Motion of your hands while running.
Ans: Oscillatory motion.
II. Motion of a horse pulling a cart on a straight road.
Ans: Straight line motion.
III. Motion of a child in a merry-go-round.
Ans: Circular motion.
IV. Motion of a child on a see-saw.
Ans: Oscillatory motion.
V. Motion of the hammer of an electric bell.
Ans: Oscillatory motion.
VI. Motion of a train on a straight bridge.
Ans: Straight line motion.
2. Which of the following are not correct:
I. The basic unit of time is second. Correct
II. Every object moves with a constant speed. Incorrect
III. Distance between two cities are measured in kilometers. Correct
IV. The time period of a given pendulum is not constant. Incorrect
V. The speed of a train is expressed in m/h. Incorrect
3. A simple pendulum takes 32 s to complete 20 oscillations. What is the time period of the pendulum?
Sol:
Number of Oscillations: 20
Time taken: 32 s
=> Time Period = Time taken
No of Oscillations
= 32/20
= 1.6 seconds.
4. The distance between two stations is 240 km. A train takes 4 hours to cover this distance. Calculate the speed of the train.
Sol:
Distance = 240 km
Time taken = 4 hrs
=> Speed of train = Distance
Time
= 240/4
= 60 km/h
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.
Sol:
Initial reading on odometer = 57321 km
Final reading on odometer = 57336 km
Distance moved by the car = 57336 - 57321
= 15 km.
Now,
Distance moved by car = 15 km
Time taken by car = 8.50 - 8.30 = 20 min
= 60 km/h
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.
Sol:
Initial reading on odometer = 57321 km
Final reading on odometer = 57336 km
Distance moved by the car = 57336 - 57321
= 15 km.
Now,
Distance moved by car = 15 km
Time taken by car = 8.50 - 8.30 = 20 min
=>Speed of Car in Km/min = 15
20
= 0.75 km/min
Speed of car in km/h = 15
20/60
= 15 × 60
20
= 45 km/h
6. Geeta 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?
Sol=
Time taken to reach school = 15 min
Speed of bicycle = 2 m/s
Hint= Since time is given in minutes, and soeed in m/s, therefore we need to convert time into seconds.
20
= 0.75 km/min
Speed of car in km/h = 15
20/60
= 15 × 60
20
= 45 km/h
6. Geeta 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?
Sol=
Time taken to reach school = 15 min
Speed of bicycle = 2 m/s
Hint= Since time is given in minutes, and soeed in m/s, therefore we need to convert time into seconds.
Time taken by geeta = 15 min = 15 × 60 = 900 s
Therefore, Distance = Speed × time
Distance = 2 × 900
Distance = 1800 metres
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 roadside.
Ans:
(i) A car moving with a constant speed:
(ii) A car parked on a roadside:
8. Which of the following relations is correct?
i. Speed = Distance × time
ii. Speed = Distance Time
iii. Speed = Time
Distance
iv. Speed= 1/Distance x time
Ans:
ii. Speed = Distance Time
9. The basic unit of speed is:
(i) km/min (ii) m/min
(iii) km/h (iv) m/s
Ans: (iv) m/s
10. A car moves with a speed of 40km/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) 100km (ii) 25 km (iii) 15km (iv) 10 km
Ans: (ii) 25 km
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 metres is shown by 1 cm in these photographs, calculate the speed of the blue car.
iii. Speed = Time
Distance
iv. Speed= 1/Distance x time
Ans:
ii. Speed = Distance Time
9. The basic unit of speed is:
(i) km/min (ii) m/min
(iii) km/h (iv) m/s
Ans: (iv) m/s
10. A car moves with a speed of 40km/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) 100km (ii) 25 km (iii) 15km (iv) 10 km
Ans: (ii) 25 km
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 metres is shown by 1 cm in these photographs, calculate the speed of the blue car.
Sol:
Since colours are not shown in the diagram of jkbose class 7th textbook. We suppose the middle car highlighted in diagram is the blue car.
Now,
Total distance travelled by our blue car in 10 seconds = 2 cm.
But according to scale, 1cm is 100 m, so, 2cm = 200m
=> Speed of Blue car = Distance
Time
= 200/10
= 20 m/s
12. Fig 13.15 shows the distance-time graph for the motion of two vehicles A and B. Which one of them is moving faster.
Time
= 200/10
= 20 m/s
12. Fig 13.15 shows the distance-time graph for the motion of two vehicles A and B. Which one of them is moving faster.
Ans:
Car A is moving faster than Car B as it is covering more distance at any point on time axis.
13. Which of the following distance time graph shows a truck moving with speed which is not constant?
Ans:
iii.
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