cambridge school noida class 9 assignment motion solutions

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NCERT Solutions for class 9 Science chapter-8 Motion

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NCERT Solutions for class 9 Science chapter 8 Motion is prepared and uploaded for reference by academic team of expert members of pw. Get solutions of all chapters of NCERT class 9 Science from pw. use as a reference of the following NCERT solutions of chapter 8 prepared by pw. Read the theory of chapter-8 Motion while before going to exam.

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Question 1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.  Solution : Yes, if an object has moved through a distance it can have zero displacement because displacement of an object is the actual change in its position when it moves from one position to the other. So if an object travels from point A to B and then returns back to point A again, the total displacement is zero.

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Question 2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?

Solution : Distance covered by farmer in 40 seconds

=4 x(10) m= 40 m

Speed of the farmer = distance/time = 40m/40s = 1 m/s.

Total time given in the Question = 2min 20seconds

= 60+60+20 =140 seconds

Since he completes 1round of the field in 40seconds so in he will complete 3rounds in 120seconds (2mins) or 120m distance is covered in 2minutes. In another 20seconds will cover another 20m so total distance covered in 2min20sec

= 120 +20 =140m.

Displacement = √10 2 + 10 2  = √ 200

= 10 √ m (as per diagram) =10 x 1.414= 14.14 m.

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Question 3. Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object. Solution : Both (a) as well as (b) are false with respect to concept of displacement.

Question 4. Distinguish between speed and velocity. Solution : Speed of a body is the distance travelled by it per unit time while velocity is displacement per unit time of the body during movement.

Question 5.Under what condition(s) is the magnitude of average velocity of an object equal to its average speed? Solution : If distance travelled by an object is equal to its displacement then the magnitude of average velocity of an object will be equal to its average speed.

Question 6. What does the odometer of an automobile measure? Solution : The odometer of an automobile measures the distance covered by that automobile.

Question 7. What does the path of an object look like when it is in uniform motion? Solution : Graphically the path of an object will be linear i.e. look like a straight line when it is in uniform motion.

Question 8. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is 3 x10 8 ms -1 Solution :

Distance = Speed x time

Question 9. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration? Solution :

(i) uniform acceleration: When an object travels in a straight line and its velocity changes by equal amount in equal intervals of time, it is said to have uniform acceleration.

(ii) non uniform acceleration: It is also called variable acceleration. When the velocity of an object changes by unequal amounts in equal intervals of time, it is said to have non uniform acceleration.

Question 11. A train starting from a railway station and moving with uniform acceleration attains a speed 40km h -1 in 10 minutes. Find its acceleration. Solution : Since the train starts from rest(railway station) = u = zero

time (t) = 10 min = 10 x 60

= 600 seconds

Question 12. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object? Solution : If an object has a uniform motion then the nature of distance time graph will be linear i.e. it would a straight line and if it has non uniform motion then the nature of distance time graph is a curved line.

Question 13. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

Solution : If the object’s distance time graph is a straight line parallel to the time axis indicates that with increasing time the distance of that object is not increasing hence the object is at rest i.e. not moving.

Question 14. What can you say about the motion of an object if its speed time graph is a straight line parallel to the time axis? Solution : Such a graph indicates that the object is travelling with uniform velocity.

Question 15. What is the quantity which is measured by the area occupied below the velocity-time graph? Solution : The area occupied below the velocity-time graph measures the distance moved by any object.

Question 16. A bus starting from rest moves with a uniform acceleration of 0.1m s -2 for 2 minutes. Find (a)the speed acquired, (b) the distance travelled. Solution :

Question 17. A train is travelling at a speed of 90km h -1 . Brakes are applied so as to produce a uniform acceleration of -0.5m s -2 . Find how far the train will go before it is brought to rest. Solution :

v =0(train is brought to rest)

0 =25 – 0.5 x

0.5t = 25, or t = 25/0.5 = 50seconds

= 1250 – 625 = 625m

Question 18. A trolley, while going down an inclined plane, has an acceleration of 2cm s -2 . What will be its velocity 3 s after the start? Solution :

v= u +at = 0 + 2 x 3 = 6 cm/s

Question 19. A racing car has a uniform acceleration of 4cm s -2 . What distance will it cover in 10 s after start? Solution :

Question 20. A stone is thrown in a vertically upward direction with a velocity of 5 cm s -2 . If the acceleration of the stone during its motion is 10 cm s -2 n the downward direction, what will be the height attained by the stone and how much time will it take to reach there? Solution :

v = 0 (since at maximum height its velocity will be zero)

v = u + at 

= 5 + (-10) x t

0 = 5 – 10t

10t = 5, or, t = 5/10 =0.5second.

= 2.5 – 1.25 = 1.25m

Question 21. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s? Solution :

circumference of circular track = 2πr

rounds completed by athlete in 2min20sec = s= 140/40 = 3.5

therefore, total distance covered =4400 / 7 x 3.5= 2200 m

Since one complete round of circular track needs 40s so he will complete 3 rounds in 2mins and in next 20s he can complete half round therefore displacement = diameter = 200m.

Question 22. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C? Solution : (a) distance = 300m

time = 2min30seconds = 150 seconds

average speed from A to B = average velocity from A to B

= 300m/150s = 2m/s

(b) average speed from A to C = (300+100)m/(150+60)sec

= 400m/210s = 1.90m/s

displacement from A to C

= (300 – 100)m =200m

time =2min30sec + 1min = 210s

velocity = displacement/time = 200m/210s = 0.95m/s

Question 23. Abdul, while driving to school, computes the average speed for his trip to be 20km h -1 . On his return trip along the same route, there is less traffic and the average speed is 40km h -1 . What is the average speed for Abdul’s trip? Solution :

If we suppose that distance from Abdul’s home to school = x kms

while driving to school :-

velocity = displacement/time

20 = x/t, or, t=x/20 hr

on his return trip :-

speed = 40 km h–1 ,

40= x /t’

or, t’ =x/40 hr

total distance travelled = x + x = 2x

total time = t + t’ = x/20 + x/40

=(2x + x)/40 = 3x/40 hr

average speed for Abdul’s trip

= 2x/(3x/40) = 80x/3x = 26.67km/hr

Question 24. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s -2 for 8.0 s. How far does the boat travel during this time? Solution :

since the motorboat starts from rest so u= 0

Question 25. A driver of a car travelling at 52 km h -1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h -1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied? Solution :

∴ Clearly the first car will travel farther (36.11 m) than the first car(4.16 m).

Question 26. Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following Questions :

(a) Which of the three is travelling the fastest?

(b) Are all three ever at the same point on the road?

(c) How far has C travelled when B passes A?

(d) How far has B travelled by the time it passes C?

(a) It is clear from graph that B covers more distance in less time. Therefore, B is the fastest.

(b) All of them never come at the same point at the same time.

(c) According to graph; each small division shows about 0.57 km.

A is passing B at point S which is in line with point P (on the distance axis) and shows about 9.14 km

Thus, at this point C travels about

9.14 - (0.57 x 3.75)km = 9.14 km – 2.1375 km

Thus, when A passes B, C travels about 7 km.

(d) B passes C at point Q at the distance axis which is

Therefore, B travelled about 5.28 km when passes to C.

Question 27. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10m s -2 , with what velocity will it strike the ground? After what time will it strike the ground? Solution : Let us assume, the final velocity with which ball will strike the ground be ‘v’ and time it takes to strike the ground be ‘t’

Initial Velocity of ball u=0

Distance or height of fall s=20m

∴ Time taken by the ball to strike= (20-0)/10

= 2 seconds

Question 28. The speed-time graph for a car is shown is Fig. 8.12.

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.

(b) Which part of the graph represents uniform motion of the car?

(a) Distance travelled by car in the 4 second

The area under the slope of the speed – time graph gives the distance travelled by an object.

In the given graph

56 full squares and 12 half squares come under the area slope for the time of 4 second.

Total number of squares = 56 + 12/2 = 62 squares

The total area of the squares will give the distance travelled by the car in 4 second. on the time axis,

5 squares = 2seconds, therefore 1 square = 2/5 seconds

on speed axis there are 3 squares = 2m/s

therefore, area of one square

= 248/15 m = 16.53 m

Hence the car travels 16.53m in the first 4 seconds.

(b) The straight line part of graph, from point A to point B represents a uniform motion of car.

Question 29. State which of the following situations are possible and give an example for each of these:

(a) an object with a constant acceleration but with zero velocity

(b) an object moving in a certain direction with an acceleration in the perpendicular direction.

(a) An object with a constant acceleration can still have the zero velocity. For example an object which is at rest on the surface of earth will have zero velocity but still being acted upon by the gravitational force of earth with an acceleration of 9.81 ms-2 towards the center of earth. Hence when an object starts falling freely can have contant acceleration but with zero velocity.

(b) When an athlete moves with a velocity of constant magnitude along the circular path, the only change in his velocity is due to the change in the direction of motion. Here, the motion of the athlete moving along a circular path is, therefore, an example of an accelerated motion where acceleration is always perpendicular to direction of motion of an object at a given instance. Hence it is possible when an object moves on a circular path.

Question 30. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth. Solution :

Let us assume an artificial satellite, which is moving in a circular orbit of radius 42250 km covers a distance ‘s’ as it revolve around earth with speed ‘v’ in given time ‘t’ of 24 hours.

Radius of circular orbit r =42250 x 1000 m

Time taken by artificial satellite t=24 hours

=24 x 60 x 60 s

Distance covered by satellite s =Circumference of circular orbit

∴ Speed of satellite v = (2πr) / t

= 3.073 km/s

Check, Class 9 Science

CHAPTER WISE NCERT SOLUTIONS OF CLASS 9 SCIENCE

Chapter-1 Matter in Our Surroundings

Chapter-2 Is Matter Around Us Pure

Chapter-3 Atoms and Molecules

Chapter-4 Structure of the Atom

Chapter-5 The Fundamental Unit of Life

Chapter-6 Tissues

Chapter-7 Diversity in Living Organisms

Chapter-8 Motion

Chapter-9 Force and Laws of Motion

Chapter-10 Gravitation

Chapter-11 Work and Energy

Chapter-12 Sound

Chapter-13 Why Do We Fall Ill

Chapter-14 Natural Resources

Chapter-15 Improvement in Food Resources

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Related Chapters

• chapter-3 Atoms and Molecules
• chapter-4 Structure of Atom
• chapter-5 The Fundamental Unit of Life
• chapter-6 Tissues
• chapter-7 Diversity in Living Organisms
• chapter-8 Motion
• chapter-9 Forces and Laws of Motion
• chapter-10 Gravitation
• chapter-11 Work and Energy
• chapter-12 Sound
• chapter-13 Why Do We Fall Ill
• chapter-14 Natural Resources
• chapter-15 Improvement in Food Resources

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NCERT Solutions for Class 9 Science Chapter 8 Motion

NCERT Solutions Class 9 Science Chapter 8 Motion – Here are all the NCERT solutions for Class 9 Science Chapter 8. This solution contains questions, answers, images, step by step explanations of the complete Chapter 8 titled Motion of Science taught in class 9. If you are a student of class 9 who is using NCERT Textbook to study Science, then you must come across Chapter 8 Motion. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 9 Science Chapter 8 Motion in one place. For a better understanding of this chapter, you should also see Chapter 8 Motion Class 9 notes , Science.

Topics and Sub Topics in Class 9 Science Chapter 8 Motion:

• Describing Motion
• Measuring the Rate of Motion
• Rate of Change of Velocity
• Graphical Representation of Motion
• Equations of Motion by Graphical Method
• Uniform Circular Motion

These solutions are part of NCERT Solutions for Class 9 Science . Here we have given NCERT Solutions for Class 9 Science Chapter 8 Motion.

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NCERT Solutions for Class 9 Science Chapter 7 Motion

• 10th June 2023

NCERT Solutions for Class 9 Science Class 9 Science Chapter 7 Motion provides detailed answers for all in-text and exercise Questions. These solutions contain an in-depth explanation of each topic involved in the chapter. Students studying in class 9 can access these solutions for free in PDF format.

All these solutions are prepared by expert teachers and updated for the current academic session. NCERT Solutions for Class 9 Science Chapter 7 Motion help students to understand the fundamental concepts given in class 9 Science textbook. We have prepared the answers to all the questions in an easy and well-structured manner. It helps students to grasp the chapter easily.

NCERT Class 9 Science Chapter 7 Motion Intext Questions (Solved)

PAGE NO. 74

Question 1: An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Answer: Yes, zero displacement is possible if an object has moved through a distance. Suppose a body is moving in a circular path and starts moving from point A and it returns back at same point A after completing one revolution, then the distance will be equal to its circumference while displacement will be zero.

Question 2: A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Answer: Given, side of the square field = 10 m Therefore, perimeter = 10 m × 4 = 40 m Farmer moves along the boundary in 40 second Time = 2 minutes 20 second = 2 × 60 + 20 = 140 s since, in 40 second farmer moves 40 m Therefore, in 1 second distance covered by farmer = 40 ÷ 40 = 1m. Hence, in 140 second distance covered by farmer = 1 × 140 m = 140 m Now, number of rotation to cover 140 along the boundary = Total distance ÷ perimeter = 140 m ÷ 40 m   = 3.5 round Thus after 3.5 round farmer will at point C (diagonally opposite to his initial position) of the field.

Question 3: Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object.

Answer: None of (a) and (b) are true.

(a) is false because the displacement of an object which travels a certain distance and comes back to its initial position is zero. Statement (b) is false because the displacement of an object can be equal to, but never greater than the distance travelled.

PAGE NO. 76

Question 1: Distinguish between speed and velocity

Answer: Speed has only magnitude while velocity has both magnitude and direction. So speed is a scalar quantity but velocity is a vector quantity.

Question 2: Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Answer: The magnitude of average velocity of an object will be equal to its average speed in the condition of uniform velocity in a straight line motion.

Question 3: What does the odometer of an automobile measure?

Answer: In automobiles, odometer is used to measure the distance.

Question 4: What does the path of an object look like when it is in uniform motion?

Answer: In the case of uniform motion, the path of an object will look like a straight line.

Question 5: During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3×10 8 ms -1 .

Answer: Here we have, speed = 3 × 10 8 m/s Time = 5 minute = 5 × 60 s = 300 s Using, Distance = Speed × Time ⇒ Distance = 3 × 10 8 × 300 m ⇒ Distance = 900 × 10 8 m ⇒ Distance = 9.0 × 10 10 m

PAGE NO. 77

Question 1: When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?

Answer: (i) A body is said in uniform acceleration when its motion is along a straight line and its velocity changes by equal magnitude in equal interval of time.

(ii) A body is said in non-uniform acceleration when its motion is along a straight line and its velocity changes by unequal magnitude in equal interval of time.

Question 2: A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.

Question 3: A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 minutes. Find its acceleration.

Answer:  Here we have, Initial velocity, u = 0 m/s

PAGE NO. 81

Question 1: What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

• The slope of the distance-time graph for an object in uniform motion is straight line.
• The slope of the distance-time graph for an object in non-uniform motion is not a straight line.

Question 2: What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

Answer: When the slope of the distance-time graph is a straight line parallel to the time axis, the object is stationary.

Question 3: What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

Answer: When the graph of a speed-time graph is a straight line parallel to the time axis, the object is moving with constant speed.

Question 4: What is the quantity which is measured by the area occupied below the velocity-time graph?

Answer: The quantity of distance is measured by the area occupied below the velocity-time graph.

PAGE NO. 82

Question 1: A bus starting from rest moves with a uniform acceleration of 0.1ms –2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

Answer:  Here we have, Initial velocity (u) = 0 m/s Acceleration (a) = 0.1ms –2   Time (t) = 2 minute = 120 seconds 

(a) The speed acquired: We know that, v = u + at ⇒ v = 0 + 0.1 × 120 m/s ⇒ v = 12 m/s Thus, the bus will acquire a speed of 12 m/s after 2 minute with the given acceleration.

(b) The distance travelled:

Question 2: A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of – 0.5 m/s 2 . Find how far the train will go before it is brought to rest.

Answer:  Here we have,

Final velocity, v = 0 m/s            Acceleration, a = – 0.5 m/s 2                      Distance travelled (s) = ? Using, v 2  = u 2 + 2as

Therefore, the train will go 625 m before it brought to rest.

Question 3: A trolley, while going down an inclined plane, has an acceleration of 2 cm/s 2 . What will be its velocity 3 s after the start?

Answer: Here we have, Initial velocity, u = 0 m/s Acceleration (a) = 2 cm/s 2 = 0.02 m/s 2 Time (t) = 3 s Final velocity, v = ? We know that, v = u + at            ⇒ v = 0 + 0.02 × 3 m/s           ⇒ v = 0.06 m/s Therefore, the final velocity will be 0.06 m/s after start.

Question 4:  A racing car has a uniform acceleration of 4 m/s 2 . What distance will it cover in 10 s after start?

Answer:  Here we have, Acceleration, a = 4 m/s 2 Initial velocity, u = 0 m/s Time, t = 10 s Distance covered (s) =?

⇒ s = 2 × 100 m           ⇒ s = 200 m Thus, the racing car will cover a distance of 200 m after the start in 10 s with given acceleration.

Question 5: A stone is thrown in a vertically upward direction with a velocity of 5 m/s. If the acceleration of the stone during its motion is 10 m/s 2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Answer: Here we have, Initial velocity (u) = 5 m/s Final velocity (v) = 0 m/s Acceleration (a) = – 10 m/s 2              Height, i.e. Distance, s =? Time (t) taken to reach the height =?

Thus, the stone will attain a height of 1.25 m and the time taken to attain the height is 0.5 s.

CBSE Class 9 Science Motion Exercise Questions and Answers

Question 1: An athlete completes one round of circular track of diameter 200 m in 40 sec.  What will be the distance covered and the displacement at the end of 2 minutes 20 sec? 

Answer:  Time taken = 2 min 20 sec = 140 sec. Radius, r = 100 m. In 40 sec the athlete complete one round.     So, in 140 sec the athlete will complete = 140 ÷ 40 = 3.5 round.     Now, Distance covered in 140 sec = 2πr×3.5

= 2200 m. At the end of his motion, the athlete will be in the diametrically opposite position. Displacement = diameter = 200 m.

Question 2: Joseph jogs from one end A to another end B of a straight 300 m road in 2 minutes and 30 sec and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging. (a) from A to B (b) from A to C?

Answer: For motion from A to B: Distance covered = 300 m Displacement = 300 m. Time taken = 150 sec.

(a) We know that,   Average speed  = Total distance covered ÷ Total time taken  = 300 m ÷ 150 sec = 2 ms -1   Average velocity = Net displacement ÷ time taken = 300 m ÷ 150 sec = 2 ms -1

(b) For motion from A to C: Distance covered = 300 + 100 = 400 m.

Displacement = AB – CB = 300 – 100 = 200 m.

Time taken = 2.5 min + 1 min = 3.5 min = 210 sec.

Therefore,       Average speed = Total distance covered ÷ Total time taken  = 400 ÷ 210 = 1.90 ms -1 .

 Average velocity = Net displacement ÷ time taken  = 200 m ÷ 210 sec = 0.952ms -1 . 

Question 3: Abdul, while driving to school, computes the average speed for his trip to be 20 kmh -1 . On his return trip along the same route, there is less traffic and the average speed is 30 km/h. What is the average speed of Abdul’s trip? 

Answer:  Let one side distance = km. Time taken for forward trip at a speed of 20 km/h = Distance / Speed = 20 h. Time taken in return trip at a speed of 30 km/h = 30 h. 

Question 4: A motor boat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms -2 for 8.0 s. How far does the boat travel during this time?

Answer:  Here, u = 0 m/s                  a = 3 ms -2                   t = 8 s Using,  s = ut + ½ at 2   ⇒ s = 0 × 8 + ½ × 3×8 2 ⇒ s =96 m.

Question 5: A driver of a car travelling at 52 kmh -1 applies the brakes and accelerates uniformly in the opposite direction. The car stops after 5 s. Another driver going at 34 kmh -1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for two cars. Which of the two cars travelled farther after the brakes were applied?  

 Answer: In in the following graph, AB and CD are the time graphs for the two cars whose initial speeds are 52 km/h(14.4 m/s) and 34 km/h(8.9 m/s), respectively.

Distance covered by the first car before coming to rest = Area of triangle AOB = ½ × AO × BO = ½ × 52 kmh -1 × 5 s = ½  (52 × 1000 × 1/3600) ms -1 × 5 s = 36.1 m 

Distance covered by the second car before coming to rest = Area of triangle COD = ½ × CO × DO = ½ × 34 km h -1 × 10 s = ½ × (34 × 1000 × 1/3600) ms -1 ×10 s = 47.2 m Thus, the second car travels farther than the first car after they applied the brakes.

Question 6: Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C?

Answer: (a) B is travelling fastest as he is taking less time to cover more distance. (b) All three are never at the same point on the road. (c) Approximately 6 km. [as 8 – 2 = 6] (d) Approximately 7 km. [as 7 – 0 = 7] 

Question 7: A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10ms -2 , with what velocity will it strike the ground? After what time will it strike the ground?

Answer:  Here, u = 0 m/s,     s = 20 m,                          a = 10 ms -2 ,                     v = ?,                                  t = ? Using v 2   –  u 2 = 2as We have,  v 2 – 0 2 = 2 × 10 × 20 ⇒ v 2 = 20 × 20 ⇒ v 2 = 400  ⇒ v = 20 ms -1 .  Again we know v = u +at ⇒ t = (v – u) ÷ a ⇒ t = (20 – 0) ÷ 10 ⇒ t = 20 ÷ 10 ⇒ t = 2 s. 

Question 8: The speed – time graph for a car is shown in Figure:

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period. (b) Which part of the graph represents uniform motion of the car?

Answer: (a) The shaded area represents the displacement of the car over a time period of 4 seconds. It can be calculated as: (1/2)×4×6 = 12 meters. Therefore the car travels a total of 12 meters in the first four seconds.

(b) After 6 seconds the car moves in uniform motion (at a speed of 6 m/s).

Question 9: State which of the following situations are possible and give an example of each of the following:

(a) an object with a constant acceleration but with zero velocity, (b) an object moving in a certain direction with an acceleration in the perpendicular direction. 

Answer: (a) Yes, a body can have acceleration even when its velocity is zero. When a body is thrown up at the highest point its velocity is zero but it has acceleration equal to the acceleration due to gravity.

(b) Yes, an acceleration moving horizontally is acted upon by acceleration due to gravity that acts vertically downwards.  

Question 10: An artificial is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hrs to revolve around the earth.   

Answer: Here,  r = 42250 km = 42250000 m T = 24 h = 24 × 60 × 60 s  Using , Speed, v = 2πr ÷ T  v = (2 × 3.14 × 42250000) ÷ (24 × 60 × 60) m/s  = 3070.9 m/s = 3.07 km/s

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• CBSE Notes For Class 9
• Class 9 Science Notes
• Chapter 8: Motion

Motion Class 9 CBSE Notes - Chapter 8

According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 7.

CBSE Class 9 Science Chapter 8 Motion Notes

In Class 9 Science Chapter 8 Motion, students learn to describe the motion of objects along a straight line and express such motions through simple equations and graphs. The chapter also discusses ways of describing circular motion.

Chapter Summary Video

Understanding Motion

Reference point and reference frame.

• To describe the position of an object, we need a reference point or origin. An object may seem to be moving to one observer and stationary to another.
• Example: A passenger inside a bus sees the other passengers to be at rest, whereas an observer outside the bus sees the passengers to be in motion.
• In order to make observations easy, a convention or a common reference point or frame is needed. All objects must be in the same reference frame.

For more information on Motion, watch the below videos

To know more about the frame of reference, visit here .

Distance and Displacement

The magnitude of the length covered by a moving object is called distance. It has no direction.

Displacement is the shortest distance between two points or the distance between the starting and final positions with respect to time. It has magnitude as well as direction.

Displacement can be zero, but distance cannot.

To know more about Distance and Displacement, visit here .

Magnitude is the size or extent of a physical quantity. In physics, we have scalar and vector quantities.

Scalar quantities are only expressed as magnitude. E.g.: time, distance, mass, temperature, area, volume

Vector quantities are expressed in magnitude as well as the direction of the object. E.g.: Velocity, displacement, weight, momentum, force, acceleration, etc.

Time, Average Speed and Velocity

Time and speed.

Time is the duration of an event that is expressed in seconds. Most physical phenomena occur with respect to time. It is a scalar quantity.

Speed is the rate of change in distance. If a body covers a certain distance in a certain amount of time, its speed is given by

The instantaneous speed is the speed of an object at a particular moment in time.

Average speed is stated as the distance covered by the object within a period of time.

Average speed = Total distance travelled / Total time taken

The below table lists the difference between Average Speed and Instantaneous Speed.

Uniform motion and Non-uniform motion

When an object covers equal distances in equal intervals of time, it is in uniform motion.

Examples of Uniform Motion

• Movement of the ceiling fan’s blades.
• Motion of Earth around the sun
• Pendulum with equivalent amplitude on either side

When an object covers unequal distances in equal intervals of time, it is said to be in non-uniform motion.

• Bouncing ball
• Running horse
• Moving train

To know more about Uniform Motion and Non-Uniform Motion, visit here .

The Rate of change of displacement is velocity. It is a vector quantity. Here the direction of motion is specified.

Instantaneous velocity is the rate of change of position for a time interval which is very small, i.e. almost zero. In more simple words, the velocity of an object at a given instant of time is known as instantaneous velocity.

Average velocity is defined as the displacement (∆x) divided by the time intervals (∆t) in which the displacement occurs.

For more information on Average Speed and Velocity, watch the below video

To know more about Average Speed and Average Velocity, visit here .

Acceleration

The rate of change of velocity is called acceleration. It is a vector quantity. In non-uniform motion, velocity varies with time, i.e., the change in velocity is not 0. It is denoted by “a”

A c c e l e r a t i o n   =  Change in Velocity / Time

Where t (time taken), v (final velocity) and u (initial velocity).

To know more about Acceleration, visit here .

Motion Visualised

Distance-time graph.

• Distance-Time graphs show the change in the position of an object with respect to time.
• Linear variation = uniform motion and non-linear variations imply non-uniform motion
• The slope gives us speed

• OA implies uniform motion with constant speed as the slope is constant
• AB implies the body is at rest as the slope is zero
• B to C is a non-uniform motion

To know more about Distance-Time Graph, visit here .

Velocity-Time Graph

• Velocity-Time graphs show the change in velocity with respect to time.
• Slope gives acceleration
• The area under the curve gives displacement
• Line parallel to x-axis implies constant velocity-

OA = constant acceleration, AB = constant velocity, BC = constant retardation

To know more about Velocity-Time Graph, visit here .

Equations of Motion

The motion of an object moving at uniform acceleration can be described with the help of three equations, namely

(i) v = u + at

(ii) v 2 – u 2 = 2as

(iii) s = ut + (1/2)at 2

where u is the initial velocity, v is the final velocity, t is the time, a is the acceleration and s is the displacement.

To know more about Equations of Motion, visit here .

Derivation of Velocity-Time Relation by Graphical Method

A body starts with some initial non-zero velocity at A and goes to B  with constant acceleration a.

From the graph BD = v (final velocity) – DC = u (initial velocity)…………..(eq 1).

BD = BC – DC……………..( eq 2 ).

Therefore   BD = at………………….(eq 3).

Substitute everything we get: at = v – u.

Rearrange to get v = u + at.

Derivation of Position-Time Relation by Graphical Method

A body starts with some initial non-zero velocity at A and goes to B  with constant acceleration a

Area under the graph gives Displacement as follows:

OA = u , OC = t and BD = at

Substituting in (eq 1) we get s = \(\begin{array}{l}ut + \frac{1}{2}at^{2}\end{array} \)

Derivation of Position-Velocity Relation by Graphical Method

Displacement covered will be the area under the curve which is the trapezium OABC.

We know the area of trapezium is 

OA = u and BC = v and OC = t

Substitute (eq 2) in (eq 1) and arrange to get

v 2 −u 2 =2as

Uniform Circular Motion

• If an object moves in a circular path with uniform speed, its motion is called uniform circular motion.
• Velocity changes as direction keeps changing.
• Acceleration is constant.
• The uniform circular velocity is given by the following formula: \(\begin{array}{l}v=\frac{2r}{t}\end{array} \)

Uniform Circular Motion Examples

• The motion of artificial satellites around the Earth is an example of uniform circular motion.
• The motion of electrons around its nucleus.
• The motion of blades of the windmills.
• The tip of the second hand of a watch with a circular dial shows uniform circular motion.

To know more about Uniform Circular Motion, visit here .

Frequently Asked Questions on CBSE Class 9 Science Notes Chapter 8 Motion

What is ‘motion’.

When any object moves from one point to another with respect to the observer, then this object is said to be in ‘motion’.

Is this chapter difficult?

If a student fully understands the concept of the chapter ‘motion’ and practises sums regularly, then they can score good marks in this chapter.

Are the derivations in this chapter difficult?

There are derivations in this chapter and these need to be practised constantly.

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Elektrostal

Elektrostal Localisation : Country Russia , Oblast Moscow Oblast . Available Information : Geographical coordinates , Population, Altitude, Area, Weather and Hotel . Nearby cities and villages : Noginsk , Pavlovsky Posad and Staraya Kupavna .

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Coordinates of elektrostal in degrees and decimal minutes, utm coordinates of elektrostal, geographic coordinate systems.

WGS 84 coordinate reference system is the latest revision of the World Geodetic System, which is used in mapping and navigation, including GPS satellite navigation system (the Global Positioning System).

Geographic coordinates (latitude and longitude) define a position on the Earth’s surface. Coordinates are angular units. The canonical form of latitude and longitude representation uses degrees (°), minutes (′), and seconds (″). GPS systems widely use coordinates in degrees and decimal minutes, or in decimal degrees.

Latitude varies from −90° to 90°. The latitude of the Equator is 0°; the latitude of the South Pole is −90°; the latitude of the North Pole is 90°. Positive latitude values correspond to the geographic locations north of the Equator (abbrev. N). Negative latitude values correspond to the geographic locations south of the Equator (abbrev. S).

Longitude is counted from the prime meridian ( IERS Reference Meridian for WGS 84) and varies from −180° to 180°. Positive longitude values correspond to the geographic locations east of the prime meridian (abbrev. E). Negative longitude values correspond to the geographic locations west of the prime meridian (abbrev. W).

UTM or Universal Transverse Mercator coordinate system divides the Earth’s surface into 60 longitudinal zones. The coordinates of a location within each zone are defined as a planar coordinate pair related to the intersection of the equator and the zone’s central meridian, and measured in meters.

Elevation above sea level is a measure of a geographic location’s height. We are using the global digital elevation model GTOPO30 .

Elektrostal , Moscow Oblast, Russia

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Out of the Centre

Savvino-storozhevsky monastery and museum.

Zvenigorod's most famous sight is the Savvino-Storozhevsky Monastery, which was founded in 1398 by the monk Savva from the Troitse-Sergieva Lavra, at the invitation and with the support of Prince Yury Dmitrievich of Zvenigorod. Savva was later canonised as St Sabbas (Savva) of Storozhev. The monastery late flourished under the reign of Tsar Alexis, who chose the monastery as his family church and often went on pilgrimage there and made lots of donations to it. Most of the monastery’s buildings date from this time. The monastery is heavily fortified with thick walls and six towers, the most impressive of which is the Krasny Tower which also serves as the eastern entrance. The monastery was closed in 1918 and only reopened in 1995. In 1998 Patriarch Alexius II took part in a service to return the relics of St Sabbas to the monastery. Today the monastery has the status of a stauropegic monastery, which is second in status to a lavra. In addition to being a working monastery, it also holds the Zvenigorod Historical, Architectural and Art Museum.

Belfry and Neighbouring Churches

Located near the main entrance is the monastery's belfry which is perhaps the calling card of the monastery due to its uniqueness. It was built in the 1650s and the St Sergius of Radonezh’s Church was opened on the middle tier in the mid-17th century, although it was originally dedicated to the Trinity. The belfry's 35-tonne Great Bladgovestny Bell fell in 1941 and was only restored and returned in 2003. Attached to the belfry is a large refectory and the Transfiguration Church, both of which were built on the orders of Tsar Alexis in the 1650s.  

To the left of the belfry is another, smaller, refectory which is attached to the Trinity Gate-Church, which was also constructed in the 1650s on the orders of Tsar Alexis who made it his own family church. The church is elaborately decorated with colourful trims and underneath the archway is a beautiful 19th century fresco.

Nativity of Virgin Mary Cathedral

The Nativity of Virgin Mary Cathedral is the oldest building in the monastery and among the oldest buildings in the Moscow Region. It was built between 1404 and 1405 during the lifetime of St Sabbas and using the funds of Prince Yury of Zvenigorod. The white-stone cathedral is a standard four-pillar design with a single golden dome. After the death of St Sabbas he was interred in the cathedral and a new altar dedicated to him was added.

Under the reign of Tsar Alexis the cathedral was decorated with frescoes by Stepan Ryazanets, some of which remain today. Tsar Alexis also presented the cathedral with a five-tier iconostasis, the top row of icons have been preserved.

Tsaritsa's Chambers

The Nativity of Virgin Mary Cathedral is located between the Tsaritsa's Chambers of the left and the Palace of Tsar Alexis on the right. The Tsaritsa's Chambers were built in the mid-17th century for the wife of Tsar Alexey - Tsaritsa Maria Ilinichna Miloskavskaya. The design of the building is influenced by the ancient Russian architectural style. Is prettier than the Tsar's chambers opposite, being red in colour with elaborately decorated window frames and entrance.

At present the Tsaritsa's Chambers houses the Zvenigorod Historical, Architectural and Art Museum. Among its displays is an accurate recreation of the interior of a noble lady's chambers including furniture, decorations and a decorated tiled oven, and an exhibition on the history of Zvenigorod and the monastery.

Palace of Tsar Alexis

The Palace of Tsar Alexis was built in the 1650s and is now one of the best surviving examples of non-religious architecture of that era. It was built especially for Tsar Alexis who often visited the monastery on religious pilgrimages. Its most striking feature is its pretty row of nine chimney spouts which resemble towers.

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The Unique Burial of a Child of Early Scythian Time at the Cemetery of Saryg-Bulun (Tuva)

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Pages:  379-406

In 1988, the Tuvan Archaeological Expedition (led by M. E. Kilunovskaya and V. A. Semenov) discovered a unique burial of the early Iron Age at Saryg-Bulun in Central Tuva. There are two burial mounds of the Aldy-Bel culture dated by 7th century BC. Within the barrows, which adjoined one another, forming a figure-of-eight, there were discovered 7 burials, from which a representative collection of artifacts was recovered. Burial 5 was the most unique, it was found in a coffin made of a larch trunk, with a tightly closed lid. Due to the preservative properties of larch and lack of air access, the coffin contained a well-preserved mummy of a child with an accompanying set of grave goods. The interred individual retained the skin on his face and had a leather headdress painted with red pigment and a coat, sewn from jerboa fur. The coat was belted with a leather belt with bronze ornaments and buckles. Besides that, a leather quiver with arrows with the shafts decorated with painted ornaments, fully preserved battle pick and a bow were buried in the coffin. Unexpectedly, the full-genomic analysis, showed that the individual was female. This fact opens a new aspect in the study of the social history of the Scythian society and perhaps brings us back to the myth of the Amazons, discussed by Herodotus. Of course, this discovery is unique in its preservation for the Scythian culture of Tuva and requires careful study and conservation.

Keywords: Tuva, Early Iron Age, early Scythian period, Aldy-Bel culture, barrow, burial in the coffin, mummy, full genome sequencing, aDNA

Information about authors: Marina Kilunovskaya (Saint Petersburg, Russian Federation). Candidate of Historical Sciences. Institute for the History of Material Culture of the Russian Academy of Sciences. Dvortsovaya Emb., 18, Saint Petersburg, 191186, Russian Federation E-mail: [email protected] Vladimir Semenov (Saint Petersburg, Russian Federation). Candidate of Historical Sciences. Institute for the History of Material Culture of the Russian Academy of Sciences. Dvortsovaya Emb., 18, Saint Petersburg, 191186, Russian Federation E-mail: [email protected] Varvara Busova  (Moscow, Russian Federation).  (Saint Petersburg, Russian Federation). Institute for the History of Material Culture of the Russian Academy of Sciences.  Dvortsovaya Emb., 18, Saint Petersburg, 191186, Russian Federation E-mail:  [email protected] Kharis Mustafin  (Moscow, Russian Federation). Candidate of Technical Sciences. Moscow Institute of Physics and Technology.  Institutsky Lane, 9, Dolgoprudny, 141701, Moscow Oblast, Russian Federation E-mail:  [email protected] Irina Alborova  (Moscow, Russian Federation). Candidate of Biological Sciences. Moscow Institute of Physics and Technology.  Institutsky Lane, 9, Dolgoprudny, 141701, Moscow Oblast, Russian Federation E-mail:  [email protected] Alina Matzvai  (Moscow, Russian Federation). Moscow Institute of Physics and Technology.  Institutsky Lane, 9, Dolgoprudny, 141701, Moscow Oblast, Russian Federation E-mail:  [email protected]

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