NCERT Solutions For Class 9 Science Chapter 8

NCERT Solutions For Class 9 Science Chapter 8 Motion

1. Exercise Questions
2. Intext Questions

NCERT Solutions For Class 9 Science Chapter 8 Motion on this step-by-step Motion answer guide. In some of State Boards and CBSE schools, students are taught thru NCERT books. As the chapter comes to an end, students are requested few questions in an exercising to evaluate their knowledge of the chapter. Students regularly want guidance dealing with those NCERT Solutions. It’s most effective natural to get stuck withinside the exercises while solving them so that you could assist students rating better marks, we have provided grade by grade NCERT answers for all exercises of Class 9 Science Motion so you can be searching for assist from them. Students have to solve those exercises carefully as questions withinside the very last exams are requested from those so these exercises at once have an impact on students’ final score. Find all NCERT Solutions for Class nine Science Motion below and prepare for your tests easily.

NCERT class 9 Chapter 8: Motion 

Exercise Questions

 

Q. 1. An athlete completes one round of 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 ?

Ans . Given that ,

Time taken 2 minutes 20 s = 140 s .

Radius , r = 100 m .

In 40 s , the athlete completes one round .

So , in 140 s , the athlete will complete 140 ÷ 40 = 3.5 round .

TextDescription automatically generated

At the end of his motion , the athlete will be in the diametrically opposite position .

Displacement = Diameter = 200 m

Q. 2. Joseph jogs from one end A to another end B of a straight 300 m road in 2 minutes and 30 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 ?

Ans . ( a ) Given that ,

For motion from A to B

Distance covered = 300 m

Displacement = 300 m .

Time taken = 150 seconds .

TableDescription automatically generated with low confidence

( b ) Given that ,

For motion A to C :

Distance covered = 300 + 100 = 400 m .

Displacement AB – CB

= 300 – 100 = 200 m .

Time taken = 2.5 + 1 = 3.5 mins

= 3.5 x 60 = 210 s

TextDescription automatically generated with medium confidence

TableDescription automatically generated with medium confidence

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

Ans . Let one side distance = x km .

TableDescription automatically generated

Q. 4. 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 ?

Ans . Given that ,

u = 0 m / s

a = 3ms ²

t = 8s

Diagram, schematicDescription automatically generated

Q. 5. A driver of a car travelling at 52 km h¯¹ applies the brakes and accelerates uniformly in the opposite direction . The car stops after 5 seconds . 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 two cars . Which of the two cars travelled farther after the brakes were applied ?

Ans . 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 3 km / h ( .83 m / s ) , respectively .

TextDescription automatically generated

Thus , the first car travels farther than the second car after they applied the brakes .

Q. 6. Following figure shows the distance – time graph of three objects A , B and C. Study the graph and answer the following questions :

Chart, line chartDescription automatically generated

( 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 ?

Ans . According the graph given above :

( 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 ]

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

Ans . Given that ,

u = 0 m / s ,

s = 20 m ,

a = 10 ms ² ,

Text, letterDescription automatically generated

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

ChartDescription automatically generated

( 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 ?

Ans . ( a ) Distance covered = Area under speed = Time graph

Shaded area representing the distance travelled is as follows :

ChartDescription automatically generated

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

Q. 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 .

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

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

Q. 10. An artificial satellite is moving in a circular orbit of radius 42,250 km . Calculate its speed if it takes 24 hours to revolve around the earth .

Ans . Given that ,

r = 42,250 km = 4,22,50,000 m

T = 24 hours = 24 x 60 x 60 seconds

Using speed ,

v = 2πr ÷ T

v = ( 2 x 3.14 x 4,22,50,000 ) ÷ ( 24 × 60 × 60 )

= 3070.9 ms – 1

= 3.07 km / s

Intext Questions

 

Page 100

Q. 1. An object has moved through a distance . Can it have zero displacement ? If yes , support your answer with an example . [ NCERT Q. 1 , Page 100 ]

Ans . 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 the same point A after completing one revolution , then the distance will be equal to its circumference while displacement will be zero .

Q.2 . A farmer moves along the boundary of a square field of side 10 m in 40 seconds . What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position ? [ NCERT Q. 2 , Page 100 ]

Ans . Given that ,

Side of the square field = 10 m

Therefore , perimeter = 10 m x 4 = 40 m

Farmer moves along the boundary in 40 seconds

Time = 2 minutes 20 seconds

= 2 x 60 seconds + 20 seconds

= 140 seconds

Since , in 40 seconds farmer moves 40 m ,

Therefore , in 1 second distance covered by farmer

= 40 ÷ 40 = 1 m .

in 140 seconds distance covered by farmer = 1 x 140 m = 140 m

Now , number of rotations 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 .

A picture containing textDescription automatically generated

Thus , after 2 minutes 20 seconds , the displacement of farmer will be equal to 10√2 m north – east from the initial position .

Q. 3 . Which of the following is true for displacement ?

( a ) It cannot be zero .

( b ) Its magnitude is greater than the distance travelled by an object . [ NCERT Q. 3 , Page 100 ]

Ans . ( a ) Not true , because if the initial and final positions of the object is the same displacement can become zero .

( b ) Not true , because , displacement is the shortest measurable distance between the initial and final positions of an object . It cannot be greater than the magnitude of the distance travelled by an object , but it can be smaller or equal to the distance travelled by an object .

Page 102

Q. 4. Distinguish between speed and velocity . [ NCERT Q. 1 , Page 102 ]

Ans . Differences between speed and velocity are :

TableDescription automatically generated

Q. 5. Under what condition ( s ) is the magnitude of average velocity of an object equal to its average speed ? [ NCERT Q. 2 , Page 102 ]

Ans . As we know ,

TextDescription automatically generated with medium confidence

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 .

Q. 6. What does the odometer of an automobile measure ? [ NCERT Q. 3 , Page 102 ]

Ans . Odometer is a device that measures the distance travelled by an automobile based on the perimeter of the wheel as the wheel rotates .

Q. 7. What does the path of an object look like when it is in uniform motion ? [ NCERT Q. 4 , Page 102 ]

Ans . Uniform motion is a motion in which an object covers equal distances in equal intervals of time and moves along a straight line .

Thus , in the case of uniform motion , the path is a straight line in uniform motion .

Q. 8. During an experiment , a signal from a spaceship reached the ground station in 5 minutes . What was the distance of the spaceship from the ground station ? The signal travels at the speed of light , that is , 3 x 108 m s ¹ . [ NCERT Q. 5 , Page 102 ]

Ans . Given that ,

Speed = 3 x 108 m s -1

Time = 5 minutes

= 5 x 60 seconds = 300 seconds

Using , Distance = Speed x Time

Distance = 3 x 108 x 300 m

= 900 × 108 m

= 9.0 × 10¹⁰ m

Page 103

Q.9 . When will you say a body is in : ( i ) Uniform acceleration ? ( ii ) Non – uniform acceleration ? [ NCERT Q. 1 , Page 103 ]

Ans . ( a ) A body is said to be in uniform acceleration when its motion is along a straight line and its velocity changes by equal magnitudes in equal intervals of time .

( b ) A body is said to be in non – uniform acceleration when its velocity changes by unequal magnitudes in equal intervals of time .

Q. 10. A bus decreases its speed from 80 km / h to 60 km / h in 5 s . Find the acceleration of the bus . [ NCERT Q. 2 , Page 103 ]

Ans . Given that ,

u = 80 km / h = [ 80 x 1000 ] / 3600 ms ¹ = 200/9 m s ¹

v = 60 km / h = [ 60 × 1000 ] / 3600 ms ¹ = 150 / 9 m s ¹

t = 5s

A picture containing tableDescription automatically generated

Therefore , acceleration is -1.1 ms – 2

Q.11 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 . [ NCERT Q.3 , Page 103 ]

Ans . Given that ,

Initial velocity , u = 0 m / s

Final velocity , v = 40 km / h = [ 40 x 1000 ] / 3600 m s – 1

Time ( t ) = 10 minute = 60 x 10 = 600 s

DiagramDescription automatically generated

Page 107

Q. 12. What is the nature of the distance – time graphs for uniform and non – uniform motion of an object ? [ NCERT Q. 1 , Page 107 ]

Ans . ( i ) The distance – time graph for an object in uniform motion is straight line , inclined with the time axis .

( ii ) The distance – time graph for an object in non uniform motion is a curve .

Q. 13 . What can you say about the motion of an object whose distance – time graph is a straight line parallel to the time axis ? [ NCERT Q. 2 , Page 107 ]

Ans . When the distance – time graph is a straight line parallel to time axis , it means that the object is not changing its position .

So , the object is stationary or at rest .

Q. 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 ? [ NCERT Q. 3 , Page 107 ]

Ans . When the graph of a speed – time graph is a straight line parallel to the time axis , it means that the speed of the object is not changing with time .

So , the object is performing uniform motion .

Q. 15. What is the quantity which is measured by the area occupied below the velocity – time graph ? [ NCERT Q. 4 , Page 107 ]

Ans . The quantity of displacement is measured by the area occupied below the velocity – time graph .

Page 109

Q. 16. A bus starting from rest moves with a uniform acceleration of 0.1 m s – 2 for 2 minutes .

Find :

( a ) the speed acquired .

( b ) the distance travelled . [ NCERT Q. 1 , Page 109 ]

Ans . Given that ,

Initial velocity ( u ) = 0 m / s

Acceleration ( a ) = 0.1 m s – 2

Time ( t ) = 2 minutes

= 2 x 60 seconds

= 120 seconds

( a ) The speed acquired :

We know that ,

v = u + at

→ v = 0 + 0.1 x 120 m / s

→ v = 12 m / s

Thus , the bus will acquire a speed of 12 m / s after 2 minutes with the given acceleration .

( b ) The distance travelled :

We know that ,

Text, letterDescription automatically generated

Thus , bus will travel a distance of 720 m in the given time of 2 minutes .

Q. 17. 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 / s² . Find how far the train will go before it is brought to rest . [ NCERT Q. 2 , Page 109 ]

Ans . Given that ,

Initial velocity , u = 90 km / h = [ 90 x 1,000 ] / 3,600 m s ¹ = 25 m s. ¹

Final velocity , v = 0 m / s

Acceleration , a = -0.5 m / s²

Using equation of motion

TextDescription automatically generated with medium confidence

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

Q. 18. A trolley , while going down an inclined plane , has an acceleration of 2 cm / s² . What will be its velocity 3 s after the start ? [ NCERT Q. 3 , Page 109 ]

Ans . Given that ,

Initial velocity , u = 0 m / s

Acceleration ( a ) = 2 cm / s² = 0.02 m / s²

Time ( t ) = 3 seconds

We know that , v = u = at

Therefore , v = 0 + 0.02 x 3 m / s

→ v = 0.06 m / s

Therefore , the final velocity of trolley will be 0.06 m / s after start .

Q. 19. A racing car has a uniform acceleration of 4 m / s² . What distance will it cover in 10 s after start ? [ NCERT Q. 4 , Page 109 ]

Ans . Given that ,

Acceleration , a = 4 m / s²

Initial velocity , u = 0 m / s

Time , t = 10 s

TextDescription automatically generated

Thus , racing car will cover a distance of 200 m after start in 10 seconds with given acceleration .

Q. 20. 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 / s² in the downward direction , what will be the height attained by the stone and how much time will it take to reach there ? [ NCERT Q. 5 , Page 109 ]

Ans . Given that ,

Initial velocity ( u ) = 5 m / s

Final velocity ( v ) = 0 m / s

Acceleration ( a ) = -10 m / s²

As we know ,

v² = u² + 2as

0 = ( 5 ) ² + 2 x ( -10 ) x s

s = 1.25 m

Now , we know that ,

v = u + at

0 = 5+ ( -10 ) x t

t = 5/10 second

t = ½ second = 0.5 second

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

Benefits of NCERT Solutions For Class 9 Science Chapter 8 Motion

NCERT Solutions For Class 9 Science Chapter 8 Motion contains extremely important points, and for each chapter, each concept has been simplified to make it easier to remember and increase your chances of achieving excellent exam results. Exam Preparation References Here are some tips on how these solutions can help you prepare for the exam.

  1. This helps students solve many of the problems in each chapter and encourages them to make their concepts more meaningful.
  2. NCERT Solutions for Class 9 Science Chapter 8 Motion encourage you to update your knowledge and refine your concepts so that you can get good results in the exam.
  3. These NCERT Solutions For Class 9 Science Chapter 8 Motion are the best exam materials, allowing you to learn more about your week and your strengths. To get good results in the exam, it is important to overcome your weaknesses.
  4. Most of the questions in the exam are formulated in a similar way to NCERT textbooks. Therefore, students should review the solutions in each chapter in order to better understand the topic.
  5. It is free of cost.

Tips & Strategies for Class 9 Exam Preparation

  1. Plan your course and syllabus and make time for revision
  2. Please refer to the NCERT solution available on the cbsestudyguru website to clarify your concepts every time you prepare for the exam.
  3. Use the cbsestudyguru learning app to start learning to successfully pass the exam. Provide complete teaching materials, including resolved and unresolved tasks.
  4. It is important to clear all your doubts before the exam with your teachers or Alex (an Al study Bot). 
  5. When you read or study a chapter, write down algorithm formulas, theorems, etc., and review them quickly before the exam.
  6. Practice an ample number of question papers to make your concepts stronger. 
  7. Take rest and a proper meal.  Don’t stress too much. 

Why opt for cbsestudyguru NCERT Solutions for Class 9 ? 

  • cbsestudyguru provide NCERT Solutions for all subjects at your fingertips.
  • These solutions are designed by subject matter experts and provide solutions to every NCERT textbook questions. 
  • cbsestudyguru especially focuses on making learning interactive, effective and for all classes.
  • We provide free NCERT Solutions for class 9 and all other classes.

Leave a Comment