NCERT Solutions Class 9 Maths Chapter 8 Motion – Here are all the NCERT solutions for Class 9 Maths Chapter 8. This solution contains questions, answers, images, explanations of the complete Chapter 8 titled The Fun They Had of Maths taught in class 9. If you are a student of class 9 who is using NCERT Textbook to study Maths, 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 Maths Chapter 8 Motion in one place.

## NCERT Solutions Class 9 Science Chapter 8 Motion

Here on **AglaSem Schools**, you can access to **NCERT Book Solutions** in free pdf for Science for Class 9 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 8 Motion , Science, Class 9.

Class | 9 |

Subject | Science |

Book | Science |

Chapter Number | 8 |

Chapter Name |
Motion |

### NCERT Solutions Class 9 Science chapter 8 Motion

Class 9, Science chapter 8, Motion solutions are given below in PDF format. You can view them online or download PDF file for future use.

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### NCERT Solutions Class 9 Science chapter 8 Motion- Video

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### Download NCERT Solutions Class 9 Science chapter 8 Motion In PDF Format

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### Question & Answer

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

Ans :Yes. An object that has moved through a distance can have zero displacement. Displacement is the shortest measurable distance between the initial and the final position of an object. An object which has covered a distance can have zero displacement, if it comes back to its starting point, i.e., the initial position. Consider the following situation. A man IS walking in a square park of length 20 m (as shown in the following figure). He staffs walking from point A and after moving along all the corners of the park (point B, C, D), he again comes back to the same point, i.e., A. In this case, the total distance covered by the man is 20 m +20 m + 20 m + 20 m = 80 m. However, his displacement is zero because the shortest distance between his initial and final position is zero.

Q.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?

Ans :The farmer takes 40 s to cover 4 x 10 =40 m. In 2 min and 20 s (140 s), he will cover a distance \( \frac { 40 } { 40 } \times 140\)=40m Therefore, the farmer completes \( \frac { 140 } { 40 } = 3.5\) rounds (3 complete rounds and a half round) of the field in 2 min and 20 s. That means, after 2 min 20 s, the farmer Will be at the opposite end of the starting point. Now, there can be two extreme cases. Case I: Starting point is a comer point of the field. In this case, the farmer will be at the diagonally opposite corner of the field after 2 min 20 s. Therefore, the displacement will be equal to the diagonal to the field. Hence, the displacement will be \( \sqrt { 10 ^ { 2 } + 10 ^ { 2 } }\)=14.1m Case II: Starting point is the middle point of any side of the field. In this case the farmer Will be at the middle point of the opposite side of the field after 2 min 20 s. Therefore, the displacement will be equal to the side of the field, i.e., 10 m. For any other starting point, the displacement will be between 14.1 m and 10 m.

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 the object.

Ans :(a) Not true Displacement can become zero when the initial and final position of the object is the same. (b) Not true 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. However, sometimes, it may be equal to the distance travelled by the object.

Q.4:Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Ans :\( \ = \frac { \text { Total distance covered } } { \text { Total time taken } }\) \( \ = \frac { \text { Displacement } } { \text { Total time taken } }\)

Q.5:What does the odometer of an automobile measure?

Ans :The odometer an automobile measures the distance covered by an automobile.

## NCERT / CBSE Book for Class 9 Science

You can download the NCERT Book for Class 9 Science in PDF format for free. Otherwise you can also buy it easily online.

- Click here for NCERT Book for Class 9 Science
- Click here to buy NCERT Book for Class 9 Science

### All NCERT Solutions Class 9

- NCERT Solutions for Class 9 English
- NCERT Solutions for Class 9 Hindi
- NCERT Solutions for Class 9 Maths
- NCERT Solutions for Class 9 Science
- NCERT Solutions for Class 9 Social Science
- NCERT Solutions for Class 9 Sanskrit

### All NCERT Solutions

You can also check out NCERT Solutions of other classes here. Click on the class number below to go to relevant NCERT Solutions of Class 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

Class 1 | Class 2 | Class 3 |

Class 4 | Class 5 | Class 6 |

Class 7 | Class 8 | Class 9 |

Class 10 | Class 11 | Class 12 |

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