Joseph jogs from one end class 9

CBSE Study Material • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • NCERT Solutions for Class 9 Science Chapter 8: Motion NCERT Solutions Class 9 Science Chapter 8 – Free PDF Download * According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 7. NCERT Solutions for Class 9 Science Chapter 8 Motion is designed with the intention of clarifying doubts and concepts easily. NCERT Solutions for Class 9 Science approaches students in a student-friendly way and is loaded with questions, activities, and exercises that are CBSE exam and competitive exam-oriented. NCERT Solutions for Class 9 Science is the contribution of our faculty, having vast teaching experience. It is developed keeping in mind the concept-based approach along with the precise answering method for CBSE examinations. Refer to Previous Next  Access Answers of Science NCERT class 9 Chapter 8: Motion (All intext and exercise questions solved) Intext Questions – 1   Page: 100 1. An object has moved through a distance. Can it have zero displacement? If ...

Answer (a) From end A to end B Distance covered A to B = 300 m Time taken to cover that distance = 2 min 30 seconds = 2 x 60 + 30 = 150 s Average speed = 300 / 150 = 2 m/s Average velocity = 300 / 150= 2 m/s The average speed and average velocity of Joseph from A to B are the same and equal to 2 m/s. (b) From end A to end C Total distance covered = Distance from A to B + Distance from B to C = 300 + 100 = 400 m Total time taken = Time taken to travel from A to B + Time taken to travel from B to C = 150 + 60 = 210 s Average speed = 400 / 210= 1.90 m/s Average velocity = 200 / 210= 0.95 m/s The average speed of Joseph from A to C is 1.90 m/s and his average velocity is 0.95 m/s. Popular Questions of Class 9 Science • Q:- 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 40 km h -1. What is the average speed for Abdul’s trip? • Q:- An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down. • Q:- A driver of a car travelling at 52 km h -1applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h -1in 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...

• Menu Toggle • CBSE Class 6 • CBSE Class 7 • CBSE Class 8 • CBSE Class 9 • CBSE Class 10 • Menu Toggle • NCERT Solutions • NCERT Books • CBSE Sample Papers • CBSE Extra Questions • CBSE Important Questions • CBSE English Summary • CBSE Syllabus • Menu Toggle • RS Aggarwal Solutions • RD Sharma Class 10 Solutions • Menu Toggle • About Us • Contact Us • Terms And Conditions • Privacy Policy • Disclaimer NCERT Solutions for Class 9 Science Chapter 8 Motion NCERT Solutions for Class 9 Science (physics) Chapter 8 Motion are given below. In these solutions, we have answered all the intext and exercise questions provided in NCERT class 9 science textbook. Class 9 Science Chapter 8 Textbook Questions and Answers INTEXT QUESTIONS PAGE NO. 100 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 alon...

Transcript NCERT Question 2 Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 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? From A to B Average Speed Total Distance = AB = 300 m Total Time = 2 mins 30 secs = 2 × 60 + 30 sec = 120 +30 = 150 s Avg. Speed = (𝑇𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒)/(𝑇𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒) = 300/150 = 2 m/s Average Velocity Total Displacement = AB = 300 m Total Time = 2 mins 30 secs = 2 × 60 + 30 sec = 120 +30 = 150 s Avg. Velocity = (𝑇𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡)/(𝑇𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒) = 300/150 = 2 m/s (b) From A to C Average Speed Total Distance = AB + BC = 300 + 100 m = 400 m Total Time = 2 min 30 sec + 1 min = 3 min 30 sec = 3 × 60 + 30 sec = 180 + 30 = 210 s Average Velocity Total Displacement = AC = 300 – 100 = 200m Total Time = 2 min 30 sec + 1 min = 3 min 30 sec = 3 × 60 + 30 sec = 180 + 30 = 210 s Avg. Speed = (𝑇𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒)/(𝑇𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒) = 400/210 = 40/21 = 1.90 m/s Avg. Velocity = (𝑇𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡)/(𝑇𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒) = 200/210 = 20/21 = 0.95 m/s Show More

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