NCERT Solutions for Class 9 Science Chapter 8
Motion
NCERT Books for Session 2022-2023
CBSE Board and UP Board
Exercises Questions
Page No-112
Questions No2
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?
Share
(a) For motion from A to B:
Distance covered = 300 m
Displacement = 300 m.
Time taken = 150 sec.
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.
For more answers visit to website:
https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-8/
1. Average speed = Total distance / Total time taken
2. Average velocity = Total displacement / Total time taken
Let’s start with the given information:
– Joseph jogs from A to B, a distance of 300 meters, in 2 minutes 30 seconds.
– Then, he turns around and jogs back 100 meters to point C in another 1 minute.
(a) Average Speed and Velocity from A to B:
1. Average Speed from A to B:
Speed = Total distance / Total time taken
Total distance from A to B = 300 meters
Total time taken from A to B = 2 minutes 30 seconds = 2.5 minutes
Speed = 300 meters / 2.5 minutes
Speed = 120 meters per minute
Therefore, Joseph’s average speed from A to B is 120 meters per minute.
2. Average Velocity from A to B:
As Joseph moves from A to B in a straight line, his displacement is the distance between the initial and final points.
Displacement from A to B = 300 meters (since he returns to the starting point, there’s no net displacement)
Total time taken from A to B = 2.5 minutes
Velocity = Displacement / Total time taken
Velocity = 300 meters / 2.5 minutes
Velocity = 120 meters per minute
Therefore, Joseph’s average velocity from A to B is 120 meters per minute.
(b) Average Speed and Velocity from A to C:
1. Average Speed from A to C:
Total distance from A to C = 300 meters + 100 meters = 400 meters
Total time taken from A to C = 2.5 minutes + 1 minute = 3.5 minutes
Speed = Total distance / Total time taken
Speed = 400 meters / 3.5 minutes
Speed ≈ 114.29 meters per minute
Therefore, Joseph’s average speed from A to C is approximately 114.29 meters per minute.
2. Average Velocity from A to C:
Joseph’s displacement from A to C accounts for the net distance covered in a straight line.
Displacement from A to C = 300 meters (distance from A to B) – 100 meters (distance from B to C)
Displacement from A to C = 200 meters (in the direction from A to C)
Total time taken from A to C = 3.5 minutes
Velocity = Displacement / Total time taken
Velocity = 200 meters / 3.5 minutes
Velocity ≈ 57.14 meters per minute
Therefore, Joseph’s average velocity from A to C is approximately 57.14 meters per minute.