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  1. Class 9
  2. Chapter 8 Class 9 - Motion

Transcript

Example 8.3 Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average Velocity of Usha. Given Length of the pool = 90 m Time taken = 1 minute = 60 seconds Usha travels from A to B, then back from B to A Distance travelled = AB + BA = 90 + 90 = 180 m Since she comes back to initial position, Displacement = 0 Average Speed Distance = AB + BA = 90 + 90 = 180 m Average speed = (๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘‘๐‘–๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’)/(๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘ก๐‘–๐‘š๐‘’) = 180/60 = 3 m/s Average Velocity Since she comes back to initial position, Displacement = 0 Average velocity = ๐ท๐‘–๐‘ ๐‘๐‘™๐‘Ž๐‘๐‘š๐‘’๐‘›๐‘ก/(๐‘‡๐‘–๐‘š๐‘’ ๐‘ก๐‘Ž๐‘˜๐‘’๐‘›) = 0/60 = 0 m/s

About the Author

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.