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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

CA Maninder Singh's photo - Founder at Teachoo
CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 11 years and a teacher from the past 11 years. He teaches Science, Accounts and English at Teachoo