Find Displacement and Acceleration from Time 0-2 seconds, 2-10 secs, 10-12 seconds, 12-16 seconds. Also tell the type of motion in all the regions. And find the total displacement

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Let’s mark the points and their coordinates

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Motion in region OA = Uniformly accelerated

Motion in region AB = Uniformly motion (No acceleration)

Motion in region BC = Uniformly retarded

Motion in region CD = Uniformly motion (No acceleration)

Total displacement

   = Displacement in OA + Displacement in AB

     + Displacement in BC + Displacement in CD

  = 8 + 64 + 12 + 16

  = 100 m

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

Transcript

In 2 โˆ’ 10 seconds (Region AB) Acceleration = Slope of AB = (๐‘ฆ_2 โˆ’ ๐‘ฆ_1)/(๐‘ฅ_2 โˆ’ ๐‘ฅ_1 ) = (8 โˆ’ 8)/(2 โˆ’ 0) = 0/2 = 0 m/s2 Displacement = Area under v โ€“ t graph = Area of rectangle ABQP = Length ร— Breadth = AP ร— AB = 8 ร— 8 = 64 m In 10 โˆ’ 12 seconds (Region BC) Acceleration = Slope of BC = (๐‘ฆ_2 โˆ’ ๐‘ฆ_1)/(๐‘ฅ_2 โˆ’ ๐‘ฅ_1 ) = (4 โˆ’ 8)/(12 โˆ’ 10) = (โˆ’4)/2 = โ€“2 m/s2 Acceleration = Slope of BC = (๐‘ฆ_2 โˆ’ ๐‘ฆ_1)/(๐‘ฅ_2 โˆ’ ๐‘ฅ_1 ) = (4 โˆ’ 8)/(12 โˆ’ 10) = (โˆ’4)/2 = โ€“2 m/s2 In 12 โˆ’ 16 seconds (Region CD) Acceleration = Slope of OA = (๐‘ฆ_2 โˆ’ ๐‘ฆ_1)/(๐‘ฅ_2 โˆ’ ๐‘ฅ_1 ) = (4 โˆ’ 4)/(16 โˆ’12) = 0/4 = 0 m/s2 Displacement = Area of rectangle CDSR = Length ร— Breadth = CD ร— CR = 4 ร— 4 = 16 m

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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.