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Chapter 8 Class 9 - Motion

Question 2 - Graph Question (Difficult) - Teachoo Questions - Chapter 8 Class 9 - Motion (Term 1)

Last updated at March 16, 2023 by

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

Β  Β = Displacement in OA + Displacement in AB

Β  Β  Β + Displacement in BC + Displacement in CD

Β  = 8 + 64 + 12 + 16

Β  = 100 m

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Transcript

In 2 βˆ’ 10 seconds (Region AB) Acceleration = Slope of AB = (𝑦_2 βˆ’ 𝑦_1)/(π‘₯_2 βˆ’ π‘₯_1 ) = (8 βˆ’ 8)/(2 βˆ’ 0) = 0/2 = 0 m/s2 Displacement = Area under v – t graph = Area of rectangle ABQP = Length Γ— Breadth = AP Γ— AB = 8 Γ— 8 = 64 m In 10 βˆ’ 12 seconds (Region BC) Acceleration = Slope of BC = (𝑦_2 βˆ’ 𝑦_1)/(π‘₯_2 βˆ’ π‘₯_1 ) = (4 βˆ’ 8)/(12 βˆ’ 10) = (βˆ’4)/2 = –2 m/s2 Acceleration = Slope of BC = (𝑦_2 βˆ’ 𝑦_1)/(π‘₯_2 βˆ’ π‘₯_1 ) = (4 βˆ’ 8)/(12 βˆ’ 10) = (βˆ’4)/2 = –2 m/s2 In 12 βˆ’ 16 seconds (Region CD) Acceleration = Slope of OA = (𝑦_2 βˆ’ 𝑦_1)/(π‘₯_2 βˆ’ π‘₯_1 ) = (4 βˆ’ 4)/(16 βˆ’12) = 0/4 = 0 m/s2 Displacement = Area of rectangle CDSR = Length Γ— Breadth = CD Γ— CR = 4 Γ— 4 = 16 m

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