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

Last updated at Dec. 16, 2024 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

slide6.jpg

Let’s mark the points and their coordinates

slide7.jpg

Velocity-Time Graph Question - Difficult - Motion Class 9 - Teachoo - Teachoo Questions part 2 - Question 2 - Graph Question (Difficult) - Teachoo Questions - Chapter 8 Class 9 - Motion - Class 9 part 3 - Question 2 - Graph Question (Difficult) - Teachoo Questions - Chapter 8 Class 9 - Motion - Class 9 part 4 - Question 2 - Graph Question (Difficult) - Teachoo Questions - Chapter 8 Class 9 - Motion - Class 9

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

Remove Ads

Transcript

In 2 − 10 seconds (Region AB) Acceleration = Slope of AB = (𝑦_2 − 𝑦_1)/(𝑥_2 − 𝑥_1 ) = (8 − 8)/(2 − 0) = 0/2 = 0 m/s2 Displacement = Area under v – t graph = Area of rectangle ABQP = Length × Breadth = AP × AB = 8 × 8 = 64 m In 10 − 12 seconds (Region BC) Acceleration = Slope of BC = (𝑦_2 − 𝑦_1)/(𝑥_2 − 𝑥_1 ) = (4 − 8)/(12 − 10) = (−4)/2 = –2 m/s2 Acceleration = Slope of BC = (𝑦_2 − 𝑦_1)/(𝑥_2 − 𝑥_1 ) = (4 − 8)/(12 − 10) = (−4)/2 = –2 m/s2 In 12 − 16 seconds (Region CD) Acceleration = Slope of OA = (𝑦_2 − 𝑦_1)/(𝑥_2 − 𝑥_1 ) = (4 − 4)/(16 −12) = 0/4 = 0 m/s2 Displacement = Area of rectangle CDSR = Length × Breadth = CD × CR = 4 × 4 = 16 m

Maninder Singh's photo - Co-founder, Teachoo

Made by

Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 14 years and a teacher from the past 18 years. He teaches Science, Economics, Accounting and English at Teachoo