graphical representation equation of motion

• Class 6 Maths
• Class 6 Science
• Class 6 Social Science
• Class 6 English
• Class 7 Maths
• Class 7 Science
• Class 7 Social Science
• Class 7 English
• Class 8 Maths
• Class 8 Science
• Class 8 Social Science
• Class 8 English
• Class 9 Maths
• Class 9 Science
• Class 9 Social Science
• Class 9 English
• Class 10 Maths
• Class 10 Science
• Class 10 Social Science
• Class 10 English
• Class 11 Maths
• Class 11 Computer Science (Python)
• Class 11 English
• Class 12 Maths
• Class 12 English
• Class 12 Economics
• Class 12 Accountancy
• Class 12 Physics
• Class 12 Chemistry
• Class 12 Biology
• Class 12 Computer Science (Python)
• Class 12 Physical Education
• GST and Accounting Course
• Excel Course
• Tally Course
• Finance and CMA Data Course
• Payroll Course

Interesting

• Learn English
• Learn Excel
• Learn Tally
• Learn GST (Goods and Services Tax)
• Learn Accounting and Finance
• GST Tax Invoice Format
• Accounts Tax Practical
• Tally Ledger List
• GSTR 2A - JSON to Excel
• NCERT Questions
• Questions from Inside the chapter
• Examples from NCERT Book
• Teachoo Questions

Graphical Derivation of Equations of Motion

Last updated at April 16, 2024 by Teachoo

Graphical Derivation of all 3 Equations of Motion

Our 3 equations of motion are

s = ut + 1 / 2at 2

v 2 - u 2 = 2as

Let's suppose an object with initial velocity u to final velocity v in time t. Let's derive all 3 equations

Initial velocity = u = OA = CD

Final velocity = v = BD

Time taken = t = OD = AC

First Equation of Motion

Second Equation of Motion

Third Equation of Motion

First equation of motionWe know that Acceleration = Slope of v-t graph Coordinates of point A x-coordinate of A = Time = 0 y-coordinate of A = Initial velocity = u ∴ Coordinates of point A = (0, u) Coordinates of point B x-coordinate of B = Time = t y-coordinate of A = Final velocity = v ∴ Coordinates of point A = (t, v) Acceleration = Slope of v-t graph a = (𝑦_2 − 𝑥_1)/(𝑥_2 − 𝑥_1 ) a = (𝑣 − 𝑢)/(𝑡 − 0) a = (𝑣 − 𝑢)/𝑡 at = v − u v = u + at Acceleration = Slope of v-t graph a = (𝑦_2 − 𝑥_1)/(𝑥_2 − 𝑥_1 ) a = (𝑣 − 𝑢)/(𝑡 − 0) a = (𝑣 − 𝑢)/𝑡 at = v − u v = u + at Second equation of motion We know that Distance = Area under v-t graph Distance = Area of OABD Distance = Area of rectangle OACD + Area of Δ ABC s = OA × OD + 1/2 × AC × BC s = u × t + 1/2 × t × (v − u) From 1st equation of motion v = u + at v − u = at Putting v − u = at in equation (1) s = ut + 1/2 t (at) s = ut + 𝟏/𝟐 at2 Third equation of motion Here, we find Distance again But this time, we find Distance of OABD using Area of Trapezium formula We know that Distance = Area under v-t graph Distance = Area of OABD Distance = 1/2 × (Sum parallel sides) × Height s = 1/2 (BD + OA) × OD s = 1/2 (v + u) t Now, From 1st equation of motion v = u + at v – u = at (𝑣 − 𝑢)/𝑎 = t t = (𝑣 − 𝑢)/𝑎 Put value of t in (2) s = 1/2 (v + u) ((𝑣 − 𝑢))/𝑎 s = 1/2𝑎 (v2 − u2) 2as = v2 − u2 v2 − u2 = 2as

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

Hi, it looks like you're using AdBlock :(

Please login to view more pages. it's free :), solve all your doubts with teachoo black.