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Derivation of kinematic equations

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Derivation of Kinematic
Equations
Sama Ehab, Mirna Amr, Nancy Mohamed, Salma Ammar
๐’—๐’Š + ๐’—๐’‡
๐œŸ๐’™ = (
)๐œŸ๐’•
๐Ÿ
Δ๐‘ฅ
•
Average velocity is represented as: ๐‘ฃ =
•
Average velocity can also be represented as: ๐‘ฃ =
•
Rearrange the equation above to solve for Δ๐‘ฅ
Δ๐‘ก
๐‘ฃ๐‘– +๐‘ฃ๐‘“
2
Δ๐‘ฅ = ๐‘ฃΔ๐‘ก
•
๐‘ฃ๐‘– +๐‘ฃ๐‘“
Substitute ๐‘ฃ with
๐‘ฃ๐‘– + ๐‘ฃ๐‘“
Δ๐‘ฅ = (
)Δ๐‘ก
2
2
in the equation above to get our first kinematic equation.
๐›ฅ๐‘ฅ =
1
๐‘๐‘Ž๐‘ ๐‘’ × โ„Ž๐‘’๐‘–๐‘”โ„Ž๐‘ก + (๐‘™๐‘’๐‘›๐‘”๐‘กโ„Ž × ๐‘ค๐‘–๐‘‘๐‘กโ„Ž)
2
๐›ฅ๐‘ฅ =
1
๐›ฅ๐‘ก ๐‘ฃ๐‘“ − ๐‘ฃ๐‘– + ๐‘ฃ๐‘– × ๐›ฅ๐‘ก
2
1
1
= ๐‘ฃ๐‘“ ๐›ฅ๐‘ก − ๐‘ฃ๐‘– ๐›ฅ๐‘ก + ๐‘ฃ๐‘– ๐›ฅ๐‘ก
2
2
1
1
= ๐‘ฃ๐‘“ ๐›ฅ๐‘ก + ๐‘ฃ๐‘– ๐›ฅ๐‘ก
2
2
Δ๐‘ฅ = (
o
We know that the area under the graph is
the displacement (Δ๐‘ฅ).
๐‘ฃ๐‘“ + ๐‘ฃ๐‘–
)๐›ฅ๐‘ก
2
๐’—๐’‡ = ๐’—๐’Š + ๐’‚๐œŸ๐’•
Δ๐‘ฃ
Δ๐‘ก
•
Acceleration is represented as: ๐‘Ž =
•
Substitute Δ๐‘ฃ with ๐‘ฃ๐‘“ − ๐‘ฃ๐‘–
๐‘ฃ๐‘“ − ๐‘ฃ๐‘–
๐‘Ž=
Δ๐‘ก
•
Rearrange the equation above to solve for ๐‘ฃ๐‘“
๐‘ฃ๐‘“ = ๐‘ฃ๐‘– + ๐‘ŽΔ๐‘ก
๐‘ฃ๐‘“ = ๐ต๐ถ
๐‘ฃ๐‘“ = ๐ต๐ท + ๐ท๐ถ
๐‘ฃ๐‘“ = ๐ต๐ท + ๐‘ฃ๐‘–
๐‘Ž=
๐ต๐ท
๐ด๐ท
๐‘Ž=
๐ต๐ท
Δ๐‘ก
๐ต๐ท = ๐‘ŽΔ๐‘ก
๐‘ฃ๐‘“ = ๐‘ฃ๐‘– + ๐‘ŽΔ๐‘ก
๐Ÿ
๐œŸ๐’™ = ๐’—๐’Š ๐œŸ๐’• + ๐’‚๐œŸ๐’•๐Ÿ
๐Ÿ
๐‘ฃ๐‘– +๐‘ฃ๐‘“
•
Start with the first kinematic equation (XVT) Δ๐‘ฅ =
•
Substitute ๐‘ฃ๐‘“ with our second kinematic equation (VAT) ๐‘ฃ๐‘“ = ๐‘ฃ๐‘– + ๐‘ŽΔ๐‘ก.
Δ๐‘ฅ =
๐‘ฃ๐‘– + (๐‘ฃ๐‘– + ๐‘ŽΔ๐‘ก)
Δ๐‘ก
2
2๐‘ฃ๐‘– ๐‘ŽΔ๐‘ก
+
Δ๐‘ก
2
2
•
Expand the equation
•
Simplify the equation Δ๐‘ฅ = ๐‘ฃ๐‘– Δ๐‘ก + ๐‘ŽΔ๐‘ก 2
Δ๐‘ฅ =
1
2
2
Δ๐‘ก.
๐›ฅ๐‘ฅ =
1
๐‘๐‘Ž๐‘ ๐‘’ × โ„Ž๐‘’๐‘–๐‘”โ„Ž๐‘ก + (๐‘™๐‘’๐‘›๐‘”๐‘กโ„Ž × โ„Ž๐‘’๐‘–๐‘”โ„Ž๐‘ก)
2
=
1
Δ๐‘ก ๐ต๐ท + (๐‘ฃ๐‘– × Δ๐‘ก)
2
๐‘Ž=
๐ต๐ท
๐ด๐ท
๐‘Ž=
๐ต๐ท
Δ๐‘ก
๐ต๐ท = ๐‘ŽΔ๐‘ก
=
1
Δ๐‘ก ๐‘ŽΔ๐‘ก + ๐‘ฃ๐‘– Δ๐‘ก
2
1
๐›ฅ๐‘ฅ = ๐‘ฃ๐‘– ๐›ฅ๐‘ก + ๐‘Ž๐›ฅ๐‘ก 2
2
๐’—๐Ÿ๐’‡ = ๐’—๐Ÿ๐’Š + ๐Ÿ๐’‚๐œŸ๐’™
•
๐‘ฃ๐‘– +๐‘ฃ๐‘“
Start with the first kinematic equation (XVT) Δ๐‘ฅ = (
•
2
)Δ๐‘ก.
Use the second kinematic equation (VAT) ๐‘ฃ๐‘“ = ๐‘ฃ๐‘– + ๐‘ŽΔ๐‘ก,
and rearrange it to solve for t.
๐‘ฃ๐‘“ − ๐‘ฃ๐‘–
Δ๐‘ก =
๐‘Ž
๐‘ฃ๐‘“ − ๐‘ฃ๐‘–
• Substitute Δ๐‘ก in our first equation with
.
๐‘Ž
๐‘ฃ๐‘– +๐‘ฃ๐‘“
Δ๐‘ฅ = (
2
๐‘ฃ๐‘“ − ๐‘ฃ๐‘–
)(
๐‘Ž
).
๐‘ฃ๐‘“2 −๐‘ฃ๐‘–2
•
Multiply the fractions and simplify Δ๐‘ฅ = (
•
We can rearrange the equation above to solve for ๐‘ฃ๐‘“2
๐‘ฃ๐‘“2 = ๐‘ฃ๐‘–2 + 2๐‘ŽΔ๐‘ฅ
2๐‘Ž
).
1
Δ๐‘ฅ = (๐‘๐‘Ž๐‘ ๐‘’1 + ๐‘๐‘Ž๐‘ ๐‘’2)(โ„Ž๐‘’๐‘–๐‘”โ„Ž๐‘ก)
2
1
Δ๐‘ฅ = ๐‘ฃ๐‘– + ๐‘ฃ๐‘“ Δ๐‘ก
2
๐‘ฃ๐‘“ − ๐‘ฃ๐‘–
๐‘Ž=
Δ๐‘ก
Δ๐‘ก =
=
๐‘ฃ๐‘“ − ๐‘ฃ๐‘–
๐‘Ž
1
๐‘ฃ + ๐‘ฃ๐‘“
2 ๐‘–
๐‘ฃ๐‘“ − ๐‘ฃ๐‘–
๐‘Ž
1 ๐‘ฃ๐‘“2 − ๐‘ฃ๐‘–2
=
2
๐‘Ž
2๐‘ŽΔ๐‘ฅ = ๐‘ฃ๐‘“2 − ๐‘ฃ๐‘–2
๐‘ฃ๐‘“2 = ๐‘ฃ๐‘–2 + 2๐‘ŽΔ๐‘ฅ
Resources
“Derivation of Equations of Motion.” Byju’s, 28 June 2018,
https://byjus.com/physics/derivation-of-equation-of-motion/#Derivation-of-ThirdEquation-of-Motion.
“Kinematic Equations.” Pasco, https://www.pasco.com/products/guides/kinematic-equations.
“What are the kinematic formulas?” Khan Academy,
https://www.khanacademy.org/science/physics/one-dimensional-motion/kinematicformulas/a/what-are-the-kinematic-formulas.
“Derivation of Kinematic Equations.”
https://www.muncysd.org/site/handlers/filedownload.ashx?moduleinstanceid=2437&data
id=4035&FileName=Kinematic%20Eqns.pdf.
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