Graphicly derive3rd equation motion

1.	Graphicly derive3rd equation motion
Answer» Answer: Derivation of Third EQUATION of Motion by Graphical Method Derivation Of Equation Of Motion The total distance travelled, S = Area of TRAPEZIUM OABC. So, S= 1/2(SumofParallelSides)×Height S=(OA+CB)×OC Since, OA = u, CB = v, and OC = t The above equation becomes S= 1/2(u+v)×t Now, since t = (v – u)/ a The above equation can be written as: S= 1/2(u+v)×(v-u)/a Rearranging the equation, we get S= 1/2(v+u)×(v-u)/a S = (v2-u2)/2a Third equation of motion is obtained by solving the above equation: v2 = u2+2aS Derivation of Third Equation of Motion by Calculus Method It is KNOWN that, Derivation Of Equation Of Motion Explanation: PLS MARK ME AS BRAINLIST PLS

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Answer:

Derivation of Third EQUATION of Motion by Graphical Method

Derivation Of Equation Of Motion

The total distance travelled, S = Area of TRAPEZIUM OABC.

So, S= 1/2(SumofParallelSides)×Height

S=(OA+CB)×OC

Since, OA = u, CB = v, and OC = t

The above equation becomes

S= 1/2(u+v)×t

Now, since t = (v – u)/ a

The above equation can be written as:

S= 1/2(u+v)×(v-u)/a

Rearranging the equation, we get

S= 1/2(v+u)×(v-u)/a

S = (v2-u2)/2a

Third equation of motion is obtained by solving the above equation:

v2 = u2+2aS

Derivation of Third Equation of Motion by Calculus Method

It is KNOWN that,

Derivation Of Equation Of Motion

Explanation:

PLS MARK ME AS BRAINLIST PLS

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