Question

derive the equation of motion using graphical method (v*-u*=2as​

Answer 1

Correct Question :

Derive the equation of motion using graphical method v² = u² + 2as

Answer :

To prove :

• v² = u² + 2as

Proof :

$\mapsto\sf{s = Area \: of \: trapezium \: OABCD} \\ \\ \mapsto \sf{s = \dfrac{1}{2} \big(sum \: of \parallel side \big) \times height} \\ \\ \sf \mapsto{s = \frac{1}{2} \big(OA + BD) \times OD } \\ \\ \mapsto \sf{s = \frac{1}{2} \big(u + v \big) \times t } \\ \\ \mapsto \sf{ s = \frac{1}{2} \big(u + v \big) \times \big( \frac{v - u}{a} \big)} \\ \\ \mapsto \sf{s = \frac{ {v}^{2} - {u}^{2} }{2a} } \\ \\ \sf \mapsto{ {v}^{2} - {u}^{2} = 2as } \\ \\ \sf \mapsto{ \boxed{ \sf \red{ {v}^{2} = {u}^{2} + 2as }}}$

_______________....

Other Equations of Motion :

• v = u + at

• s = ut + ½ at²

• v² = u² + 2as

• v stands for Final velocity

• u stands for Initial velocity

• t stands for Time

• a stands for Acceleration

• s stands for Distance

Answer 2

Question given,

Derive the equation of motion using graphical method (v²-u²=2as)

$\bf\underline\red{Solution:}$

$S=ut+\frac{1}{2}a²\:and\:v=u+at,(time\:variant\:t)$

S(displacement)=Area of ABCD

$S=\frac{1}{2}(AB+CD)(AD)$

$S=\frac{1}{2}(U+V)(A)$

$\bf\underline\blue{Solving:}$

By using $V=u+as$

We will get$&#x21AA$$S=\frac{1}{2}(u+u+at)t$

$&#x21AA$$S=\frac{1}{2}(2u+at)t$

$&#x21AA$$S=ut+\frac{1}{2}at²$

$&#x21AA$$S=\frac{1}{2}(u+v)\frac{v-u}{a}$

$&#x21AA$$S=\frac{v²-u²}{2a}$

$&#x21AA$$\bf\underline\pink{v²=u²+2as}$

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