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2nd Equation of Motion

Derive second...

Question

Derive second and third equation of motion(mathematically

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Solution

$L e t t h e d i s tan c e t r a v e l l e d b y t h e b o d y b e “ S ” . W e k n o w t h a t D i s tan c e = A v e r a g e v e l o c i t y \times T i m e A l s o, A v e r a g e v e l o c i t y = \frac{u + v}{2} ∴ D i s tan c e = \frac{u + v}{2} \times t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1) A g a i n w e k n o w t h a t v = u + a t s u b s t i t u t i n g t h i s v a l u e o f “ v ” i n e q . (1), w e g e t D i s tan c e = \frac{u + u + a t}{2} \times t S = \frac{2 u t + a t^{2}}{2} S = u t + \frac{1}{2} a t^{2} t h i s i s s e c o n d e q u a t i o n o f m o t i o n . N o w l e t' s d e r i v e t h i r d e q a t i o n o f m o t i o n F r o m t h e f i r s t e q u a t i o n o f m o t i o n v = u + a t ∴ t = \frac{v - u}{a} A v e r a g e v e l o c i t y i s g i v e n a s = \frac{v + u}{2} D i s tan c e = A v e r a g e v e l o c i t y \times T i m e S = \frac{v + u}{2} \times \frac{v - u}{a} S = \frac{v^{2} - u^{2}}{2 a} v^{2} = u^{2} + 2 a s T h i s i s t h i r d e q u a t i o n o f m o t i o n$

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