Question

derive the equation of motion (i) v = u +at (ii) v×2 = u×2 +2as​

Answer 1

$\bf \to \: \green{{ \underline{explanation \div}}}$

$\rm \to \: by \: using \: calculas \: method$

$\rm \to \: 1) = \: v = u \: + at \\ \\ \rm \to \: a \: = \frac{dv}{dt} \\ \\ \rm \to \: dv \: = adt \\ \\ \rm \to \: \int \limits_{u} {}^{v} (dv) = \int \limits_{0} {}^{t} (adt) \\ \\ \rm \to \: ( \: v \: - \: u \: ) = at \: - \: 0 \\ \\ \rm \to \: v \: = u \: + \: at$

$\rm \to \: 2) = {v}^{2} = {u}^{2} + 2as \\ \\ \rm \to \: from \: first \: equation \: of \: motion \\ \\ \rm \to \: v \: = u \: + at \\ \\ \rm \to \: t \: = \frac{v \: - \: u}{a} \\ \\ \rm \to \: from \: second \: equation \: of \: motion \\ \\ \rm \to \: s \: = ut \: + \: \frac{1}{2} a {t}^{2} \\ \\ \rm \to \: s \: = u \: ( \frac{v \: - \: u}{a} ) + \frac{1}{2} a( \frac{v \: - \: u}{a} ) \\ \\ \rm \to \: 2as \: = 2u \: ( \: v \: - \: u \: ) + \: ( \: v \: - \: u \: ) {}^{2} \\ \\ \rm \to \: 2as \: = 2vu \: - 2 {u}^{2} + {v}^{2} + {u}^{2} - 2vu \\ \\ \rm \to \: 2as \: = \: {v}^{2} - \: {u}^{2} \\ \\ \rm \to \: {v}^{2} = {u}^{2} + \: 2as$