Question

derive calculusly S=ut+1/2at^2

Answer 1

Hope u like my process
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=> Let the initial velocity at time t= 0 be u.

=> at time = 0, displacement will also be 0.

=> we know,
$a = \green{\frac{dv}{dt} }\: \: \: and \: \: v = \green{ \frac{ds}{dt} }$

Now,
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$= > \blue{ \frac{dv}{dt}} = \blue{a} \\ \\ = > \blue{dv} = \blue{a.dt} \\ \\ \underline{ \: integrating \: \: both \: \: sides \: } \\ \\ = > \int \blue{dv} = \blue{a}\int \blue{dt} \\ \\ = > \bf \blue{v} = \blue{at + u} \\ \\ = > \blue{\frac{ds}{dt} }= \blue{ u + at} \\ \\ = > \blue{ds} = \blue{u.dt + a.tdt} \\ \\ \underline{integrating \: \: both \: \: sides \: \: again} \\ \\ = > \int \blue{ds} = \blue u \int \blue{dt} + \blue{a} \int \blue{t.dt} \\ \\ = > \bf \blue{s} = \blue{ut + a \frac{ {t}^{2} }{2} } \\ \\ \boxed{ = > s = \orange{ut + \frac{1}{2}a {t}^{2} }}$

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Answer 2

Hello!

$\sf v = \dfrac{\sf ds}{\sf dt}$

$\implies \sf ds = \sf vdt$

$\implies \int \sf ds = \int \sf vdt$

$\implies \int \sf ds = \int (\sf u+at) \: \sf dt$

$\implies \int^{s}_{0} \sf ds = \int^{t}_{0} \sf uat + \int^{t}_{0} \sf atdt$

$\implies \sf (s-0) = \sf u(t-0) + \sf a(\dfrac{t^{2}}{2}-0)$

$\implies \boxed{\red{\bf{s = \sf ut + \dfrac{1}{2}at^{2}}}}$

Cheers!