Question

derive the equation s = ut+1at^2/2​

Answer 1

Explanation:

velocity = displacement/time

Vavg =s/t

s = Vavg.t

= (v+u/2)t

= (u+at +u/2)t

= (2u + at /2)t

= 2ut + at²/2

= ut +1/2at² -----> hence proved

Answer 2

Solution:-

$\rm \: to \: prove : s = ut \: + \dfrac{1}{2} at ^{2} \: \: \: \: \: \: \: \: first \: of \: all \: we \:have \: to \: prove \: v = u \: + at$

Now

=> For one dimensional motion with a = constant

$\rm \: as \: \: \: \: \: a = \dfrac{dv}{dt}$

We can write

$\rm \: dv \: = a \: dt$

Integrating both side we have

$\rm \: \int dv = a \int dt$

At t=0 , velocity is u and at t = t velocity is v hence

$\rm \: \int ^{v} _{u}dv \: = a \int^{t} _{0}dt$

$\rm \: [v]^{v}_{u} = a[t]^{t}_{0}$

So

$\boxed{ \rm \: v \: - u \: = at \: } \\ \rm \: or \\ \boxed{ \rm \: v \: = u \: + at}$

Further , we can write

$\rm \: v \: = \dfrac{ds}{dt} \implies \: ds \: = vdt$

$\rm \: = (u + at \: )dt$

At time t = 0 suppose s = 0 and at t = t , displacement is s then

$\rm \: \int ^{s} _{0}ds = \int^{t}_{0}(u + at) dt$

$\rm \: [s] ^{s}_{0} = \bigg[ut + \dfrac{1}{2} a {t}^{2} \bigg]^{t} _{0}$

$\boxed{\rm \: s \: = ut \: + \dfrac{1}{2} a {t}^{2} }$

Hence proved