Question

prove that:(1) v=u+at (2) s=it+1/2at² (3)2as=v²-u²​

Answer 1

Derivation of equation of motion

For One dimensional motion with a = constant

We can write,

$\rm \implies \: dv \: = \: adt \: \: \implies \: a \: = \dfrac{dv}{dt}$

Where a = constant

Integrating both side , we get

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

At t = 0 , Velocity is u and at t = t velocity is v Hence

$\rm \implies \: \int ^{v} _{u} dv = a \int ^{t} _0dt$

$\rm \implies \: \bigg[v \bigg]^{v} _u = a \bigg[t \bigg]^{t} _0$

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

Hence proved

Now we can write

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

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

At time t = 0 Suppose s=0 and at t = t , displacement is a

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

$\rm \implies \bigg[s \bigg]^{s} _0 = \bigg[ut + \dfrac{1}{2} a {t}^{2} \bigg] ^{t} _0$

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

Hence proved

Again we can write as

$\rm \implies \: vds = ads \implies \: a = v \dfrac{dv}{ds}$

When s = 0 , v is u and at s = s , velocity is v . Therefore

$\rm \implies \: \int^{v} _uv.dv = a \int ^{s} _0ds$

$\implies \rm \: \bigg[ \dfrac{ {v}^{2} }{2} \bigg]^{v} _u = a \bigg[s \bigg]^{s} _0$

$\rm \implies \: \dfrac{ {v}^{2} }{2} - \dfrac{ {u}^{2} }{2} = as$

$\implies \boxed{ \rm \: {v}^{2} = {u}^{2} + 2as}$

Hence proved