Derive equations of uniformly acceleration motion by calculus method.

Explanation

(i) Consider an object moving in a straight line with uniform or constant acceleration 'a'.
(ii) Let 'u' be the initial velocity at time t = 0 and 'v' be the final velocity at time t.
Velocity - time relation :
(i) Acceleration, `a=(dv)/(dt) or dv=adt`
(ii) By integrating both sides, we get,
`int_(u)^(v)dv=int_(0)^(t)adt=aint_(0)^(t)dt=a[t]_(0)^(t)`
`v-u=at`
`v=u+at`
(b) Displacment - time relation. :
(i) Velocity, `v=(ds)/(dt)`
`"or "ds=v dt=(u+at)dt`
`" "[because v=u+at]`
(ii) By integrating both sides, we get,
`int_(0)^(s) dx=int_(0)^(t)(u+at)dt, int_(0)^(s)ds=u int_(0)^(t)dt+a int_(0)^(t)tdt`
`s=ut+(1)/(2)at^(2)`
(c) Velocity displacement relation
(i) Acceleration `a=(dv)/(dt)=(dv)/(ds)(ds)/(dt)=(dv)/(ds)v`
`ds=(1)/(a)vdv`
(ii) By interating both sides, we get
`int_(0)^(s)ds=(1)/(a)int_(a)^(v)vdv=(1)/(a)[(v^(2))/(2)]_(u)^(v)`
`s=(1)/(2a)(v^(2)-u^(2)), v^(2)-u^(2)=2as`
`v^(2)=u^(2)+2as`
Alternative method :
`a=vdv(dv)/(ds), a ds=vdv`
Integrating both sides, `int_(0)^(s)ads=int_(u)^(v)vdv`
If acceleration is constant, `aint_(0)^(s)ads=int_(u)^(n)vdv`
`a[s]_(0)^(s)={(v^(2))/(2)}(u)^(v)=((v^(2)-u^(2)))/(2),2as =v^(2)-u^(2)`
`v^(2)=u^(2)+2as`
(d) Displacement - average velocity relation :
(i) First Velocity `v=u+at`
`at=v-u" ...(1)"`
(ii) We know displacement `s=ut+(1)/(2)at^(2)`
(iii) Substituting equation (1), we get,
`s=ut+(1)/(2)(v-u)t, s=ut+(1)/(2)vt-(1)/(2)ut`
`s=((u+v)t)/(2)`