Obtain equations of motion for constant a acceleration using the method of calculus.

Explanation

(i) By definition `a=(dv)/(dt)` i.e., `dv=adt`
On intergrating both sides, `int_(v_(0))^(v)dv=int_(0)^(t)adt`
`(v-v_(0))=at`
`therefore v-v_(0)+at`
(ii) Consider `v=(dx)/(dt)` i.e., `dx=vdt`
Integrating we get `int_(x_(0))^(x)dx=int_(0)^(t)vdt=int_(0)^(t)(v_(0)+at)dt`
i.e., `(x-x_(0))=v_(0)t+1//2 at^(2)`
`therefore" "x=x_(0)+v_(0)t+1//2at^(2)`
Note : For vertical motion,
`y=y_(0)+v_(0)t pm g t^(2)`
Consider `a=(dv)/(dt)=(dv)/(dx).(dx)/(dt)=(vdv)/(dx)`
`therefore adx= vdx`. Integrating both sides,
`a int_(x_(0))^(x)dx=int_(v_(0))^(v)vdv`
`a(x-x_(0))=(1)/(2)[v^(2)-v_(0)^(2)]`
`therefore" "v^(2)=v_0)^(2)+2a(x-x_(0))` [Note `v_(0)=u, x-x_(0)=s`]
Note : For verical motion,
`v^(2)=v_(0)^(2)-2g(y-y_(0))` and for a free fall `v_(0)=0`.