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

Explanation

Acceleration `a = (dv)/(dt)`
`therefore dv =a dt`
By intergrating both side,
`int _(v _(0)) ^(v) dv = int _(0) ^(t) a dt`
`[v] _(v _(0)) ^(v) = a [t] _(0) ^(t)`
`therefore v - v _(0) = at `
`therefore v =v _(0) + at " "...(1)`
Now, `(dx)/(dt)=v`
`therefore dx = v dt`
By integrating both sides,
`int _(x _(0)) ^(x) dx = int _(0) ^(t) v dt`
`[x] _(x _(0)) ^(x) = int_(0) ^(t) (v _(0) + at) dt` [from equation (1)]
`x-x_(0) =[ v _(0) t + (at ^(2))/( 2) ] _(0) ^(t)`
` = v _(0) t + 1/2 at ^(2)`
`therefore x=x_(0) +v_(0) t + 1/2 at _(2) " "...(2)`
Now `a = (dv)/(dt) = (dv)/(dx) xx (dx)/(dt)`
`therefore a = v (dv)/(dx) [ because (dx)/(dt) =v]`
`therefore v dv = a dx`
By intergrating both sides,
`int _(v _(0)) ^(v) v dv = int _(x _(0)) ^(x) a dx `
`[ (v ^(2))/( 2) ] _(v_(0)) ^(v) = a [x] _(x _(0)) ^(x)`
` therefore (v ^(2) - v _(0) ^(2))/(2) = a (x-x_(0))`
`therefore v ^(2) -v_(0) ^(2) = 2a (x -x _(0))`
`therefore v ^(2) = v _(0) ^(2) + 2a (x-x_(0)) " "...(3)`
Equation (1), (2) and (3) are equation of uniformly accelerated motion.