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

Explanation

By definition
`a=(d upsilon)/(dt)`
`d upsilon = a dt`
Integrating both sides
`int_(upsilon_(0))^(upsilon)d upsilon = int_(0)^(t) a dt`
`= a int_(0)^(t) dt " "` (a is constant)
`upsilon - upsilon_(0)=at`
`upsilon=upsilon_(0)+at`
Further, `upsilon = (dx)/(dt)`
`dx = upsilon dt`
Integrating both sides
`int_(x_(0))^(x)dx=int_(0)^(t)upsilon dt`
`=int_(0)^(t)(upsilon_(0)+at)dt`
`x-x_(0)=upsilon_(0)t+(1)/(2)at^(2)`
`x=x_(0)+upsilon_(0)t+(1)/(2)at^(2)`
We can write
`a=(d upsilon)/(dt)=(d upsilon)/(dx)(dx)/(dt)=upsilon (d upsilon)/(dx)`
or `upsilon d upsilon = a dx`
Integrating both sides,
`int_(upsilon_(0))^(upsilon)upsilon d upsilon = int_(x_(0))^(x)a dx`
`(upsilon^(2)-upsilon_(0)^(2))/(2)=a(x-x_(0))`
`upsilon^(2)=upsilon_(0)^(2)+2a(x-x_(0))`
The advantage of this method is that it can be used for motion with non-uniform acceleration also. Now, we shall use these equations to some important cases. t