Let dv/dt= a, Then integration yields the equation v = at + C
But at time t=0 the equation becomes v = C = u. Thus v = u + at
And, also, let ds/dt= v, after integration you get the equation s = ut + 1/2at^2 + C (Note that during integration you have to subsitute 'v' for 'u + at' since 'v' cannot always be considerd a constant)

So now that you have s = ut + 1/2at^2 + C, set the time to '0' to give the equation s = C = s'(which I chose to denote the initial distance coverd already)
Then you have the two equations;

s - s' = ut + 1/2at^2
v = u + at

Solve the two equations by eliminating in turn by eliminating the constants the accleration, a, and the the time; t.
and then you have the other equations

v^2 = u^2 + 2a(s-s')
s-s' = 1/2(v + u)t
Swell huh? ;D