Theory:

Position-time relation:
Assume an object covering a distance (\(s\)) in time (\(t\)) in an uniform acceleration (\(a\)). The area enclosed within \(OABC\) gives the distance travelled by the object.
 
Fig86.svg
Velocity time graph
 
Distance (\(s\)) travelled by the object,

s=areaOABC(atrapezium)=areaoftherectangleOADC+areaofthetriangleABD=OA×OC+12 (AD×BD)
 
Substituting \(OA\) \(=\) \(u\), \(OC\) \(=\) \(AD\) \(=\) \(t\) and \(BD\) \(=\) \(at\), we get

s=u×t+12t×ats=ut+12at2
Position-velocity relation:
The area enclosed within the trapezium \(OABC\) gives the distance (\(s\)) travelled by the object in time (\(t\)), moving under uniform acceleration (\(a\)).
 
s=areaofthetrapeziumOABCs=OA+BC×OC2
 
Substituting \(OA\) \(=\) \(u\), \(BC\) \(=\) \(v\) and \(OC\) \(=\) \(t\), we get
\(\)
s=(u+v)t2(a)

From the velocity-time relation (from the second equation of motion),

t=vua(b)

Substituting equation (b) on (a) we get,

s=(v+u)(vu)2a
 
(or)
 
2as=v2u2v2=u2+2as