Derive s = ut + 1/2gt^2

Let u be the initial velocity of an object.

Under constant acceleration g,  the velocity of the object after time t is u+at.

Let us represent this in a graph. Let t be  on the X axis and velocity be shown along the y axis.

Mark the origin, (0,0) as O.

At time 0, velocity of the object is u. Mark the (0,u) and let this point be A on Y axis.

At time t, the velocity is  u+gt because of constant acceleration g.Mark the point B. Also mark the point (t , 0) on X axis as P.

Join the line PB

Draw a parallel line to X axis from A to meet PB at A'.

Now see that the displacement of the object due to the variable  velocity  is represented by the area under the line AB and and x axis and enclosed between the ordinates OA and PB.

Under uniform velocity the object would have made a displacement equal to the area OAA'P = (ordinate length u)(OP)= ut.

Under the effect of acceleration it has an additional displacement represented by the area of the triangle AA'B= (1/2)(AA')(A'P)

=(1/2)(OP)(PB-PA')=(1/2)(t)(gt)=(1/2)gt^2.

Thus the  displacemet, s of a body with an initial velocity u and constant acceleration g, in time t is given by s= ut+(1/2)gt^2. Proved geometrically.

Hope this helps.