If the area below the line on a velocity time graph is the displacement (12(v-u) t) doesnt that contradict with the formula of motion that is (1 2 (v + u) t)?

Velocity-time graph formula

You have to add the area for when it was moving at speed u initially, which is u*t.

Adding (v-u)*t/2+u*t=(v*t-u*t+2*u*t)/2=(v*t+u*t)/2=(v+u)*t/2.

,The triangle area (v-u)*t/2 sits on top of the rectangle u*t, so the total displacement is the entire area, the sum, not just the triangle for accelerated motion.

What is displacement

Sometimes you just have to endure ambiguity and work with inconsistent conventions.

In situations where motion is restricted to one direction, itu2019s efficient and reasonable to just measure u201cdisplacementu201d along that line, an explicitly scalar quantity.

A more general version is when you define some starting point as the origin of coordinates and u201cwherever the object is at some later timeu201d as the displacement u2014 which in more than one dimension is intrinsically a vector quantity.

One could also imagine referring to the displacement along a fixed curved path (like a circle), again a scalar quantity even though the position of the object is now a vector.

,Just make sure you understand which version you are working with; sometimes itu2019s helpful to write out the current definition explicitly.