What is the relationship between displacement and time squared (d = vt -.5at2)?

Question

just to clarify, that is the relationship of a displacement vs time squared graph?

Answer 1

isnt that accerlation?

d = vt +/- 0.5at^2

Answer 2

The distance an object travels under constant acceleration, and with an initial velocity, is given by the equation:

d = vt + ½at²

Say you are moving steadily at 10 m/s, then accelerate at 5 m/s² for 2 seconds. Then

d = 10m/s x 2 + ½ (5 x 2²)

d = 20 m + 10 m = 30 m