If an object covers distance in direct proportion to the square of the time elapsed, then the acceleration is constant. Why?
In the given scenario,
d `alpha` t^2
where, d is the distance covered and t is the time taken.
Let us try to prove that if an object has a constant acceleration then the distance covered is directly proportional to the square of time taken.
Now, acceleration = rate of change of velocity
or, a = (final velocity - initial velocity)/time = (Vf - Vi)/t
or, Vf = Vi + at
and distance traveled = average velocity x time = (Vinitial + Vfinal)/2 x t
= (Vi + Vf)/2 x t
or, d = (Vi + Vi + at)/2 x t = Vi x t + 1/2 at^2
If the motion started from rest, that is, the object accelerated from zero velocity, Vi = 0 and,
d = 1/2 at^2
Thus, we can see that if the object starts from rest and accelerates at a constant acceleration, the distance covered is directly proportional to the square of the time elapsed.
Free fall of an object is a good example of such a motion.
Hope this helps.