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 =...

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.