# Why does something constant have a parabolic shape?

Consider an object dropped from a certain position, and the only force is acceleration due to gravity. The object accelerates the same throughout the free fall; not speeding up or slowing down. So this is constant motion, right? Why, then, does the total distance (position) covered show a parabolic shape?

The total distance covered (final position, final height) is:

$$S_f=S_0-\frac{1}{2}at^2.$$

This is a parabolic shape. Why does this constancy take on a parabolic shape?

Is it because the initial velocity is horizontal? Is the initial velocity horizontal?

1. For example: An object is dropped from the top of a cliff $630$ meters high. It's height above ground $t$ seconds after it is dropped is $630-4.9t^2.$

The object falls vertically, and so you would consider acceleration due to gravity, which is constant.
I'm confused on the parabolic shape of this position function. Does it have anything to do with $t=0?$

I guess this is a pretty basic question for MSE but I would really like to un-confuse myself.
Thank you!

• The distance is a parabolic function with respect to time. If the acceleration is constant and the acceleration is the rate of change of velocity, then the velocity grows linearly. For example, if I am accelerating at 2 m/s^2 and I start from rest, then 2 seconds later, I am travelling at 4 m/s. Now, if the velocity grows linearly, then position grows parabolically. – muffle Nov 14 '13 at 3:53
• If you take the derivative of the function $x(t) = x_0 + \frac{1}{2}at^2$, which is the general position function for a body moving with no inital velocity and at constant accleration, you'll get $$x'(t) = at$$ which is a linear function since $a$ is constant. So the slope of the tangent line to the object's position at any point is increasing, meaning that the function gets steeper as time goes on, giving it a parabolic shape. – user71641 Nov 14 '13 at 3:57

Do you know calculus? If so, this is very easily answerable. Acceleration is the second derivative of position, so if acceleration is constant, we have $x''(t) = a$. Thus by integrating twice, we have $x(t) = x_0+v_0t+\frac{1}{2}at^2$.