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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!

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  • $\begingroup$ 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. $\endgroup$ – muffle Nov 14 '13 at 3:53
  • $\begingroup$ 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. $\endgroup$ – user71641 Nov 14 '13 at 3:57
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The object accelerates the same throughout the free fall; not speeding up or slowing down.

That right there is a contradiction. If something has a nonzero acceleration, it is either speeding up or slowing down.

When you're talking about motion, you can't just say "constant motion." You have to be careful to identify what is constant about the motion. For instance, there is constant velocity motion, in which an object neither speeds up nor slows down (nor does it change direction). But there is also constant acceleration motion, in which the rate of change of velocity is fixed, but the velocity itself changes. This all stems from the fact that the antiderivative of a constant function is not a constant.

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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 $x(t) = x_0+v_0t+\frac{1}{2}at^2$.

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  • $\begingroup$ I haven't learned about integration/derivatives yet, but what would be the "constant of integration" in that final equation? thanks. $\endgroup$ – Emi Matro Nov 14 '13 at 4:18

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