Question

How does displacement change with time for a falling object?

Answer 1

1. How does displacement change with time for a falling object? How might you describe the mathematical relationship between the distance and time of a falling object? Explain the answer using your data, graphs, and kinematic equations.

*Posted full question.

Answer 2

d.t. = distance traveled

d.t. = total units moved

Lets say Sammi walks three feet east, and then two feet west. What is her distance traveled?

To calculate d.t., you would simply add three feet by two feet. Sammi moved a total of five feet, and that is here distance traveled. Since d.t. is a scalar property (a quantity that does not depend on direction), we do not need to add a direction.

\[\Delta x = Displacement = x _{f} - x _{i}\]

This is essentially saying change in x = displacement, or final position minus initial position.

Lets take our same example.
Lets say Sammi walks three feet east, and then two feet west. What is her distance traveled? She starts at the origin



Sammi has been displaced by one foot. This is because she initial position is 0, and her final position is one foot to the east.

Displacement = 1 - 0 = 1ft East (You give a direction since it is a vector property).

Answer 3

So for a displacement in terms of a falling object, we would use the kinematic:

\[\Delta x = V _{i} t + \frac{ 1 }{ 2 } a t^2\]

As you would guess, an objects final position will drastically change from its initial position if it is falling. So its displacement would increase over time. This relationship is quadratic, as the displacement of an object changes with the square of time (t^2).

Answer 4

Will keep this open for a bit to see if anyone passing by has something interesting to add to this question, or a better explanation about something.