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?

isnt displacement dependent on the square of time>

d=1/2 g t^2

The displacement of a falling object changes with time in a specific way. Initially, as the object starts falling from rest, its displacement increases at a constant rate. This is because the object is accelerating due to the force of gravity.

To describe the mathematical relationship between the distance and time of a falling object, we can use the equations of motion. Specifically, we can use the equation for displacement:

displacement = initial velocity * time + (1/2) * acceleration * time^2

In the case of a falling object from rest, the initial velocity is 0, so the equation simplifies to:

displacement = (1/2) * acceleration * time^2

Here, the acceleration is the acceleration due to gravity, which is approximately 9.8 m/s^2 on Earth.

Therefore, the mathematical relationship between the distance (displacement) and time of a falling object can be described by a quadratic equation, where the displacement increases with the square of time.