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.