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.

the object experiences constant acceleration due to gravity. That means that its speed steadily increases. That means that each second, it falls farther than it did in the previous second.

distance is proportional to the square of the time falling.

in fact, s = 1/2 at^2

where a is the magnitude of the acceleration. On earth, that's 9.8 m/s^2, meaning that each second the speed increases by 9.8 m/s.

Well, when it comes to falling objects, the relationship between displacement and time is a real classic. It's just like when you drop your sandwich on the ground – it's all about the distance it travels while it's on its way down.

Now, in terms of math, we can describe this relationship using the kinematic equations. For a falling object, we can use the equation:

d = (1/2)gt^2

Where "d" represents displacement, "g" is the acceleration due to gravity (approximately 9.8 m/s^2), and "t" stands for time.

This equation tells us that the displacement of a falling object is proportional to the square of the time. So, as time goes on, the displacement increases quite rapidly – kind of like how your heart rate goes up when you see your crush.

If you were to graph this relationship, you would end up with a parabolic curve. The steepness of the curve represents how quickly the displacement changes with time. At first, the curve is shallow, but as time goes on, it becomes steeper and steeper, just like the slope of your driveway when it gets icy.

So, in a nutshell, the displacement of a falling object increases with time and can be mathematically described by the equation d = (1/2)gt^2. Just remember, the key ingredient in this relationship is gravity – it always has your back, even when it comes to falling objects.

To describe how displacement changes with time for a falling object, we can analyze the motion using kinematic equations and plot the data on a graph. Let's assume that the falling object is experiencing free fall (neglecting air resistance).

The kinematic equation that relates distance, time, initial velocity, and acceleration is:

d = ut + (1/2)at^2

where d represents the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.

In the case of a falling object, the initial velocity is usually zero (as the object starts from rest), and the acceleration is due to gravity, which is approximately 9.8 m/s^2.

Taking these values into account, the equation simplifies to:

d = (1/2)gt^2

Now, let's plot the distance traveled as a function of time on a graph, assuming we are measuring the distance in meters and time in seconds.

As time increases, the distance traveled by the falling object will also increase, but at an increasingly greater rate. This is because the object is accelerating due to gravity.

The graph will be a curved line, resembling a parabola, with time on the x-axis and distance on the y-axis. The slope of the graph increases as time goes on, indicating an increasing rate of displacement.

It's important to note that for the object in free fall, the relationship between distance and time is quadratic (d ∝ t^2), as seen in the kinematic equation. This means that the distance traveled by an object in free fall is proportional to the square of the time.

Overall, the mathematical relationship between distance and time for a falling object can be described using an equation that is quadratic in nature, and the graph of this relationship will be a curved line (parabolic).

To understand how displacement changes with time for a falling object, we can start by analyzing the mathematical relationship between distance and time for such an object. This can be done using kinematic equations and basic principles of physics.

The kinematic equation that relates distance (displacement), initial velocity, time, and acceleration for an object in free fall is:

d = v0t + (1/2)at^2

Where:
- d is the displacement (distance) traveled by the object
- v0 is the initial velocity of the object
- t is the time elapsed
- a is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth)

Let's break this equation down to better understand its implications:

1. The term v0t represents the displacement caused by the initial velocity. This term is proportional to time, meaning that the distance increases linearly with time if there is an initial velocity. For example, if an object is initially moving upward, this term can result in a decreasing displacement over time until the velocity reaches zero.

2. The term (1/2)at^2 represents the displacement caused by the acceleration. This term is proportional to the square of time (t^2), meaning that the displacement increases quadratically over time due to acceleration.

When we consider a falling object starting from rest (v0 = 0), the equation simplifies to:

d = (1/2)at^2

This equation tells us that the distance (displacement) traveled by a falling object is directly proportional to the square of the time elapsed.

Now, let's consider the data and graphs to support these findings. We can plot the displacement of a falling object over time and observe the relationship.

If we perform an experiment to measure the distance traveled by a falling object at different time intervals, we can collect data points. For example, we can drop the object from a known height and measure the displacement at specific intervals using an appropriate measuring instrument.

With the collected data, we can plot a graph with time on the x-axis and displacement on the y-axis. If the relationship is indeed quadratic, we would expect a curved graph, resembling a parabolic shape. By fitting the data points to this curve, we can confirm that the relationship between displacement and time follows a quadratic pattern.

In summary, using the kinematic equation for free fall, we can describe the relationship between distance and time for a falling object. The equation tells us that displacement is the sum of the displacement caused by initial velocity (v0t) and the displacement caused by acceleration ((1/2)at^2). When considering a falling object starting from rest, the equation simplifies to d = (1/2)at^2, indicating that the displacement is directly proportional to the square of time. By conducting experiments, collecting data, and plotting graphs, we can verify this quadratic relationship between displacement and time.