The curve of displacement vs time squared graph?

Question

I'm just starting Physics this year. We have a lab where we collected data of a falling object.

Here is my graph of displacement vs time squared.
http://i5.photobucket.com/albums/y182/luvlaws/graph.gif?t=1190494426 (Ignore the black 'best fit' line).

I'm trying to figure out "what the shape of the graph indicates" and "the significance of the slope of the graph".

I've never come across a graph with time squared before.

My answer is that the falling object accelerates downwards until it hits the velocity -9.8 m/s (gravity)? So the curve in the beginning is the acceleration; the rest of the graph is more straight, meaning constant velocity as the object falls?

Is this right?
Thanks a ton if you can help.

Answer 1

When a graph is a straight line that passes through the origin, it means that the quantity measured in the x-axis is "proportional" to the quantity measured in the y-axis. (In your experiment it doesn't quite pass through the origin, but pretty close.)

Mathematically:

(y-axis-quantity) = (k)(x-axis-quantity)

where "k" is some constant. That's the definition of proportionality, and anything that satisfies that (for any constant "k") will have a graph that's a straight line.

In your case:
x-axis-quantity = t²
y-axis-quantity = d

So, the straight line means that d is proportional to t².

That means there is some constant "k" such that:

d = (k)t²

> "the significance of the slope of the graph"

Well, for a straight line that passes through the origin, slope = (y-axis-value)/(x-axis-value).

In your case, slope = d/t²

Ah, but remember that constant "k"? It's also true that k = d/t² (see the previous equation). And that means the constant "k" is just the slope of the graph. That's the significance of the slope.

> ...the rest of the graph is more straight, meaning constant velocity as the object falls?

Let's think about that. If the velocity were constant, we'd have:

v = d/t = constant

or:

d = (constant v)(t)

And that means d would be proportional to t. But your graph shows that d is proportional to t², NOT to t.

If something is proportional to t², it can't also be proportional to t. You can convince yourself of this by drawing another graph; this time of d vs. t. (Use the same raw data as the first graph. It may help to first make a table with 3 columns, for d, t², and t.) You will find that the d vs. t graph is NOT a straight line. This means d is not proportional to t, and therefore v is not constant.

Answer 2

For an object falling from rest:

displacement = 1/2 at^2

So the slope of displacement vs. t^2 is a/2. So your acceleration is twice the slope.

Ideally, the slope shouldn't change. Early on, though, your measurements may be somewhat less accurate, so you can probably ignore them. Just take the slope once the curve flattens and double it to get a.

Answer 3

START with Basic REASONING If the “Y” axis was in UNITS of “mm” and the “X” axis was in “mm” too THEN the “AREA” given by the MULTIPLICATION of both would be mm^2 To a Velocity vs Time graph a defined area would be:- Velocity (metres/second) X Time (second) = Metres [ distance travelled ] To a HORIZONTAL LINE emanating from the Velocity axis it represents a constant Velocity to “something” which has neither Acceleration or deceleration.

Answer 4

parabola since quadratic