Calculating acceleration using Experimental equation

[Artel]

Homework Statement

Determine the y intercept and slope of the linearization equation of the graph. Find an experimental equation. From this and the theoretical equation given in the introduction, determine the acceleration on the object.

Context: I recorded data for the distance vs velocity of an object. With that distance, I plotted it a graph and found the linearization equation to it.

Homework Equations

The theoretical equation relating distance and velocity:
v = $\sqrt{2a}$Δx^1/2

The Attempt at a Solution

I found the log10 x = log10 y and x = lny graphs of my data.
The log10 x = log10 y graph has the best R² value, so I will use this in finding my experimental equation.
According to excel, the linearization equation is
y = .5018x - .1117
with R² being .9999
According to my question, I have to convert this to an experimental equation.
Experimental equation for power function is: y = bx^m

What I don't know how to do is, 1. Converting it to an experimental equation and 2. Finding the acceleration of the object.

My attempt at converting:
y = 10^-.1117*x^.5018

As to determining the acceleration of the glider from this using the experimental and theoretical equation, I have no idea how.

 

[Redbelly98]

Welcome to Physics Forums.

The idea is to compare the equations:
$$\begin{align}<br /> v & = \sqrt{2a} \cdot \Delta x^{1/2} \\<br /> \text{and} &\\<br /> y & = 10^{-.1117} \cdot x^{.5018}<br /> \end{align}$$
From looking at those two equations, it is evident that
$$\sqrt{2a} = \text{ _____?}$$

 

[Artel]

Thank you for responding!

Alright, comparing the two equations:
I set both equations equal to each other.

$\sqrt{2a}$ = 10^-.1117 • x^.5018/x^1/2

$\sqrt{2a}$ = .773215 x^.0018

squaring both sides

2a = .597861 x^.0036

dividing by 2

a = .298931 x^.0036

I still don't understand. I have a value for a, but it is dependent on x. What does this mean? Isn't acceleration a constant?

 

[frogjg2003]

When you are plotting data, and obtain a fit, you will get a list of parameters.
In your case, you plotted y=b*x^m and obtained values for b and m,b = 10^-0.1117 and m = 0.508.
Your next step is to compare your data (b and m) to the expected values ($\sqrt{2a}$ and 1/2).
The important part of this is to recognize that you are not comparing the experimental equation b*x^m to the theoretical $\sqrt{2a}$*x^1/2, but b to $\sqrt{2a}$ and m to 1/2.

 

[Artel]

How do I compare b to $\sqrt{2a}$ and m to 1/2 and find a? What do you mean by compare?

 

[Redbelly98]

a = .298931 x^.0036

I still don't understand. I have a value for a, but it is dependent on x. What does this mean? Isn't acceleration a constant?

Yes, acceleration should be a constant.

The fact that your best fit line gives x^.5018 instead of x^.5000 can be attributed to experimental error. So x^.5018 is essentially (approximately) equal to x^1/2, and you may treat them as canceling out where you had x^.5018/x^1/2 in your work.

EDIT: you may ignore the rest of this post, it is probably just going to confuse the issue...

Or looked at another way ...

Assuming you are using meters for x and m/s for v, I would bet that your x-values are mostly somewhere between 0.1 m and 10 m. How would this change the value of a in

a = .298931 x^.0036 ?​

I.e., what is 10^.0036, and what is 0.1^.0036?

 

[Redbelly98]

The important part of this is to recognize that you are not comparing the experimental equation b*x^m to the theoretical $\sqrt{2a}$*x^1/2, but b to $\sqrt{2a}$ and m to 1/2.

How do I compare b to $\sqrt{2a}$ and m to 1/2 and find a? What do you mean by compare?

He means:

1. Compare b to $\sqrt{2a}$

and

2. Compare m to 1/2

So there are two separate, independent comparisons to make.

Also, this all assumes consistent length and time units for x, v, and a. For example meters and seconds everywhere (or cm and seconds would work too).

 

[Artel]

Yes, in that case x is very close to 1. So my a is .298931, which is an acceptable result. Thank you!

 

[frogjg2003]

The value of a does not depend on x.
When I said compare, i mean set equal to each other.
Like Redbelly98 said, 1/2 is pretty close to .5018, probably within your experimental error, so you get 1/2≈.5018
The important part is that you set $\sqrt{2a}=10^{-.1117}$ and solve for a.
No need to assume x is 1 or close to 1 or any other value.