Converting Curved Graph to Linear for Acceleration Calculation

[Akbar123]

Homework Statement

For my physics coursework I need to convert a curved graph into a linear graph. This will allow me to calculate the acceleration of the object.

The experiment was repeated multiple times with the independent variable being the mass of the object. The object was dropped from a height and the time taken for the object to fall to the ground was measured.

The object that was dropped was a paper helicopter.

Homework Equations

a = v/t

The Attempt at a Solution

http://s1.postimg.org/6ot1kfg4f/Helicopter.jpg

I attempted to solve this problem by using the inverse of time and applying to the graphs. However I get the value for acceleration to be 10m/s^2 which is inaccurate. This value is greater than the value for freefall.

 

[lewando]

A couple questions for now:

1) The velocity data that you collected: is that the final velocity of each event? How did you get this data?
2) The data labeled "2s/m": what do you think that represents?
3) Two columns with identical data: one labeled "Velocity/m/s" and the other "2s/t". Why the different labels and what are you trying to mean with "2s/t"?

 

[Akbar123]

1) This is the final velocity of the paper helicopter. The velocity at which it hits the ground. The velocity was calculated by using the equation 2s/t = v. Where 2s is "two multiplied by displacement (5.10m - drop height)" The t represents the time taken for the helicopter to drop 5.10m. v is the final velocity.

2) 2s means "two multiplied by displacement." The m stands for meters

3) 2s/t and velocity are the same thing. It represents the final velocity of the helicopter.

 

[lewando]

Okay thanks. Now I see what you are doing. If you know displacement, time, v_i and v_f, you can directly determine the acceleration for each helicopter, right?

 

[Akbar123]

I want to calculate the acceleration on a linear graph by using the formula a=v/t.

However one of the values in the equation above needs to be manipulate to obtain a linear graph.

I tried plotting a graph with v on the y-axis and the inverse of t on the x axis. The acceleration (of the paper helicopter) was 10m/s^2. However I know this is wrong.

 

[lewando]

You realize that you will have 5 different accelerations for each of the 5 helicopters, right?

 

[Akbar123]

A trend line will be drawn on Microsoft Excel to obtain an acceleration for the paper helicopter.

Yes, 5 different accelerations can be obtained.

 

[lewando]

Why wouldn't you just directly calculate the accelerations for each helicopter and call it a day? You already have enough information to do this now.

The only thing I can think of is that you are trying to get the slope of a linear trend line using v=at with v as the y-axis and t as the x-axis, using only one helicopter at multiple different heights. The slope being "a" for that helicopter of that particular mass. But that is not what you have done.

You keep saying you are getting an acceleration of 10m/s². For which helicoper is that? How did you get that value?

 

[Akbar123]

I am using a basic paper helicopter.

I guess you are right, there must have been a flaw with my method.