Calculating acceleration of a falling orange

1. Nov 22, 2013

mr.plow

1. The problem statement, all variables and given/known data
Using the data found in this chart, create a time-velocity table and draw a velocity-time graph. Once you have done that, calculate the acceleration of the orange as it falls to the ground. (table attached as image).

2. Relevant equations
I'm not even sure about this... I don't understand how to convert the time-position table into a time-velocity table, and it wasn't clearly explained in the lesson.

3. The attempt at a solution
Another attachment will show my attempt..
The time-position graph mentioned in this attempt turned out to be a 17mb jpeg, and I'm sure I didnt even need this graph to get this answer now that I look at it again... I feel like this should be a really easy solution. Any help would be GREATLY appreciated!

2. Nov 22, 2013

mr.plow

Is the attachment showing?

3. Nov 22, 2013

Staff: Mentor

I'm not seeing it. Try again?

4. Nov 24, 2013

mr.plow

OK, I guess I'll try uploading that data again. I also figured out how to compress the graph data into a .zip file. I end up getting 8.4m/s, which is strange because I don't see where I could have made an error... again, any help is so greatly appreciated.

Attached Files:

File size:
16.7 KB
Views:
145
• time-position to time-velocity table.zip
File size:
56.7 KB
Views:
94
5. Nov 24, 2013

mr.plow

oops, zipped the wrong file, let's try this againnn hah..

Attached Files:

• jesse's graph unit 2.jpg
File size:
23.2 KB
Views:
153
6. Nov 24, 2013

Staff: Mentor

So have you been able to make progress on the problem? The latest attachment looks like you have been able to do some graphing? (it's hard to read though...)

7. Nov 24, 2013

mr.plow

The graphing in the latest attachment is where I got the data for the time-instantaneous velocity table for points A,B, and C by calculating the slopes for each tangent line. I was hoping that it would be more easily read from the thumbnail, but I'm new on this forum. If you open the graph in a new tab/window, you can see where I got my data from for the first thumbnail photo a little more clearly. You may have to increase the zoom on the page you view it in, but you should be able to tell if I goofed up on the slope calculations.

8. Nov 24, 2013

Staff: Mentor

I can see what they did in the attachment. They used the data from 0.25 and 0.5 to get the average velocity over this interval. The same goes for the data from 0.75 to 1.0, and from 1.25 to 1.5. The question is, at what time should the velocity be plotted for each of these intervals, if the calculated velocity is the average over these intervals? The answer is, to get good accuracy on the instantaneous velocity, it should be plotted at the half-way point for each interval: 0.375, 0.875, and 1.375 sec. This gives "second order accuracy" on the instantaneous velocity, and is exact if the distance is varying as a quadratic with time. This is about the best you are going to do.

You used a different method, by drawing tangents to the smooth curve you drew, and taking the slope. If you plot your results and the results obtained using the method I described above, your method should agree pretty well with the results in the attachment. You can also do the calculations for the other time intervals in the table using the method I described. This will give you a more fleshed-out plot of how the instantaneous velocity versus time.

The same time of approach can be used to take the results obtained from the velocity calculations and using them to estimate the instantaneous acceleration versus time.