Kinematics/calculating a line from a curve?

  • Thread starter Thread starter fruitbowl
  • Start date Start date
  • Tags Tags
    Curve Line
Click For Summary
To produce a linear relationship from the given distance and time data, the student should manipulate the equation D = 1/2gt². By rearranging it into the form Y = MX + C, the student can plot distance (D) against time squared (t²) to achieve a straight line. This transformation allows for the determination of gravitational acceleration (g) from the slope of the line. The student is seeking clarification on which quantities to graph to effectively linearize the data. Ultimately, plotting D versus t² will yield the desired linear relationship.
fruitbowl
Messages
5
Reaction score
0

Homework Statement


"If only the variables D and t are used, what quantities should the student graph in order to produce a linear relationship between the two quantities?"

The data:
distances (m): .1, .5, 1, 1.7, 2
Time(s): .14, .32, .46, .59, .63

Basically, a student is dropping a ball from several points and recording how long the fall took from each point. They're trying to experimentally determine g. So the data yields a curve. The full problem is here, it's the first one after the reference sheet: http://apcentral.collegeboard.com/apc/public/repository/_ap06_frq_physicsb_fo_51783.pdf


Homework Equations


n/a

The Attempt at a Solution


I'm not exactly sure what the question is asking. Since the data gives a curve, but I need a line, The "quantities" they want are the end points of the line segment that approaches all the points in the curve the most, right? How do I get that?

I tried graphically, this is the best I could get:
j9n02q.jpg


But there's got to be a math way to do it. Please help... :rolleyes:
 
Last edited by a moderator:
Physics news on Phys.org
Well you know that

D=ut+1/2at2 which simplifies to D=1/2gt2

You can plot appropriately to get a straight line.
 
I'm sorry to seem so dense but- what do you mean? what do I plot? D=1/2gt^2 is still a curve...
 
fruitbowl said:
I'm sorry to seem so dense but- what do you mean? what do I plot? D=1/2gt^2 is still a curve...

If you make the equation into the form Y=MX+C and plot Y vs. X you will get a straight line. Put Y=D, what should X be to get a straight line?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K