Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Kinematics/calculating a line from a curve?

  1. Jan 19, 2010 #1
    1. The problem statement, all variables and given/known data
    "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


    2. Relevant equations
    n/a

    3. 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 gotta be a math way to do it. Please help... :uhh:
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Jan 19, 2010 #2

    rock.freak667

    User Avatar
    Homework Helper

    Well you know that

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

    You can plot appropriately to get a straight line.
     
  4. Jan 19, 2010 #3
    I'm sorry to seem so dense but- what do you mean? what do I plot? D=1/2gt^2 is still a curve...
     
  5. Jan 19, 2010 #4

    rock.freak667

    User Avatar
    Homework Helper

    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?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook