1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Need Help!: Drag force on a skater question ?

  1. Sep 22, 2011 #1
    1. The problem statement, all variables and given/known data

    Here, insted of trying to type it out, I will link a picture of the question from the book.
    photo.jpg






    2. Relevant equations



    3. The attempt at a solution

    The 3 questions I am having trouble answering are these:

    1: How would you plot a graph of the data to obtain a straight line??

    2:Describe how to find the value for the constant c, using the slope of the graph?

    3: What Physical quantity is represented by the y-intercept of the graph? Would the physical quantity be the mass of the skater???

    Can anyone help me out?
     
  2. jcsd
  3. Sep 22, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    What manipulations of the formula have you tried?
     
  4. Sep 22, 2011 #3
    I just dont quite understand how I would change up that forumla, in order to be able to plot it as a straight line.

    because I need it so y = mx + b correct?
     
  5. Sep 22, 2011 #4

    gneill

    User Avatar

    Staff: Mentor

    Correct. Note that t is in the denominator on the RHS. What would happen if you took the reciprocal of both sides?
     
  6. Sep 22, 2011 #5
    What would happen if I took the reciprocal of t from both sides?
     
  7. Sep 22, 2011 #6

    gneill

    User Avatar

    Staff: Mentor

    I don't know, what would happen if you took the reciprocal of t from both sides? Probably nothing helpful! :smile:

    Invert both sides of the equation. What does it look like?
     
  8. Sep 22, 2011 #7
    lol...

    How would you invert this? you mean top from bottom?
     
  9. Sep 22, 2011 #8

    gneill

    User Avatar

    Staff: Mentor

    Yup. Reciprocal: reciprocal of z is 1/z.
     
  10. Sep 22, 2011 #9
    Have not done this much. so,

    so the top: 1/Ms 1/Vo

    bottom: 1/Ms + 1/CVo 1/t

    ???
     
  11. Sep 22, 2011 #10

    gneill

    User Avatar

    Staff: Mentor

    Not quite, but almost. You want to preserve the equality, so invert the left hand side and invert the whole right hand side: just turn the fraction upside down.
     
  12. Sep 22, 2011 #11
    Oh, so...

    The top would be: Ms + CVo t

    and bottom would be Ms Vo ?

    So how does this explain how I would plot this on a graph?

    is Ms + CVo t now basically y = MX + b?
     
  13. Sep 22, 2011 #12

    gneill

    User Avatar

    Staff: Mentor

    Here, let me demonstrate:
    [tex] v = \frac{M_s V_o}{M_s + c v_o t} [/tex]
    Taking the reciprocal:
    [tex] \left( \frac{1}{v} \right) = \frac{M_s + c v_o t}{M_s v_o}[/tex]
    [tex] \left( \frac{1}{v} \right) = \frac{c}{M_s} t + \left( \frac{1}{v_o} \right) [/tex]

    Now, what happens if you plot (1/v) on the y-axis and t on the x-axis?

    EDIT: You could also multiply through by the mass:
    [tex] \left( \frac{M_s}{v} \right) = c t + \left( \frac{M_s}{v_o} \right) [/tex]
    and plot Ms/v versus t.
     
    Last edited: Sep 22, 2011
  14. Sep 22, 2011 #13
    OHHH....!

    so if I plot 1v, t on the graph, I get a data point correct? So, this is how I would manipulate it in order to graph it ?
     
  15. Sep 22, 2011 #14

    gneill

    User Avatar

    Staff: Mentor

    Yes. Take your data points and apply the indicated operation on the velocities (reciprocals) for plotting. So the units on the y-axis will be s/m or kg*s/m.
     
  16. Sep 22, 2011 #15
    Thank you very much!

    Just one moere here if you can?

    What Physical quantity is represented by the y-intercept of the graph?

    Would the physical quantity be the mass of the skater??? Or is the Physical quantity be M/v ?
     
  17. Sep 22, 2011 #16

    gneill

    User Avatar

    Staff: Mentor

    The equation indicates that it's 1/vo (or Ms/vo). So it's either the reciprocal of the initial velocity, or that reciprocal scaled by the mass of the skater. Which one you find more useful may dictate which version you want to plot :smile:
     
  18. Sep 22, 2011 #17
    Ohhh, ok. Wow, great thanks! You helped me big time. And thanks for not just GIVING me the answer! :)

    If your still around, can you maybe point me in the right direction with the last one?

    Describe how to find the value for the constant c, using the slope of the graph?

    I dont quite understand that....?
     
  19. Sep 22, 2011 #18

    gneill

    User Avatar

    Staff: Mentor

    What's the slope in the equation of the line, y = mx + b? How do you find the slope of a line when it's drawn on the XY plane? (hint: remember "rise over run"?)
     
  20. Sep 22, 2011 #19
    its just y2 -y1 / x2 - x1, and this gives us the slope. Is that all they are asking here???
     
  21. Sep 22, 2011 #20

    gneill

    User Avatar

    Staff: Mentor

    Presumably. That's the slope of a straight line, and if your data is plotted as discussed that slope will in fact be the value of c (can you work out its units?).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Need Help!: Drag force on a skater question ?
Loading...