# Need Help!: Drag force on a skater question ?

1. Sep 22, 2011

### nukeman

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.

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. Sep 22, 2011

### Staff: Mentor

What manipulations of the formula have you tried?

3. Sep 22, 2011

### nukeman

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?

4. Sep 22, 2011

### Staff: Mentor

Correct. Note that t is in the denominator on the RHS. What would happen if you took the reciprocal of both sides?

5. Sep 22, 2011

### nukeman

What would happen if I took the reciprocal of t from both sides?

6. Sep 22, 2011

### Staff: Mentor

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

Invert both sides of the equation. What does it look like?

7. Sep 22, 2011

### nukeman

lol...

How would you invert this? you mean top from bottom?

8. Sep 22, 2011

### Staff: Mentor

Yup. Reciprocal: reciprocal of z is 1/z.

9. Sep 22, 2011

### nukeman

Have not done this much. so,

so the top: 1/Ms 1/Vo

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

???

10. Sep 22, 2011

### 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.

11. Sep 22, 2011

### nukeman

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?

12. Sep 22, 2011

### Staff: Mentor

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

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:
$$\left( \frac{M_s}{v} \right) = c t + \left( \frac{M_s}{v_o} \right)$$
and plot Ms/v versus t.

Last edited: Sep 22, 2011
13. Sep 22, 2011

### nukeman

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 ?

14. Sep 22, 2011

### 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.

15. Sep 22, 2011

### nukeman

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 ?

16. Sep 22, 2011

### 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

17. Sep 22, 2011

### nukeman

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....?

18. Sep 22, 2011

### 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"?)

19. Sep 22, 2011

### nukeman

its just y2 -y1 / x2 - x1, and this gives us the slope. Is that all they are asking here???

20. Sep 22, 2011

### 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?).