Need Help: Drag force on a skater question ?

[nukeman]

Homework Statement

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

Homework Equations

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?

 

[gneill]

What manipulations of the formula have you tried?

 

[nukeman]

What manipulations of the formula have you tried?

I just don't quite understand how I would change up that formula, in order to be able to plot it as a straight line.

because I need it so y = mx + b correct?

 

[gneill]

I just don't quite understand how I would change up that formula, in order to be able to plot it as a straight line.

because I need it so y = mx + b correct?

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

 

[nukeman]

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

 

[gneill]

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

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?

 

[nukeman]

lol...

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

 

[gneill]

lol...

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

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

 

[nukeman]

Have not done this much. so,

so the top: 1/Ms 1/Vo

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

?

 

[gneill]

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.

 

[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?

 

[gneill]

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.

 

[nukeman]

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?

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 ?

 

[gneill]

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.

 

[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 ?

 

[gneill]

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 ?

The equation indicates that it's 1/v_o (or M_s/v_o). 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

 

[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 don't quite understand that...?

 

[gneill]

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"?)

 

[nukeman]

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"?)

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

 

[gneill]

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

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?).

 

[nukeman]

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?).

What do you mean "can you work out its units"

We are not to draw a graph or anthing, just explaining everytthing.

 

[gneill]

What do you mean "can you work out its units"

We are not to draw a graph or anthing, just explaining everytthing.

If in an actual experiment you were to plot the data as indicated and measured the slope of the line to determine c, what units would you assign to the constant c?

The slope of a line y = mx + b, when y and x are just real numbers without units, is just a real number without units. In a physical situation though, most constants will have units associated with them in order to make the equations meaningful.

 

[nukeman]

Oh ok, yea I see what you are saying.

But yea, all the info given to me is in the image of the question back in my origianl post.