Weird algebra and logic in a mechanics cart quesiton

[aspodkfpo]

Homework Statement

Q 12 C)
Page 13

https://www.asi.edu.au/wp-content/uploads/2015/08/NQE_2009_Physics_solutions.pdf

Relevant Equations

−Fd = ma3
vb = le^-k/m*0
v = vb e^-kx/m

Logic and equations seem to have come out of nowhere in this question. I have been unable to understand where these equations come from and why they are used.

Can someone describe the logic for the steps in the question?

 

[BvU]

Relevant Equations:: −Fd = ma3

If the link is not permanent, this thread becomes worthless at some point.

Even then, please learn to properly render and explain all variables used
Rendering: use buttons for subscript and superscript, or better: learn a litttle $\LaTeX$
(There's a good tutorial on LaTeX here.)

The equation reads $$-F_d = ma_3$$
and further up in this puzzle you dump on us it says $F_d = \kappa v^2$ with $\kappa = 0.030$ kg m ^-1
What exactly, is unclear ?

Logic and equations seem to have come out of nowhere in this question

They claim it comes from the 'useful information'. Did you miss that ?

 

[BvU]

You ask about 12 (c) ; does that mean you did and understood 12 (b) completely ?

I must admit that I find 12 (b) rather demanding: they intended you to ignore the drag, solely because at $\approx$0.2 m/s² it is much smaller than $g\sin\theta$ at 4.9 m/s² .

(plus a cryptic chunk of 'useful information'

In physics it is often useful to make approximations. This can simplify your calculations, and if the approximation you make is appropriate, it won’t change your result appreciably. For example, if you know that A = B + C and that C is much, much smaller than B, you may be able to say that A = B and get the same result as you would have using A = B + C. If you make an approximation you must demonstrate that it is valid.​

but who is to say 4% is acceptable ?)

I know it's the science olympics, but asking this kind of judgment is unscientific to me.

I assume the 'useful information' is provided together with the exercise, not only with the solutions ?

In that case, what part of the logic escapes you ?

$\$

 

[aspodkfpo]

You ask about 12 (c) ; does that mean you did and understood 12 (b) completely ?

I must admit that I find 12 (b) rather demanding: they intended you to ignore the drag, solely because at $\approx$0.2 m/s² it is much smaller than $g\sin\theta$ at 4.9 m/s² .

(plus a cryptic chunk of 'useful information'

In physics it is often useful to make approximations. This can simplify your calculations, and if the approximation you make is appropriate, it won’t change your result appreciably. For example, if you know that A = B + C and that C is much, much smaller than B, you may be able to say that A = B and get the same result as you would have using A = B + C. If you make an approximation you must demonstrate that it is valid.​

but who is to say 4% is acceptable ?)

I know it's the science olympics, but asking this kind of judgment is unscientific to me.

I assume the 'useful information' is provided together with the exercise, not only with the solutions ?

In that case, what part of the logic escapes you ?

$\$

12b) was fine, I could do it, was not sure about the -k/m x becoming - k/m 0 or how they get the equations for the first few lines. Will look at it again.

How would I observe latex code in non-linear form when typing? i.e. \ frac a b isn't in text.

 

[haruspex]

How would I observe latex code in non-linear form when typing? i.e. \ frac a b isn't in text.

Use the preview button.

 

[aspodkfpo]

Use the preview button.

12b) was fine, I could do it, was not sure about the -k/m x becoming - k/m 0 or how they get the equations for the first few lines. Will look at it again.

How would I observe latex code in non-linear form when typing? i.e. \ frac a b isn't in text.

$\frac {-b \pm \sqrt{b^2 -4ac}} {2a}$

- why does this not work?

 

[BvU]

enclose in $...$ for inline math, in $$... $$ for displayed math

 

[haruspex]

\frac {-b \pm \sqrt{b^2 -4ac}} {2a}

- why does this not work?

Because you did not wrap it up in either a pair and of double hashes (# #...# #, but without the spaces) or a pair of double dollar signs:
$\frac {-b \pm \sqrt{b^2 -4ac}} {2a}$
$$\frac {-b \pm \sqrt{b^2 -4ac}} {2a}$$