# Thread: Modulus (Brainteaser question)

1. ## Modulus (Brainteaser question)

a) This I can do:

$\sum F = ma$

$9g - 0.01v^2 = 9a$

$a = -\frac{(v^2-900g)}{900}$

b) My working is as follows:

$\frac{dv}{dx} = -\frac{(v^2-900g)}{900v}$

$\frac{dx}{dx} =-\frac{900v}{(v^2-900g)}$

$x = -900 \int \frac{v}{(v^2-900g)} dv$

$= -450\log_e|v^2-900g| + c$

When $x = 0$ $v = 0$

$\therefore c = 450\log_e(900g)$

$\implies x = 450\log_e(900g) - 450\log_e|v^2-900g|$

$\implies x = 450\log_e|\frac{900g}{v^2-900g}|$

$x = 450\log_e|\frac{8820}{v^2-8820}|$

Now how do I get rid of the mod signs to get their form? I did the following:

$x = 450\log_e|\frac{8820}{-(8820-v^2)}|$

$x = 450\log_e\left(\frac{8820}{|-1||8820-v^2|}\right)$

$x = 450\log_e\left(\frac{8820}{|8820-v^2|}\right)$

But then how to get rid of the mods in the denominator? Since it still could be negative and you need the mods to ensure it's positive?

Next part says:

$x = 450\log_e\left(\frac{8820}{|8820-v^2|}\right)$

When $x = 150$

$\therefore 150 = 450\log_e\left(\frac{8820}{|8820-v^2|}\right)$

Solving these on the TI-89 gives: $v = \pm 50.002, \pm 123.044$

Now since the question asks for the speed we require $|v| = 50.002, 123.044$

However which one do you choose?

Now if you solve their equation which is $x = 450\log_e\left(\frac{8820}{8820-v^2}\right)$

you get $|v| = 50.002$ only and not $123.044$

So which formula do you use and most importantly when you use my formula how do you dismiss $123.044$ as a solution?

2. The equation of motion $a = -\frac{v^2-900g}{900}$ tells you that a=0 when $v = \sqrt{900g}$. This is the terminal velocity $v_\infty$, the speed at which gravitational pull is exactly balanced by air resistance. If $v < v_\infty$ then the acceleration will be positive, and the velocity will increase towards $v_\infty$. If $v>v_\infty$ then a<0 and the velocity will decrease towards $v_\infty$.

In this problem, the initial value of v is 0, so v can never exceed $v_\infty$. That is why you have to take $v^2-8820$ to be negative. If the problem told you that the sandbag had been hurled downwards from the balloon with an initial speed greater than $\sqrt{900g}$, then your solution would have needed to take $v^2-8820$ to be positive.