Author Topic: VCAA Exam 1 2007 (Read 1123 times) Tweet Share

Hooligan · « **on:** October 26, 2009, 10:26:39 pm »

Hey Everyone,

I did the VCAA 2007 Exam 1, and I had a couple of problems, which I tried to resolve myself after looking at the examiner's report, and the itute solutions, however, was still unable to understand how to do the question/s. Any help is much appreciated.

Question 5b.
I looked at the itute solutions, but I don't get how they get: friction = ma.... they don't consider the force? $:-\$ Shouldn't it be: $Force - \mu N = ma$ ? Is the 'force' absent because it is only an initial force, and hence, we don't consider it?
I totally don't get in the itute working, this bit:
$\[R = ma$
$\therefore -\mu N = 6\left (-\frac{8}{3} \right )$

Question 6c.
When checking for the range, why do we randomly pick to sub in $\frac{\pi }{4}$ ? 'cause I just sub in the end value of $\frac{\pi }{2}$ bringing me to the incorrect range of $\begin{bmatrix} 0, 0 \end{bmatrix}$ (if there is even such thing!!

)

Question 10
In the itute solutions, I don't get any of the working, past the first line... what did they do to the equation? $:-\$

TrueTears · « **Reply #1 on:** October 26, 2009, 10:36:14 pm »

5b.

The initial push is not sustained through out the whole motion. Thus the only force acting on the box the whole time is the friction.

6c.

Range of cartesian = range of y(t)

Domain of cartesian = range of x(t)

With that you can't go wrong

10.

My working for this question:

$\tan(2x) = \frac{4\sqrt{2}}{7}$

Let $X = 2x$

$\tan(X) = \frac{4\sqrt{2}}{7}$

Draw your right angle triangle.

Let hypotenuse = a

$7^2 + (4\sqrt{2})^2 = a^2$

$a = \sqrt{7^2 + (4\sqrt{2})^2}$

Thus $\sin(X) = \frac{4\sqrt{2}}{a}$

Now $\tan(2x) = \frac{\sin(2x)}{\cos(2x)}$

Sub in your values to work out $\cos(2x)$

$\cos(2x) = 1 - 2\sin^2(x)$

Work out $\sin(x)$ only positive solution would work since $x \in [0, \frac{\pi}{4})$

Hooligan · « **Reply #2 on:** October 26, 2009, 10:54:21 pm »

First off, thank you sooooo much TrueTears. You are truly amazing. Thank you soo much for your extremely prompt replies and your willingness to help others, people like me!! I truly appreciate it, and don't take it for granted. =)

Quote from: TrueTears on October 26, 2009, 10:36:14 pm

5b.
The initial push is not sustained through out the whole motion. Thus the only force acting on the box the whole time is the friction.

Okay, thanks. That's what I thought... I've never come across one of these questions where there is just an initial force and therefore it disappears later on.

Quote from: TrueTears on October 26, 2009, 10:36:14 pm

6c.

Range of cartesian = range of y(t)

Domain of cartesian = range of x(t)

With that you can't go wrong

Ahahahaha... too easy! Thanks!!!!!

Quote from: TrueTears on October 26, 2009, 10:36:14 pm

10.

My working for this question:

$\tan(2x) = \frac{4\sqrt{2}}{7}$

Let $X = 2x$

$\tan(X) = \frac{4\sqrt{2}}{7}$

Draw your right angle triangle.

Let hypotenuse = a

$7^2 + (4\sqrt{2})^2 = a^2$

$a = \sqrt{7^2 + (4\sqrt{2})^2}$

Thus $\sin(X) = \frac{4\sqrt{2}}{a}$

Now $\tan(2x) = \frac{\sin(2x)}{\cos(2x)}$
______________________________________
Sub in your values to work out $\cos(2x)$

$\cos(2x) = 1 - 2\sin^2(x)$

Work out $\sin(x)$ only positive solution would work since $x \in [0, \frac{\pi}{4})$

I get everything, right up until the red line... after that, you've lost me.... $:-\$

Hooligan · « **Reply #3 on:** October 26, 2009, 11:02:51 pm »

My working from the red line....

$\frac{7}{\sqrt{57}} = 1 - 2 sin^{2}(x) $

$ \[2 sin^{2}(x)= \frac{\sqrt{57}}{\sqrt{57}} - \frac{7}{\sqrt{57}} \]$

$sin(x)= \sqrt{\frac{\sqrt{57}-7}{2\sqrt{57}}} \approx 0.19082368 \neq \frac{1}{3} $ $:-\$

TrueTears · « **Reply #4 on:** October 26, 2009, 11:14:22 pm »

haha np glad to help.

Okay from the red line.

we know $\tan(2x) = \frac{4\sqrt{2}}{7}$

and we found out $\sin(2x) = \frac{4\sqrt{2}}{a} = \frac{4\sqrt{2}}{9}$

Now $\tan(2x) = \frac{\sin(2x)}{\cos(2x)}$

$\frac{4\sqrt{2}}{7} = \frac{\frac{4\sqrt{2}}{9}}{\cos(2x)}$

Then you can work out $\cos(2x)$ and use double angle formula

Hooligan · « **Reply #5 on:** October 26, 2009, 11:23:25 pm »

Quote from: TrueTears on October 26, 2009, 11:14:22 pm

haha np glad to help.

Okay from the red line.

we know $\tan(2x) = \frac{4\sqrt{2}}{7}$

and we found out $\sin(2x) = \frac{4\sqrt{2}}{a} = \frac{4\sqrt{2}}{9}$

Now $\tan(2x) = \frac{\sin(2x)}{\cos(2x)}$

$\frac{4\sqrt{2}}{7} = \frac{\frac{4\sqrt{2}}{9}}{\cos(2x)}$

Then you can work out $\cos(2x)$ and use double angle formula

ignore my post before this post of yours... LOL... I did a BIG boo boo on the calc. to get $\sqrt{57}$ :idiot2:

I got it now! Thank you soooooooooooo much TrueTears!! xoxoxoxx

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