HSC 2012 MX1 Marathon #2 (archive) (1 Viewer)

Sy123 · Jun 13, 2012

Re: HSC 2012 Marathon

$\\ \\ x''=0 \\ x'=10\sqrt{2} \\ x=10\sqrt{2}t \ \ \ \ [1] \\ \\ y''=-10 \\ y'=-10t+10\sqrt{2} \\ y=-5t^2+10sqrt{2}t \ \ \ \ [2] \\ \\ [1]=[2] \\ y=-5(\frac{x}{10\sqrt{2}})^2+x \\ \\ y=x-\frac{x^2}{40} \\ \\ \frac{dy}{dx}=1-\frac{x}{20} \\ \\ \\ \tan{-15}=\frac{m-0}{1+0} \\ \\ m=tan{-15} \\ \\ \tan{-15}=1-\frac{x}{20} \\ \\ x=-20(\tan{-15}-1) \ \ \ \text{Into} \ \ [1] \\ \\ t=\frac{-20(\tan{-15}-1)}{10\sqrt{2}} \\ \\ t=1.79$

Im sorry if the question is a bit ambiguos, Im not a pro question maker, Im just a guy trying to keep a marathon going.

bleakarcher · Jun 13, 2012

Re: HSC 2012 Marathon

yeh dude doesn't sound like you worded the question right.

Sy123 · Jun 13, 2012

Re: HSC 2012 Marathon

That is what I was aiming for.
But anyway, I will probably stay away from word questions for a while, they confuse me as much as they confuse others.
So someone else ask a question!

Timske · Jun 13, 2012

Re: HSC 2012 Marathon

Rates of change ploxx

bleakarcher · Jun 13, 2012

Re: HSC 2012 Marathon

I had a feeling you wanted it this way when I saw your working. You should said it like this:

A point mass is projected from the origin O with velocity of 20m/s at an angle of 45 degrees to the horizontal. Find the time at which the velocity of the mass is at an angle of 15 degrees below the horizontal.

Carrotsticks · Jun 13, 2012

Re: HSC 2012 Marathon

Timske said:
Rates of change ploxx

A classic:

Consider a 3m ladder leaned up against a wall at an angle of 60 degrees from the floor.

It slides down slowly at a rate of 1cm/second.

When the angle is less than 40 degrees, the weight of the person standing on it will overcome the friction holding the ladder up against the wall, and it will fall/collapse completely.

It takes 5 seconds for somebody to get off the ladder.

If a person stands on the ladder when it is at 60 degrees from the floor, how long (in seconds) do they have to stand on the ladder before it becomes dangerous to stand on?

funnytomato · Jun 13, 2012

Re: HSC 2012 Marathon

Carrotsticks said:
A classic:

Consider a 3m ladder leaned up against a wall at an angle of 60 degrees from the floor.

It slides down slowly at a rate of 1cm/second.

When the angle is less than 40 degrees, the weight of the person standing on it will overcome the friction holding the ladder up against the wall, and it will fall/collapse completely.

It takes 5 seconds for somebody to get off the ladder.

If a person stands on the ladder when it is at 60 degrees from the floor, how long (in seconds) do they have to stand on the ladder before it becomes dangerous to stand on?

maths teachers NEED better ladders

Carrotsticks · Jun 13, 2012

Re: HSC 2012 Marathon

funnytomato said:
maths teachers NEED better ladders

Haha I am very confused.

Timske · Jun 13, 2012

Re: HSC 2012 Marathon

Carrotsticks said:
A classic:

Consider a 3m ladder leaned up against a wall at an angle of 60 degrees from the floor.

It slides down slowly at a rate of 1cm/second.

When the angle is less than 40 degrees, the weight of the person standing on it will overcome the friction holding the ladder up against the wall, and it will fall/collapse completely.

It takes 5 seconds for somebody to get off the ladder.

If a person stands on the ladder when it is at 60 degrees from the floor, how long (in seconds) do they have to stand on the ladder before it becomes dangerous to stand on?

I have no idea what to do lol, i found dx/dt given dy/dt = 1

funnytomato · Jun 13, 2012

Re: HSC 2012 Marathon

Carrotsticks said:
Haha I am very confused.

what I picture in my mind when I see the word 'ladder'

when was the last time you've seen one of these:

I guess one could make a somewhat a variation of the question, using the 1st one

Timske · Jun 13, 2012

Re: HSC 2012 Marathon

funnytomato said:
what I picture in my mind when I see the word 'ladder'

when was the last time you've seen one of these:

I guess one could make a somewhat a variation of the question, using the 1st one

Ladder + wall = right-angled triangle

Nooblet94 · Jun 13, 2012

Re: HSC 2012 Marathon

The answer I'm getting is that it'll collapse after ~67 seconds, so it'd be unsafe to stand on after ~62 seconds. I'm fairly sure that's not right though...

bleakarcher · Jun 13, 2012

Re: HSC 2012 Marathon

^ That's what I got I didn't even end up using calculus.

Nooblet94 · Jun 13, 2012

Re: HSC 2012 Marathon

bleakarcher said:
^ That's what I got I didn't even end up using calculus.

Yeah, I used a long method that involved differentiating stuff and then integrating some different stuff... Only to realise I could've done it with basic geometry. Got the same answer both ways though, so maybe it is right.

Timske · Jun 13, 2012

Re: HSC 2012 Marathon

bleakarcher said:
^ That's what I got I didn't even end up using calculus.

how did u do it i dont understand what its asking for min time?

Sy123 · Jun 13, 2012

Re: HSC 2012 Marathon

Yes I got that too, I didnt need to use calculus either. Simply find initial height, then the height at 40 degrees, find their difference, times by 100 to get cm and you get 66.971 cm, at 1cm/s. It means ~67 seconds, minus the 5, then ~62 seconds

Leffife · Jun 13, 2012

Re: HSC 2012 Marathon

Here comes a relatively easy question,

A cylinder open at one end has a volume of 125π cubed units.

a) Find the height of the cylinder is 'h' metres and its radius is 'r' metres, prove that the external SA, S m² is given by:

S = πr² + (250π/r) => Hint: πr²h = 125π

b) Hence, find the dimensions of the cylinder with the least surface area.

Just keep it going...

Kingportable · Jun 17, 2012

Re: HSC 2012 Marathon

John Fitzpatrick 3 Unite Mathematics
25(C)
A particle is moving in a straight line. If x is its displacement at time t and v^2=5(4-x^2) Find the acceleration in terms of x only, Show that the motion is simple harmonic and find its period and amplitude.

Sy123 · Jun 17, 2012

Re: HSC 2012 Marathon

Kingportable said:
John Fitzpatrick 3 Unite Mathematics
25(C)
A particle is moving in a straight line. If x is its displacement at time t and v^2=5(4-x^2) Find the acceleration in terms of x only, Show that the motion is simple harmonic and find its period and amplitude.

$v^2=5(4-x^2) \\ \frac{1}{2}v^2=\frac{5}{2}(4-x^2) \\ \left(\frac{1}{2}v^2\right)\frac{d}{dx}=-5x \\ \therefore \ddot{x}=-5x$

$\therefore \text{Motion is Simple Harmonic, since acceleration is in the form} \ \ \ \ddot{x}=-n^2x$

$Period=\frac{2\pi}{n} \\ \\ n=\sqrt{5} \\ \therefore Period=\frac{2\pi}{\sqrt{5}}$

$x=a \Rightarrow v=0 \\ \\ 0=5(4-a^2) \\ 4-a^2=0 \\ a=\pm 2 \\ \\ \therefore Amplitude=2$

My question is:

Find:

$\int^1_0 x \ dx+\int^1_0 x^3 \ dx+\int^1_0 x^7 \ dx+\int^1_0 x^{15} \ dx+...$

Difficulty $\rightarrow$ Easy

barbernator · Jun 17, 2012

Re: HSC 2012 Marathon

Sy123 said:
$v^2=5(4-x^2) \\ \frac{1}{2}v^2=\frac{5}{2}(4-x^2) \\ \left(\frac{1}{2}v^2\right)\frac{d}{dx}=-5x \\ \therefore \ddot{x}=-5x$

$\therefore \text{Motion is Simple Harmonic, since acceleration is in the form} \ \ \ \ddot{x}=-n^2x$

$Period=\frac{2\pi}{n} \\ \\ n=\sqrt{5} \\ \therefore Period=\frac{2\pi}{\sqrt{5}}$

$x=a \Rightarrow v=0 \\ \\ 0=5(4-a^2) \\ 4-a^2=0 \\ a=\pm 2 \\ \\ \therefore Amplitude=2$

My question is:

Find:

$\int^1_0 x \ dx+\int^1_0 x^3 \ dx+\int^1_0 x^7 \ dx+\int^1_0 x^{15} \ dx+...$

Difficulty $\rightarrow$ Easy

1 lol

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