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

Sy123

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Re: HSC 2012 Marathon :)



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.
 
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bleakarcher

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Re: HSC 2012 Marathon :)

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

Sy123

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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

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Re: HSC 2012 Marathon :)

Rates of change ploxx
 

bleakarcher

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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

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Re: HSC 2012 Marathon :)

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

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Re: HSC 2012 Marathon :)

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
 

Timske

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Re: HSC 2012 Marathon :)

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

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Re: HSC 2012 Marathon :)

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

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Re: HSC 2012 Marathon :)

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

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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

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Re: HSC 2012 Marathon :)

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

Nooblet94

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Re: HSC 2012 Marathon :)

^ 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.
 

Sy123

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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

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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...
 
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Kingportable

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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

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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.









My question is:

Find:



Difficulty Easy
 
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