More worried than ever before about FSMQ....... Watch

fubsadoo
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#81
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#81
(Original post by GHOSH-5)
Perhaps one that hasn't been mentioned, and isn't awfully hard to derive, but somewhat useful. We notice that each of the SUVAT equations omits one of the five quantities, of displacement, initial velocity, time, final velocity and acceleration; however, you may not have one that omits the initial velocity. The ones you probably have are:

(I)  v = u+at

(II)  s = ut + \frac{1}{2}at^2

(III)  v^2 = u^2 + 2as

(IV)  s = \left(\dfrac{v+u}{2}\right)t

The other one I speak of, which you might have seen before, is:

(V)  s = vt - \frac{1}{2}at^2

But it's certainly not necessary to know that.
I was wondering why we only had 4... It makes sense now I think about it, though.
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fubsadoo
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#82
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(Original post by GHOSH-5)
Yes and no: we have to use calculus when we are dealing with variable acceleration; SUVAT equations only hold for constant acceleration. So yes, when we have constant acceleration, calculus is a long route, and SUVAT is somewhat a shortcut, but it's certainly not a shortcut, but more a mislead detour from the answer when the acceleration is variable.
You have to remember I was saying this from the point of view of explaining it. I could have gone into detail about what SUVAT equations actually are (well... I couldn't, I don't that much yet :p:), but I was just trying to get across the general idea to someone who may not have understood.
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fubsadoo
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#83
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(Original post by GHOSH-5)
A girl who likes maths = awesome :woo:
Yes, I rather think I am. :yep:
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fubsadoo
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#84
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(Original post by GHOSH-5)
:curious: the general idea is: SUVAT equations are a shortcut for constant acceleration, but you seemed to imply in the post that they were a shortcut form calculus. Not really.
I meant that bit. Apologies if I was unclear.
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Amalie_J
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#85
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Hi, does anyone know where I can please find the mark scheme for OCR Add maths 2004? I can't seem to find it
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fubsadoo
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(Original post by Amalie_J)
Hi, does anyone know where I can please find the mark scheme for OCR Add maths 2004? I can't seem to find it
Is it not on the OCR website?
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Amalie_J
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#87
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(Original post by fubsadoo)
Is it not on the OCR website?
Nope only the 2006-2008 marks schemes unfortunately!
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fubsadoo
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#88
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(Original post by Amalie_J)
Nope only the 2006-2008 marks schemes unfortunately!
In that case I have no idea. Sorry. I hope somebody else can help you. :yep:
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addylad
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#89
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(Original post by roosel4)
just like to add that it is 70 for an A - is this OCR?

The kinematics is defiantly the most challenging part so i wouldn't worry too much.

Maybe look though a few a level revision sites that include this section of mechanics

Good Luck for the 5th of June and keep doing those past papers when possible
Our teacher has raped Section A, and expects us to 'pick up a couple of marks' on the Section B. How unfair is that?!

Also, does anyone know anything about the equation of a circle and working out where points intersect (question on 06 paper)?
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Introverted moron
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#90
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(Original post by addylad)
Our teacher has raped Section A, and expects us to 'pick up a couple of marks' on the Section B. How unfair is that?!

Also, does anyone know anything about the equation of a circle and working out where points intersect (question on 06 paper)?
I don't know what the question was, but the equation of a circle is:

(x-a)^2 + (y-b)^2 = r^2

Can't help on where the points where they intersect. If you already knew the equation of a circle, then I'm sorry to have wasted your time spent reading this post. :p:
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addylad
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(Original post by Introverted moron)
I don't know what the question was, but the equation of a circle is:

(x-a)^2 + (y-b)^2 = r^2

Can't help on where the points where they intersect. If you already knew the equation of a circle, then I'm sorry to have wasted your time spent reading this post. :p:
Haha, no I didn't! That's the worrying thing!

We should start a new thread, entitled, 'My teacher failed me - fill the holes in my knowledge'. :p:

Edit: and what are a and b in the equation, please?
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Introverted moron
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#92
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(Original post by addylad)
Haha, no I didn't! That's the worrying thing!

We should start a new thread, entitled, 'My teacher failed me - fill the holes in my knowledge'. :p:

Edit: and what are a and b in the equation, please?
:lol:
a and b are the co-ordinates of the centre of the circle.
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addylad
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(Original post by Introverted moron)
:lol:
a and b are the co-ordinates of the centre of the circle.
So if two points of the circle were:
7,5
3,2

The equation would be...?

Edit: might have written that wrong, but let's see :P
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Introverted moron
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#94
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(Original post by addylad)
So if two points of the circle were:
7,5
3,2

The equation would be...?

Edit: might have written that wrong, but let's see :P
(x-7)^2 + (y-5)^2 = r^2

(x-3)^2 + (y-2)^2 = r^2
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addylad
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(Original post by Introverted moron)
(x-7)^2 + (y-5)^2 = r^2

(x-3)^2 + (y-2)^2 = r^2
Ah I see, thanks a lot! So you only need one point to work it out?

By the way, does that need to be expanded, and will I have to do anything with it?

Thanks :p:
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Introverted moron
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#96
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(Original post by addylad)
Ah I see, thanks a lot! So you only need one point to work it out?

By the way, does that need to be expanded, and will I have to do anything with it?

Thanks :p:
Well, I imagine that they'll give you the radius, in which case, you just substitute that for r and there you'll have your equation.

By 1 point, do you mean 1 co-ordinate?

Heck, I was just looking at the equation that I'd posted and was wondering whether it needed solving and, if so, how? :hmmmm2:
I'm not sure whether we'll have to do anything with it (my teacher has only taught us the equation); I'm not an expert in this field. Maybe we should call in the experts i.e. fubsadoo!
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addylad
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(Original post by Introverted moron)
Well, I imagine that they'll give you the radius, in which case, you just substitute that for r and there you'll have your equation.

By 1 point, do you mean 1 co-ordinate?

Heck, I was just looking at the equation that I'd posted and was wondering whether it needed solving and, if so, how? :hmmmm2:
I'm not sure whether we'll have to do anything with it (my teacher has only taught us the equation); I'm not an expert in this field. Maybe we should call in the experts i.e. fubsadoo!
Haha, we should!

Yeah I mean with a point, (x,y) so one co-ordinate. Hmm I'll check the paper, see what it says... oh I think it asks where the line intersects with the circle (two points obviously). I'll post the question later.
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Introverted moron
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#98
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(Original post by addylad)
Haha, we should!

Yeah I mean with a point, (x,y) so one co-ordinate. Hmm I'll check the paper, see what it says... oh I think it asks where the line intersects with the circle (two points obviously). I'll post the question later.
To answer your original question, yes, you only need 1 point to work it out.

Yeah, I think, if given the centre and a co-ordinate, then you could work out the radius. If you had to find out 2 points of intersection, then I think you'd have to re-arrange 1 equation to fit the other, and work it out from there........meh, I'm waffling here. I'll wait until you post the question. Then we can analyse it in detail. :p:
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addylad
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I have one of the questions:

(i) Find the distance between the points (2,3) and (7,9)
(ii) Hence find the equation of the circle with centre (2,3) and passing through the point (7,9)

Have fun :p:
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Introverted moron
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#100
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OK, I wasn't expecting you to post it THAT soon lol. :p:

By the way, is this a question from the 2006 paper by any chance? It's just that I remember doing that for my mock...now let me go and get my answer sheet....

Seriously though, I'll try and do it now for practice and then I can go and check later. If I'm wrong, I'll tell you so that you don't go working it out in a flawed way.

So:
(i) Find the distance between the points (2,3) and (7,9)

The difference between the x co-ordinates is 5 and the difference between the y co-ordinates is 6. Drawing this out on paper will help you to work it out, as we can create a right-angled triangle. This means that we can use Pythagoras' theorem, which is:

a^2 + b^2 = c^2

5^2 + 6^2 = c^2

= 25 + 36 = c^2
= 71 = c^2

Now just work out the square root of 71 on your calculator to find the distance (this is also the radius). Like fubs, I'm lazy when it comes to using a calculator, so you can do the dirty work for me. :p:

For the 2nd part:

(ii) Hence find the equation of the circle with centre (2,3) and passing through the point (7,9)

Ah, this is nice and easy. Using the equation of a circle:

(x-2)^2 + (y-3)^2 = square root of 71 squared (or to the power of 2). :p:

EDIT: thanks addylad. It should be 61, not 71.
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