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    (Original post by nukethemaly)
    Has anyone got a prediction paper they can post for C1?
    What's a prediction paper?
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    (Original post by joostan)
    Because the u would be different.
    ah yea that makes since

    thanks man
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    (Original post by TH3-FL45H)
    What's a prediction paper?
    take a wild guess
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    (Original post by upthegunners)
    ah yea that makes since

    thanks man
    no prob
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    (Original post by TH3-FL45H)
    What's a prediction paper?
    Someone called TheGoldenRatio uploaded one in Jan the night before the exam, so I was just wondering if someone's done anything like that again for this exam.
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    For C1 it's basically the same questions ever-so-slightly changed? Sometimes they're even in the same order.
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    (Original post by joostan)
    no prob
    do you have any m1 tips?

    I often find i over complicate things

    I see an 8 mark question and just think it cannot be as simple as i think..

    how do I over come this? thanks, oliver
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    (Original post by upthegunners)
    do you have any m1 tips?

    I often find i over complicate things

    I see an 8 mark question and just think it cannot be as simple as i think..

    how do I over come this? thanks, oliver
    Diagram. Always draw a diagram with all the forces displayed, and work out what you're interested in.
    DJMayes posted some tips, there's a link in the OP, I recommend you take a look
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    nerves have kicked in for M1

    Need help with this:

    A particle P is moving so that its velocity, v ms^{-1}, after t seconds is given by v=3t^3 - 4t

    Initially P is at rest and is a displacment of 3m from a fixed point O.

    Find v when t=1.. okay so v=-1.

    ii) Find an expression for the displacment of P at any time t.
    d= \int (3t^2 - 4t)dt
    d=t^3 - 2t^2 +3


    Now this is where I need help!

    Find the distance travelled by the particle before it returns to its intial position. thanks

    EDIT: 69th post ROFL
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    (Original post by upthegunners)
    nerves have kicked in for M1

    Need help with this:

    A particle P is moving so that its velocity, v ms^{-1}, after t seconds is given by v=3t^3 - 4t

    Initially P is at rest and is a displacment of 3m from a fixed point O.

    Find v when t=1.. okay so v=-1.

    ii) Find an expression for the displacment of P at any time t.
    d= \int (3t^2 - 4t)dt
    d=t^3 - 2t^2 +3


    Now this is where I need help!

    Find the distance travelled by the particle before it returns to its intial position. thanks

    EDIT: 69th post ROFL
    Initial position => displacement = 0. Find t at this point to start with.
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    (Original post by CD315)
    Initial position => displacement = 0. Find t at this point to start with.
    So you set the expression for displacment equal to zero and solve?
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    anyone got any resources for c3 and c4? apart from solomon/elmwood ?
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    How would you find the area of region R?


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    (Original post by upthegunners)
    So you set the expression for displacment equal to zero and solve?
    You made a mistake in your integration by the way - you integrated 3t^2 instead of 3t^3

    EDIT: In my last post - I mean set the displacement equal to 3 instead of 0, as this is where it began.
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    (Original post by kashagupta)
    How would you find the area of region R?


    Posted from TSR Mobile
    Calculate area under triangle...

    Then remove area under curve (integrate)
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    (Original post by kashagupta)
    How would you find the area of region R?


    Posted from TSR Mobile
    Area of a triangle with third side directly below N, then subtract the area beneath the curve between M and directly below N.
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    (Original post by CD315)
    You made a mistake in your integration by the way - you integrated 3t^2 instead of 3t^3

    EDIT: In my last post - I mean set the displacement equal to 3 instead of 0, as this is where it began.
    so far I have got:

    when v=0 t=4/3 amd t=0

    when t=0 d=3

    when t=4/3 d=49/27

    what other times do we take? thanks man
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    (Original post by L'Evil Fish)
    Calculate area under triangle...

    Then remove area under curve (integrate)
    Ahh got it, thanks! Don't know why I just wasn't seeing it!


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    (Original post by joostan)
    Area of a triangle with third side directly below N, then subtract the area beneath the curve between M and directly below N.
    Thanks


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    (Original post by upthegunners)
    so far I have got:

    when v=0 t=4/3 amd t=0

    when t=0 d=3

    when t=4/3 d=49/27

    what other times do we take? thanks man
    From your integration, you should have d=\dfrac{3}{4}t^4-2t^2+3

    Now, find out the time at which the displacement is the same as it's original - i.e;

    \dfrac{3}{4}t^4-2t^2+3=3

    When I solve this, I get t=\sqrt{\dfrac{8}{3}} (correct me if I've made a mistake).
 
 
 
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