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OCR AS Maths (A) Pure and Mechanics - 23/05/18 [Exam Discussion] watch

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    Leave any thoughts, feelings and answers about the exam here. How do you think you did?
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    I think I might move to Australia and never look at my grade
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    Not enough time to get it all done
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    I know!! I was rushing all of it so I couldn’t concentrate properly and get the right answer
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    (Original post by Stressedoutxxx)
    I think I might move to Australia and never look at my grade
    That's how I feel about my Further Mechanics exam.
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    :backstab:
    I got the exponential one correct at first and then spent 20 more minutes on it and changed the value of A

    Already lost 9 marks now

    OH one more now

    the pulley one, I got the answer and then crossed it out.

    FINE 17 MARKS
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    (Original post by vivienne0402)
    :backstab:
    I got the exponential one correct at first and then spent 20 more minutes on it and changed the value of A

    Already lost 9 marks now

    OH one more now

    the pulley one, I got the answer and then crossed it out.

    FINE 17 MARKS
    What were the answers?
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    (Original post by Thanu123)
    What were the answers?
    a=100, k=-ln 0.6 and something... I can't remember
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    (Original post by vivienne0402)
    a=100, k=-ln 0.6 and something... I can't remember
    a was 75, not 100.

    100 = 25+ae^-kt
    t= 0 therefore
    100 = 25+ae^0
    100 = 25+ a*1
    100= 25 + a
    a= 75
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    (Original post by Name Name123)
    a was 75, not 100.

    100 = 25+ae^-kt
    t= 0 therefore
    100 = 25+ae^0
    100 = 25+ a*1
    100= 25 + a
    a= 75
    I got 75 too
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    (Original post by Name Name123)
    a was 75, not 100.

    100 = 25+ae^-kt
    t= 0 therefore
    100 = 25+ae^0
    100 = 25+ a*1
    100= 25 + a
    a= 75
    OMG
    then i get it right now...
    ALL my friends told me it was 100
    yeah!!! thank you so much
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    does anyone remember the equations of the tangent one?
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    What did you get for the value of k for the exponential?
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    (Original post by vivienne0402)
    does anyone remember the equations of the tangent one?
    I think it was 5y + x=k and 4y +10 = x^2 ?
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    (Original post by Thanu123)
    I think it was 5y + x=k and 4y +10 = x^2 ?
    How did you find the answer to that?
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    (Original post by Stressedoutxxx)
    How did you find the answer to that?
    Dunno if this is right but it was my best guess..
    5y= k-x
    y= 1/5(k-x)
    y= -1/5(x-k) (in the form of y-y1 = m(x=x1)) therefore gradient = -1/5

    x^2+4y= 10
    4y= 10-x^2
    y= 10/4 - 1/4 x^2
    dy/dx = -1/2x


    -1/5 = -1/2x
    x= -2/5
    which also = k since k is the x coordinate of the intersection
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    (Original post by Name Name123)
    Dunno if this is right but it was my best guess..
    5y= k-x
    y= 1/5(k-x)
    y= -1/5(x-k) (in the form of y-y1 = m(x=x1)) therefore gradient = -1/5

    x^2+4y= 10
    4y= 10-x^2
    y= 10/4 - 1/4 x^2
    dy/dx = -1/2x


    -1/5 = -1/2x
    x= -2/5
    which also = k since k is the x coordinate of the intersection
    Guess i lost 5 marks then😧😊
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    (Original post by Thanu123)
    Guess i lost 5 marks then😧😊
    To be honest, I don't think what I did was correct.
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    For the exponential question what did you get for k?

    Since ky = gradient, I got 75k = 15 so k = 0.2

    For the tangent question, I got k = -12.7 by using the following method:

    

5y = k-x so y = \dfrac{k-x}{5}

    sub in to other equation:

    

x^{2} + 4(\dfrac{k-x}{5}) = 10

    

x^{2} + -\dfrac{4}{5}x + \dfrac{4}{5}k -10 = 0

    then since it is a tangent, there is one point of intersection so just one root hence b^2 - 4ac = 0

    

(-\dfrac{4}{5})^{2} - 4(1)(\dfrac{4}{5}k-10) = 0

    

\dfrac{16}{25} - \dfrac{16}{5}k + 40 = 0

    

-\dfrac{16}{5}k = -\dfrac{1016}{25}

    

k = 12.7
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    (Original post by rikkiardo)
    For the exponential question what did you get for k?

    Since ky = gradient, I got 75k = 15 so k = 0.2

    For the tangent question, I got k = -12.7 by using the following method:

    

5y = k-x so y = \dfrac{k-x}{5}

    sub in to other equation:

    

x^{2} + 4(\dfrac{k-x}{5}) = 10

    

x^{2} + -\dfrac{4}{5}x + \dfrac{4}{5}k -10 = 0

    then since it is a tangent, there is one point of intersection so just one root hence b^2 - 4ac = 0

    

(-\dfrac{4}{5})^{2} - 4(1)(\dfrac{4}{5}k-10) = 0

    so k = -12.7
    I think this is the right solution, cheers.
 
 
 

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