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Could someone verify my answers?

  1. Jan 31, 2005 #1
    Hey guys, I've been given a maths assignment thats due tomorrow and I'd just like to verify the answers before submitting. I've put the question in bold and have just put the answer where I feel confident. The rest I have shown my workings.


    Q.1
    The length, d m(meters), of a retangular field is 40m greater than the width.
    The perimeter of the field is 400m
    i) Write this info in the form of an equation for d
    ii) Solve the equation and so find the area of the field


    Ans i) 4d - 80 = 400

    Ans ii) 4d = 480
    d = 480 / 4
    d = 120m
    120 * 80 = 9600m2

    Q.2
    A) In the Formula S = ut+ 1/2 at2 make 'u' the subject
    B) In the Formula T = 2Pi SQRT l/g make 'l' the subject


    Ans A) u = s - 1/2 at

    Ans B) T = 2Pi SQRT l/g
    T / 2Pi = SQRT l /g
    T2/4Pi2 = l/g
    l = T2 g / 4Pi2
    (sorry this looks so complicted, but I don't know how to add the right symbols)

    Q.3
    Write the following as simple fractions
    i) x/3 + x/4
    ii) 3/x + 4/x
    iii) x2/x * x4/x3 - this is supposed to read xsquared over x multiplyed by x to the power of 4 divided by x to the power of 3


    i) 7x/12
    ii) 7/x
    iii) (Not sure about this one) x2/x * x4/x3 = x6/x4


    Q.4
    Two resistors; r1 and r2 are placed in parallel so that their combined resistance R is given by:

    1/R = 1/r1 + 1/r2

    If r1 = 2x and r2 = 3x, find the formula for R in terms of x (show all your workings)


    1/R = 1/r1 + 1/r2

    1/R = 1/2x + 1/3x

    1/R = 3/6x + 2/6x

    1/R = 5/6x

    1 = 5/6x *R

    R = 6x/5

    Q4.
    Solve the following quadratic equations by either factorising or using the "b2-4ac" formula (to 3 S.F.)

    i) x2 + x - 12 = 0
    ii) 6x2 + x -2 = 0


    Ans i) x2 + x - 12 = 0
    (x -3 )(x +4) = 0
    x-3 = 0 x+4 = 0
    x=-4 x=-3

    Ans ii) Using the formula x = -b +/- SQRTb2 - 4ac
    _____________________
    2a

    When a = 6
    b = 1
    c = -2

    (I can't really write the full workings because it looks too complicated without using the correct symbols ect.) The answers are:

    x = -1 + 7 = 6 = 1
    ______ ___ __
    12 12 2

    x = -1 - 7 -8 -2
    ______ ___ __
    12 12 3

    Q5.
    A stone this thrown into the aid and its height, h metres above the ground, is given by the equation:

    h = pt - qt2

    A) Where p and q are constants and t seconds is t time is has been in the air. Given that h = 40 when t = 2 and that h = 45 when t = 3, show that

    p - 2q = 20

    and

    p-3q = 15

    B) Use these equations to calculate the values of p and q Hence show that the equation for h can be expressed in the form 5t2 - 30t + h = 0

    C) Use this equation to find the values of t when h = 17, gving your answers correct to two decimals places. Explain the significance of the two values of t


    Ans A)
    h = pt - qt2
    when h = 40 and t = 2

    40 = 2p - 4q

    40/2 = 2p - 4q/2

    p - 2q = 20

    When h = 45 and t = 3

    45 = 3p - 9q

    45/3 = 3p - 9q/3

    p - 3q = 15

    Ans B)

    (p- 2q = 20) - (p-3q = 15)
    q = 5
    p-2q=20
    p-2 * 5 = 20
    p - 10 = 20
    p = 30

    therefore h = 30t - 5t2 and 5t2 - 30t + h = 0

    Ans C) (this is similar to Q4 ii using the formula
    x = -b +/- SQRTb2 - 4ac
    _____________________
    2a

    Therefore I've minimised my workings here)

    5t2 - 30t + 17 = 0

    t = -30 +/- 23.6
    _____________
    10

    t = -30 - 23.6 = 0.63
    __________
    10

    t = -30 + 23.6 = 5.37
    __________
    10

    My own Question
    I'm trying to explain the significance of the two values of t but I'm stugling to see what they are. Could someone help explain?

    Well, thats it! I hope someone can give them the quick once over :)

    Matt
     
  2. jcsd
  3. Jan 31, 2005 #2

    HallsofIvy

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    I don't have the time to check them all!

    1. Answer is correct though you skipped a few steps I would prefer to see written down.

    2. S = ut+ 1/2 at2
    You say u = s - 1/2 at. Did you forget to divide s by t?

    2B. l = T2 g / 4Pi2
    Yes, that's correct. (you can get the 2 by [ sup ] 2 [/ sup ] without the spaces.)

    3. i and ii are correct. To do (x2/x)(x4/x3) I recommend cancelling first!

    To answer your very last question: t1 is when the rock passes 17 m on the way up, t2 is when it passes that height on its way down.
     
  4. Jan 31, 2005 #3

    dextercioby

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    Everything is okay so far...

    Wrong,think again...It's not difficult at all...

    If it's
    [tex] l=\frac{T^{2}g}{4\pi^{2}} [/tex]

    ,then it's okay...

    Okay so far...
    Me neither...It should be [itex]x^{2} [/itex] and that's it...




    Okay.It's good... :smile:

    Perfect...

    Correct your typo...

    Okay...Very good.

    Perfect...

    Okay...Very good...
    HINT:Plot the parabola and then draw the horizontal line y=17...

    Daniel.
     
  5. Jan 31, 2005 #4
    Thanks Daniel, Hallsofivy!

    Regarding the question rearranging the formula s = ut 1/2 at2

    would this be the correct method:

    s = ut 1/2 at2

    /t] s = u 1/2 at

    t = s / u 1/2a

    u = t / 1/2a * s
     
  6. Jan 31, 2005 #5

    dextercioby

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    You forgot the signs...
    [tex] h=ut+\frac{1}{2}at^{2}[/tex]

    Express u=u(t,a,h).

    Daniel.
     
  7. Jan 31, 2005 #6
    ahh, ok, if I add the signs does that make it correct (sorry, I don't understand u=u(t,a,h)
     
  8. Jan 31, 2005 #7

    dextercioby

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    Of course,because "+" and "*"(multiplication) are not equivalent...

    So separate "u" from that relation...

    Daniel.
     
  9. Jan 31, 2005 #8
    s = ut + 1/2 at2
    /2] s = u + 1/2 at
    t*s = u + 1/2 a
    t = s / u+ 1/2 a
    u = t / 1/2 a * s
     
  10. Jan 31, 2005 #9

    dextercioby

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    No.
    [tex] s=ut+\frac{1}{2}at^{2} \Rightarrow ut=s-\frac{1}{2}at^{2}[/tex]

    Can u take it from there...?Merely a division...

    Daniel.
     
  11. Jan 31, 2005 #10
    so u divide both sides by t to end up with

    u = s - 1/2 a
    ________
    t
     
  12. Jan 31, 2005 #11

    dextercioby

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    I sincerely hope that u meant
    [tex] u=\frac{s}{t}-\frac{1}{2}at [/tex]

    Daniel.
     
  13. Jan 31, 2005 #12
    Hi Daniel,

    sorry, I got it wrong again! Thanks for all your help though.. I'm going to ask my tutor for more help in this area. Again, many thanks

    Regards

    Matt
     
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