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Thread: proportion and modulus function

  1. #1
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    proportion and modulus function

    1.The braking distance of a car is directly proportional to the square of its speed. When the speed is p metres per second, the braking distance is 6m. When the speed is increased by 300%, find
    a)an expression for the speed of the car
    b)the braking distance
    c)the percentage increase in the braking distance

    2.Solve $\displaystyle |x^2+3x-4|=6$

    for 1(a), what i tried to do was,
    let braking distance be d, speed be s and constant be k.
    $\displaystyle d=ks^2$
    $\displaystyle 6=kp^2$
    $\displaystyle k=\frac{6}{p^2}$
    speed increased by 300%,
    $\displaystyle d=ks^2$
    $\displaystyle s=\sqrt{\frac{d}{\frac{6}{p^2}}}$
    this doesn't seem right..........==

    for question2, is there a way to solve it without drawing the graph??
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  2. #2
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    for question 2, yes you can solve it with calculations
    it isnt much different to a regular quadratic without the absolute values, you just have to tweak it a little to get rid of them

    $\displaystyle |x^2 + 3x - 4| = \sqrt{(x^2 + 3x - 4)^2}$

    did you get that far?

    = 6 , so then square both sides

    $\displaystyle (x^2 + 3x - 4)^2 = 36$ , now root both sides (get consent first, always)

    $\displaystyle (x^2 + 3x - 4)$ = 6 or -6 (you understand that bit?)

    so we now have 2 sets of solutios, which ill split up
    $\displaystyle (x^2 + 3x - 4) = 6$ , $\displaystyle (x^2 + 3x - 4) = -6$
    $\displaystyle x^2 + 3x - 10 = 0$ , $\displaystyle x^2 + 3x + 2 = 0$
    $\displaystyle (x + 5)(x - 2) = 0$ , $\displaystyle (x + 2)(x + 1) = 0$

    so all up

    x = -5 , 2, -2 and -1
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by wintersoltice View Post
    1.The braking distance of a car is directly proportional to the square of its speed. When the speed is p metres per second, the braking distance is 6m. When the speed is increased by 300%, find
    a)an expression for the speed of the car
    b)the braking distance
    c)the percentage increase in the braking distance
    $\displaystyle d=kv^2$

    So if $\displaystyle d_1$, $\displaystyle d_2$ denote two stopping distances corresponding to speeds $\displaystyle v_1$ and $\displaystyle v_2$ respectivly:

    $\displaystyle \frac{d_1}{d_2}=\frac{v_1^2}{v_2^2}$

    We are told that if the speed is $\displaystyle p$ m/s then the breaking distance is $\displaystyle 6$m and are asked to find the speed when the speed is $\displaystyle 3\times 6=18$ m.

    So put $\displaystyle d_1=6$, $\displaystyle v_1=p$, $\displaystyle v_2=3p$ then:

    $\displaystyle \frac{6}{d_2}=\frac{p^2}{9p^2}$

    So:

    $\displaystyle d_2=\frac{9}{6}$

    or:

    ....

    CB
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