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Thread: vectors

  1. #1
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    vectors

    Find the values of t for which the velocity of the car is parallel to the vector (i + j)

    v = (3r^2 - 2t + 8)i + (5t + 6)j ms^-1

    Its doing my head in not being able to work this out
    Last edited by Carl Feltham; Sep 26th 2008 at 08:10 AM. Reason: wrong thing written
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  2. #2
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    For vectors to be parallel they need to have the same slope.

    vector i+j has slope: 1/1= 1

    so slope of v has to be 1 which is:



    (5t+6)/(3t^2-2t+8)=1

    So it is a second degree polynomial equation to solve.

    civodul
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  3. #3
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    Hello, Carl!

    Find the values of $\displaystyle t$ for which the velocity of the car
    is parallel to the vector $\displaystyle \vec i + \vec j$

    . . $\displaystyle \vec v \;= \;(3t^2 - 2t + 8)\vec i + (5t + 6)\vec j$

    A vector parallel to $\displaystyle \vec i + \vec j$ has the form: .$\displaystyle \vec v \:=\:a\vec i + a\vec j$
    . . That is, the coefficients are equal.

    So we have: .$\displaystyle 3t^2 - 2t + 8 \;=\;5t + 6 \quad\Rightarrow\quad 3t^2 - 7t + 2 \:=\:0$

    Factor: .$\displaystyle (t - 2)(3t - 1) \:=\:0 \quad\Rightarrow\quad\boxed{ t \;=\;2,\:\frac{1}{3}}$

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  4. #4
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    Quote Originally Posted by Carl Feltham View Post
    Find the values of t for which the velocity of the car is parallel to the vector (i + j)

    v = (3r^2 - 2t + 8)i + (5t + 6)j ms^-1

    Its doing my head in not being able to work this out
    Another way may be to use dot products.

    Recall that

    $\displaystyle \mathbf{a}.\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos{\theta}$.

    Since the vectors will be parallel, the angle between them is 0, and so $\displaystyle \cos{\theta}=\cos{0} = 1$.

    So in other words, in this case...

    $\displaystyle \mathbf{a}.\mathbf{b} = |\mathbf{a}||\mathbf{b}|$

    So evaluate the dot product, evaluate the product of the moduli, and solve for t.
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