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    (1) A pedestrian moves with the constant velocity [(2q^2 - 3)i + (q + 2)j] ms^-1.
    Given that the velocity of the pedestrian is parallel to the vector (i - j), show that one possible valu of q is -1 and find the other possible value of q.

    (2) The velocity of a car at time t seconds is given by:

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

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


    I have noticed that in the markschemes for "parallel to (i + j)" they just make the i component = j component, and for "parallel to (i - j) they make the i component = - (j component). Can someone please explain why that is?

    Also if it said the velocity was parallel to something else...e.g. (-7i + 10j)...then what is the method there? Is there a general way of doing the above questions for when the velocity vector is parallel to (xi + yj) ?

    Thx in advance!
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    If it's parallel to (i-j) that means it's moving in that direction. So, it can have any magnitude, as long as its directed along the vector (i-j). So it follows that:
    2q^2 - 3 = -(q+2)
    and solve...

    This is what I do for a general 'parallel to (xi+yj)' question:
    If it's parallel, it can have a magnitude 'k' but must be directed along that vector. And let's for example take the velocity in (1).

    => |k(xi+yj)| = |[(2q^2 - 3)i + (q + 2)j|
    k^2(x^2 + y^2) = (2q^2 - 3)^2 + (q+2)^2

    And carry on from there to find 'k'
    Then you say v = k(xi+yj), and sub in the value of k.
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    (Original post by mockel)
    If it's parallel to (i-j) that means it's moving in that direction. So, it can have any magnitude, as long as its directed along the vector (i-j). So it follows that:
    2q^2 - 3 = -(q+2)
    and solve...

    This is what I do for a general 'parallel to (xi+yj)' question:
    If it's parallel, it can have a magnitude 'k' but must be directed along that vector. And let's for example take the velocity in (1).

    => |k(xi+yj)| = |[(2q^2 - 3)i + (q + 2)j|
    k^2(x^2 + y^2) = (2q^2 - 3)^2 + (q+2)^2

    And carry on from there to find 'k'
    Then you say v = k(xi+yj), and sub in the value of k.
    Thanks mate...but lol I still don't quite understand. I understood that "parallel to" means "in the direction of"...and I have used the method you've shown for a general (xi + yj) question before for questions that for example ask:

    "Given that the magnitude of the velocity is 4(sqrt13) and the velocity is in the direction (2i + 3j), find the velocity in i and j form"
    ...in which case the velocity would be 4(2i + 3j)

    But if you do the same in question 1 you end up with a messy polynomial to solve...

    Hmm the main thing I don't get is why is it that if it's parallel to (i + j) you make the i component = j component? And if it's parallel to (2i + 3j) would you make (2*i component) = (3*j component)?
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    Nah, don't do it for that question. Like you said, messy polynomial.

    Basically, they've already given you the velocity with i and j components. And you have the direction that its going in, in i and j components. So it's like equating coefficients. Say the velocity was k(i-j) = ki - kj. Equating coeffs:

    k = 2q^2 - 3

    -k = q+2
    k = -(q+2)

    Now, since they're both equal to 'k' we can equate them, thus:
    2q^2 - 3 = -(q+2)

    etc...


    Say it was k(2i+3j), then you can do the same thing:

    2k = 2q^2 - 3

    3k = q+2

    You can divide the first by 2 and the second by 3, and then equate.
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    (Original post by mockel)
    Nah, don't do it for that question. Like you said, messy polynomial.

    Basically, they've already given you the velocity with i and j components. And you have the direction that its going in, in i and j components. So it's like equating coefficients. Say the velocity was k(i-j) = ki - kj. Equating coeffs:

    k = 2q^2 - 3

    -k = q+2
    k = -(q+2)

    Now, since they're both equal to 'k' we can equate them, thus:
    2q^2 - 3 = -(q+2)

    etc...


    Say it was k(2i+3j), then you can do the same thing:

    2k = 2q^2 - 3

    3k = q+2

    You can divide the first by 2 and the second by 3, and then equate.
    Ahhhhh there now I get it...lol :laugh:....cheers mate...really appreciate it :top:
 
 
 
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