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    Tomorrow is M1 day and i've been doing past papers and got stuck at part d) of this question:

    M1 January 2006 paper:
    6.
    [In this question the horizontal unit vectors i and j are due east and due north
    respectively.]
    A model boat A moves on a lake with constant velocity (–i + 6j) m s–1.
    At time t = 0, A is at the point with position vector (2i – 10j) m. Find:

    (a) the speed of A,
    (2)
    (b) the direction in which A is moving, giving your answer as a bearing.
    (3)
    At time t = 0, a second boat B is at the point with position vector
    (–26i + 4j) m.

    Given that the velocity of B is (3i + 4j) ms-1

    (c) show that A and B will collide at a point P and find the position vector of P.

    Given instead that B has speed 8 m s–1 and moves in the direction of the vector (3i + 4j),

    (d) find the distance of B from P when t = 7 s.


    I get everything until part d), I don't understand how I would turn the speed given into a velocity to be able to work out the position vector, and then the distance.


    many thanks
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    im pretty sure u need to work out the PV'S of both when t= 7 then minus them
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    well speed is the magnitude of velocity, so you can find the magnitude of your direction vector by doing the square root of (3squared + 4squared). this'll give 5, and divide 8/5 to get the number of times you need to multiply the direction vector to get the right speed.

    sooo the velocity will be 1.6(3i + 4j), and i think you can work it out from there.
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    (Original post by BrightGirl)
    well speed is the magnitude of velocity, so you can find the magnitude of your direction vector by doing the square root of (3squared + 4squared). this'll give 5, and divide 8/5 to get the number of times you need to multiply the direction vector to get the right speed.

    sooo the velocity will be 1.6(3i + 4j), and i think you can work it out from there.
    wooo i understand! :woo:
    thanks
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    (Original post by bob'syouruncle)
    wooo i understand! :woo:
    thanks
    most welcome
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    (Original post by BrightGirl)
    well speed is the magnitude of velocity, so you can find the magnitude of your direction vector by doing the square root of (3squared + 4squared). this'll give 5, and divide 8/5 to get the number of times you need to multiply the direction vector to get the right speed.

    sooo the velocity will be 1.6(3i + 4j), and i think you can work it out from there.
    i know its 2years late but any chance of anyone explaining the bold bit to me in a bit more detail?
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    is this OCR (not MEI) or is this MEI??
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    (Original post by BrightGirl)
    well speed is the magnitude of velocity, so you can find the magnitude of your direction vector by doing the square root of (3squared + 4squared). this'll give 5, and divide 8/5 to get the number of times you need to multiply the direction vector to get the right speed.

    sooo the velocity will be 1.6(3i + 4j), and i think you can work it out from there.
    Sorry I stiil don’t get the part that you need to divide it by 8/5
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    (Original post by Maggie07)
    Sorry I stiil don’t get the part that you need to divide it by 8/5
    Doubt you'll get a reply from a 5 year old thread...

    Make a new thread with your question if you need help with it.
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    (Original post by RDKGames)
    Doubt you'll get a reply from a 5 year old thread...

    Make a new thread with your question if you need help with it.
    thank you so much😊
 
 
 
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