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    Hi its me again ^^ - finding this work on vectors really hard. Teacher set us about 20 questions all of which are easy apart from the following (i have the answers but dont know how to acheive these )

    9) A particle has postion vector 2i + j initially and is moving with speed 10ms^-1 in the direction 3i - 4j. Find its postion vector when t = 3 and the distance it has travelled in those 3 seconds.

    (the distance travelled is easy - 10*3 = 30m. I have tried formulae for hte first part but dont understand it
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    v = 10 * (3i - 4j)/sqrt(3^2 + 4^2) = 6i - 8j

    so r = r0 + vt = (2+6t)i + (1-8t)j

    so at t = 3:
    r = 20i - 23j m
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    so what is the formula you used for v? speed * direction / magnitude?

    And the second formula is final position = original position + vector * time?

    I dont understand how from the second line you got the answer ?!?!?!
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    so what is the formula you used for v? speed * direction / magnitude?

    And the second formula is final position = original position + vector * time?
    ya that is true. sorry that i didnt make it clear.

    Edit: just to make it more correct: velocity = speed x unit vector of direction of velocity.


    so r = r0 + vt = (2+6t)i + (1-8t)j

    so at t = 3:
    r = 20i - 23j m
    r = 2i + j + 6ti - 8tj
    substitute t=3
    that will give u :
    r = 2i + j + 6x3i - 8x3j
    r = 20i - 23j
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    (Original post by yazan_l)
    ya that is true. sorry that i didnt make it clear.



    r = 2i + j + 6ti - 8tj
    substitute t=3
    that will give u :
    r = 2i + j + 6x3i - 8x3j
    r = 20i - 23j
    Thanks a lot - i would rep you but its worth nothing!
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    I have another question i need help on - i have answered part A but am stuck on B and C

    11 (in this question the unit vectors i and j are due east and due north respectively)

    At noon a ship S is 600m due north of an observation point O and aa speedboat B is 120m due north of the same point.
    The ship S has constant velocity (7i +8j)ms^-1 and the speedboat B has a constant velocity of (7i +24j)ms^-1

    a) write down the position vectors of S and B at time t seconds afternoon

    I was able to do this and got the right answers :

    S = 600j + (7i + 8j)t
    B = 120j + (7i + 24j)t

    now i am stuck on the following so can you please explain how to do these:

    b) show that S and B will collide and find the time when this collision occurs and the position vector of the point of collision

    c) In order to prevent a collision, 15s after noon S changes its velocity to (7i + 30j)ms^-1.
    Find the distance between S and B 30s after noon

    Help will be greatly appreciated
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    (Original post by asadtamimi)
    I have another question i need help on - i have answered part A but am stuck on B and C

    11 (in this question the unit vectors i and j are due east and due north respectively)

    At noon a ship S is 600m due north of an observation point O and aa speedboat B is 120m due north of the same point.
    The ship S has constant velocity (7i +8j)ms^-1 and the speedboat B has a constant velocity of (7i +24j)ms^-1

    a) write down the position vectors of S and B at time t seconds afternoon

    I was able to do this and got the right answers :

    S = 600j + (7i + 8j)t
    B = 120j + (7i + 24j)t

    now i am stuck on the following so can you please explain how to do these:

    b) show that S and B will collide and find the time when this collision occurs and the position vector of the point of collision

    c) In order to prevent a collision, 15s after noon S changes its velocity to (7i + 30j)ms^-1.
    Find the distance between S and B 30s after noon

    Help will be greatly appreciated
    b) if they collide S = B i.e.

    600j + (7i + 8j)t = 120j + (7i + 24j)t
    480j = 16t
    t = 480/16 = 30s

    120j + 30(7i+24j) = 330i + 840j

    c) you know where B will be (above). 15 is half way between 0 and 30, so you can find an average of the velocities, and find the position from that.
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    (Original post by chewwy)
    b) if they collide S = B i.e.

    600j + (7i + 8j)t = 120j + (7i + 24j)t
    480j = 16t
    t = 480/16 = 30s

    120j + 30(7i+24j) = 330i + 840j

    c) you know where B will be (above). 15 is half way between 0 and 30, so you can find an average of the velocities, and find the position from that.
    Thanks a lot, be was much easier then i thought it would be
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    wait chewwy you got B wrong - its 210i + 840j - meh small error but just thought to tell you
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    I dont understand how to do c) i keep getting a different answer to the book
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    (Original post by asadtamimi)
    I have another question i need help on - i have answered part A but am stuck on B and C
    Here is the complete solution.
    Attached Images
     
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    (Original post by steve2005)
    Here is the complete solution.
    Thanks a lot - i understand it now
 
 
 
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