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    A particle A starts at position vector 12i +12j. The initial velocity of A is (-i + j). Constant acceleration (2i - 4j). Another particle B has initial velocity i, and constant acceleration 2j. After 3 seconds the 2 particles collide.

    Find the position vector of the point where the two particles collide.

    How are you supposed to do this? The answer says to use s = ut + 0.5at^2 for A. but then what is s? Is this the displacement vector of s? How do you know that this is where the particles collide?

    And also how would you find the position vector of B's starting point?

    I'm quite confused as to what position vectors are and how to use them...
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    (Original post by lollage123)

    How are you supposed to do this? The answer says to use s = ut + 0.5at^2 for A. but then what is s? Is this the displacement vector of s? How do you know that this is where the particles collide?
    Yes s is the displacement
    So you add it to the initial position to get the position vector at time t
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    (Original post by lollage123)

    And also how would you find the position vector of B's starting point?
    Why do you want that
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    (Original post by TenOfThem)
    Why do you want that
    okay thanks. its part of the questio
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    (Original post by lollage123)
    okay thanks. its part of the questio
    Well they have the same position vector

    So you do the same and figure out what has been added on


    Wow ... negged for helping ... great
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    (Original post by TenOfThem)
    Well they have the same position vector

    So you do the same and figure out what has been added on


    Wow ... negged for helping ... great
    Thanks. I didn't neg you!
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    (Original post by lollage123)
    Thanks. I didn't neg you!
    No, I know


    DId you solve the question?
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    (Original post by TenOfThem)
    No, I know


    DId you solve the question?
    Yes I have another q though. If you have 2 particles A and B and want to find when B is due east of A, why is it that the i coefficients are equal? And why is it that the j coefficients are equal if you want to find when B is due north of A? (If you are given their position vectors) I can't get my head around this.
    And how do you find velocity of the particles if you are only given position vectors of them?
    Finally, if 2 vector quantities C and D collide (e.g. if C intercepts D at a certain time) how do you find the resultant vector velocities involved? is it that the vector CD = 0, or DC = 0?
    And to calculate CD, is it position vector D - position vector C?


    Suppose it would help if I gave a question for context of the intercepting thing:

    at 0900 a boat is travelling at constant speed, with position vector -2i - 4j km relative to the origin.
    0940 S is at 4i - 6j.

    Find an expression for s in terms of t

    At 1100 a motorboat leaves O and travels with constant velocity (pi + qj) kmh^-1. If it intercepts the first boat at 1130, calculate p and q.
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    (Original post by lollage123)
    Yes I have another q though. If you have 2 particles A and B and want to find when B is due east of A, why is it that the i coefficients are equal? And why is it that the j coefficients are equal if you want to find when B is due north of A? (If you are given their position vectors) I can't get my head around this.
    That depends on the definition of i and j

    And how do you find velocity of the particles if you are only given position vectors of them?
    Velocity is (change in position)/time

    Moving from 3i+5j to 7i+15j in 2 seconds
    Change in position = 4i+10j
    Divide by time = 2i+5j = velocity

    Finally, if 2 vector quantities C and D collide (e.g. if C intercepts D at a certain time) how do you find the resultant vector velocities involved? is it that the vector CD = 0, or DC = 0?
    And to calculate CD, is it position vector D - position vector C?


    Suppose it would help if I gave a question for context of the intercepting thing:

    at 0900 a boat is travelling at constant speed, with position vector -2i - 4j km relative to the origin.
    0940 S is at 4i - 6j.

    Find an expression for s in terms of t

    At 1100 a motorboat leaves O and travels with constant velocity (pi + qj) kmh^-1. If it intercepts the first boat at 1130, calculate p and q.
    Boat has velocity = (6i-2j)/{4/6} = 9i-3j
    r = (-2i-4j) + (9i-3j)t
    So at 11 r = (-2i-4j) + (9i-3j)2 = 16i-10j

    SO

    Starting at 1100
    B = (6i-10j) + (9i-3j)t
    M = (0i+0j) + (pi+qj)t

    They meet at t = 0.5

    i gives 6 + 4.5 = 0 + 0.5p

    etc
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    (Original post by TenOfThem)
    That depends on the definition of i and j



    Velocity is (change in position)/time

    Moving from 3i+5j to 7i+15j in 2 seconds
    Change in position = 4i+10j
    Divide by time = 2i+5j = velocity



    Boat has velocity = (6i-2j)/{4/6} = 9i-3j
    r = (-2i-4j) + (9i-3j)t
    So at 11 r = (-2i-4j) + (9i-3j)2 = 16i-10j

    SO

    Starting at 1100
    B = (6i-10j) + (9i-3j)t
    M = (0i+0j) + (pi+qj)t

    They meet at t = 0.5

    i gives 6 + 4.5 = 0 + 0.5p

    etc
    I would really appreciate it if you could help me out here.
    To find the position vector of where the two particles collide, can you not use r=p + vt where p is the position vector when t=0 and r is the new position vector? Youve already worked out velocity in the previous question and know the rest of the values. Ive tried this but I dont get the right answer.
 
 
 
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