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    Just wondering (something like this was on my exam)

    if you have a position vector: (making this up)

    r = (2t-0.04t^2)i + (3t+0.06t^2)j

    and the question says find the time that the particle is heading south east..how would go about doing it?

    thanks
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    South east means the angle it makes with the +ve i coord is 45. So...

    tan45 = -j/i = -(3t+0.06t^2) / (2t-0.04t^2)

    2t-0.04t^2 = -3t-0.06t^2 [since tan45=1]
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    When it's heading south east, it will be travelling in the direction k(i - j) where k is a scalar so you put 2t-0.04t^2 = -(3t+0.06t^2) and solve
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    right..nice one bezza and shift3, i think i got that right then..phew.

    thanks.
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    (Original post by ShOcKzZ)
    Just wondering (something like this was on my exam)

    if you have a position vector: (making this up)

    r = (2t-0.04t^2)i + (3t+0.06t^2)j

    and the question says find the time that the particle is heading south east..how would go about doing it?

    thanks
    At this time i = -j, so:
    2t - 0.04t^2 = -(3t + 0.06t^2)
    2t - 0.04t^2 = -3t - 0.06t^2
    5t + 0.02t^2 = 0
    t(5 + 0.02t) = 0
    5 = -0.02t
    t = -250 s

    Huh? Should this not be positive? :confused:
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    (Original post by mik1a)
    At this time i = -j, so:
    2t - 0.04t^2 = -(3t + 0.06t^2)
    2t - 0.04t^2 = -3t - 0.06t^2
    5t + 0.02t^2 = 0
    t(5 + 0.02t) = 0
    5 = -0.02t
    t = -250 s

    Huh? Should this not be positive? :confused:
    t(5+0.02t) = 0
    t = 0 or t = -250, in which case you'd take t=0.
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    (Original post by ShOcKzZ)
    Just wondering (something like this was on my exam)

    if you have a position vector: (making this up)

    r = (2t-0.04t^2)i + (3t+0.06t^2)j

    and the question says find the time that the particle is heading south east..how would go about doing it?

    thanks
    Wait a sec, when it is travelling south east you need to differentiate wrt. time:

    r = (2t-0.04t^2)i + (3t+0.06t^2)j
    dr/dt = (2 - 0.08t)i + (3 + 0.12t)j

    Velocity south east = k(i-j), so i = -j
    (2 - 0.08t) = -(3 + 0.12t)
    2 - 0.08t = -3 -0.12t
    5 + 0.04t = 0
    t = -125 s

    What you all found (and me admittedly) was when the particle was on the line (i-j) from the origin, but it was probably travelling in a different direction! :cool:
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    (Original post by mik1a)
    What you all found (and me admittedly) was when the particle was on the line (i-j) from the origin, but it was probably travelling in a different direction! :cool:
    :eek:

    Good point
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    (Original post by ShOcKzZ)
    Just wondering (something like this was on my exam)

    if you have a position vector: (making this up)

    r = (2t-0.04t^2)i + (3t+0.06t^2)j

    and the question says find the time that the particle is heading south east..how would go about doing it?

    thanks
    When particle is heading SE:

    i = -j
    2t - 0.04t^2 = -(3t + 0.06t^2)
    2t - 0.04t^2 = -0.06t^2 - 3t
    0.02t^2 + 5t = 0
    t(0.02t + 5) = 0

    t = 0 seconds OR:

    0.02t = -5
    t = -5/(0.02)
    t = -250 seconds

    But as time can't have a -ve value:

    When particle is heading SE:

    t = 0 seconds
 
 
 
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