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    Could someone please show me the method for part (c) of this question.:

    14. A particle is initially at the origin. The particle has a constant acceleration of (2i-3j) m/s^2. Two seconds after it has left the origin, the velocity of the particle is (2i + 14j) m/s. The unit vectors i and j are perpendicular.

    c) The speed of the particle is 10 m/s for the first time T seconds after it has left the origin. Find T.

    The answer given in the book is 4 seconds but I'm not sure how to get to this. Help is appreciated
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    (Original post by vedderfan94)
    Could someone please show me the method for part (c) of this question.:

    14. A particle is initially at the origin. The particle has a constant acceleration of (2i-3j) m/s^2. Two seconds after it has left the origin, the velocity of the particle is (2i + 14j) m/s. The unit vectors i and j are perpendicular.

    c) The speed of the particle is 10 m/s for the first time T seconds after it has left the origin. Find T.

    The answer given in the book is 4 seconds but I'm not sure how to get to this. Help is appreciated
    what info have you collected?
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    There is something missing from your information.

    The particle can not have both constant acceleration (2i-3j) and constant speed (10m/s).
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    (Original post by msmith2512)
    There is something missing from your information.

    The particle can not have both constant acceleration (2i-3j) and constant speed (10m/s).
    I didn't say anything about constant speed.
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    You need to find the initial velocity at the origin - which you can using v=u+at.

    Then using it again to find a different velocity vector (v) which must have a magnitude of 10 units.

    This new vector will have to be written using t (your unknown) but by equating its magnitude to 10 you will get a quadratic in t which you can solve.
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    (Original post by baphomet)
    You need to find the initial velocity at the origin - which you can using v=u+at.

    Then using it again to find a different velocity vector (v) which must have a magnitude of 10 units.

    This new vector will have to be written using t (your unknown) but by equating its magnitude to 10 you will get a quadratic in t which you can solve.
    The initial velocity is (-2i + 20j) m/s (that was part (a)). I used v = u + at like you said to get:

    v = u + at
    10 = (-2i + 20j) + (2i - 3j)t

    What do I do now?
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    u = (-2i + 20j)
    a = (2i - 3j)

    For the speed of particle to equal 10 then i^2 + j^2 = 10^2

    i^2 + j^2 = 100

    The i component at time t = -2 + 2t
    The j component at time t = 20 -3t

    Therefore (-2+2t)^2 + (20-3t)^2 = 100

    Expand, factorise and find t.
 
 
 
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