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1 December, 13:38

Two particles move along an x axis. the position of particle 1 is given by x = 9.00t2 + 6.00t + 2.00 (in meters and seconds); the acceleration of particle 2 is given by a = - 8.00t (in meters per seconds squared and seconds) and, at t = 0, its velocity is 23.0 m/s. when the velocities of the particles match, what is their velocity?

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  1. 1 December, 16:04
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    For particle 1, we can find for velocity by taking the 1st derivative of the equation.

    x = 9 t^2 + 6 t + 2

    dx / dt = 18 t + 6

    v1 = 18 t + 6

    For particle 2, we are given the acceleration a, and initial velocity v0, from this, we can get the velocity v using the formula:

    v = v0 + a t

    v2 = 23 - 8 t^2

    So when v1 = v2, we get t:

    18 t + 6 = 23 - 8t^2

    8t^2 + 18t = 17

    t^2 + 2.25 t = 2.125

    Completing the square:

    (t + 1.125) ^2 = 3.390625

    t + 1.125 = ± 1.84

    t = - 2.965, 0.715

    Since time cannot be negative,

    t = 0.715 seconds

    v1 = 18 t + 6 = 18.87 m/s = v2

    Answer:

    18.87 m/s
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