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28 January, 11:21

The velocity of a particle moving along the x axis is given by vx = a t - b t3 for t > 0, where a = 27 m/s 2, b = 3.1 m/s 4, and t is in s. What is the acceleration of the particle when it achieves its maximum displacement in the positive x direction? Answer in units of m/s 2.

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  1. 28 January, 11:34
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    -54 m/s²

    Explanation:

    Acceleration is defined as the change in velocity of a body with respect to time. Mathematically,

    Acceleration A = change in velocity/time

    A = dv/dt

    Given Vx = at - bt³

    The time at which the particle reaches its maximum displacement is at when vx = 0.

    0 = at-bt³

    t (a-bt²) = 0

    a-bt² = 0

    a = bt²

    t² = a/b ... (1)

    A = dvx/dt = a - 3bt² (by differentiating)

    Acceleration = a - 3bt² ... (2)

    Substituting t² = a/b into equation 2 will give;

    Acceleration = a - 3b (a/b)

    Acceleration = a-3a

    Acceleration = - 2a

    Substituting the value of a = 27m/s into the resulting equation of acceleration gives;

    Acceleration = - 2 (27)

    Acceleration = - 54m/s²

    Therefore at maximum displacement in the positive x direction, the acceleration of the particle will be - 54m/s²
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