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16 February, 22:37

A rocket sled for testing equipment under large accelerations starts at rest and accelerates according to the expression a = (3.6 m/s 3) t + (5.6 m/s 2). How far does the sled move in the time interval t = 0 to t = 1.6 s? Answer in units of m.

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  1. 16 February, 23:20
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    9.6 m

    Explanation:

    This is a case of motion under variable acceleration. So no law of motion formula will be applicable here. We shall have to integrate the given equation.

    a = 3.6 t + 5.6

    d²x / dt² = 3.6 t + 5.6

    Integrating on both sides

    dx / dt = 3.6 t² / 2 + 5.6 t + c

    where c is a constant.

    dx / dt = 1.8 t² + 5.6 t + c

    when t = 0, velocity dx / dt is zero

    Putting these values in the equation above

    0 = 0 + 0 + c

    c = 0

    dx / dt = 1.8 t² + 5.6 t

    Again integrating on both sides

    x = 1.8 t³ / 3 + 5.6 x t² / 2 + c₁

    x = 0.6 t³ + 2.8 t² + c₁

    when t = 0, x = 0

    c₁ = 0

    x = 0.6 t³ + 2.8 t²

    when t = 1.6

    x =.6 x 1.6³ + 2.8 x 1.6²

    = 2.4576 + 7.168

    = 9.6256

    9.6 m
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