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28 June, 00:13

The force F on an object is the product of the mass m and the acceleration a. In this problem, assume that the mass and acceleration both depend on time t, hence so the does the force. That is, F (t) = m (t) a (t) At time t=10 seconds, the mass of an object is 47 grams and changing at a rate of - 6 / frac{g}{s}. At this same time, the acceleration is 17 / frac{m}{s^2} and changing at a rate of - 7 / frac{m}{s^3}. By the product rule, the force on the object is changing at the rate of

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  1. 28 June, 00:36
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    D (F) / dt = - 431 grs*m/s³

    Step-by-step explanation:

    F (t) = M (t) * a (t)

    Taking derivatives on both sides of the equation we get

    D (F) / dt = DM (t) / dt * a (t) + Da (t) / dt * M (t) (1)

    At time t = 10 s

    M (t) = M (10) = 47 grs and DM (t) / dt = - 6 grs/s

    a (t) = a (10) = 17 m/s² and Da (t) / dt = - 7 m / s³

    Plugging these values in equation (1) we get

    D (F) / dt = DM (t) / dt * a (t) + Da (t) / dt * M (t)

    D (F) / dt = - 6 grs/s * 17 m/s² + (-7) m/s³ * 47 grs

    D (F) / dt = - 102 grs*m/s³ - 329 grs*m/s³

    D (F) / dt = - 431 grs*m/s³
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