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6 May, 01:28

A 1.4 kg particle moves along an x axis, being propelled by a variable force directed along that axis. Its position is given by x = 3.0 m + (4.0 m/s) t + ct2 - (1.6 m/s3) t3, with x in meters and t in seconds. The factor c is a constant. At t = 3.0 s, the force on the particle has a magnitude of 36 N and is in the negative direction of the axis. What is c?

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  1. 6 May, 04:51
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    The value of c is 27.4 m/s²

    Explanation:

    Hi there!

    Let's write the position function:

    x = 3.0 m + (4.0 m/s) · t + c · t² - (1.6 m/s³) · t³

    The velocity of the particle is given by the derivative of the position function with respect to time:

    dx/dt = v = 4.0 m/s + 2 · c · t - 4.8 m/s³ · t²

    The acceleration of the particle is the derivative of the velocity function with respect to t:

    dv/dt = a = 2 · c - 9.6 m/s³ · t

    The applied force at t = 3.0 s is calculated as follows:

    F = m · a

    Where:

    F = applied force.

    m = mass of the particle.

    a = acceleration.

    Then:

    F = m · a

    36 N = 1.4 kg · a

    36 N / 1.4 kg = a

    a = 26 m/s²

    We have derived the equation of the acceleration above:

    a = 2 · c - 9.6 m/s³ · t

    Then, using a = 26 m/s² and t = 3.0 s, we can solve the equation for c:

    26 m/s² = 2 · c - 9.6 m/s³ · 3.0 s

    26 m/s² + 9.6 m/s³ · 3.0 s = 2 · c

    54.8 m/s² = 2 · c

    54.8 m/s² / 2 = c

    c = 27.4 m/s²

    The value of c is 27.4 m/s²
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