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Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

math question?

a particle moves along the x-axis so that at time t its position is given by

x(t) = { t*cos2pi*t, 0 ≤t<1

ln(kt), t ≥1

a) write but do not solve an equation using a single trig function to determine the values of t for which the particle is at rest.

b) what is the value of k such that the velocity of the particle is continuous?

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  • 2 months ago

    a) The particle is at rest when it is not moving (that is, the velocity of the particle is zero).

    The velocity is given by the rate of change of distance over time, which is dx/dt

    dx/dt = t*(-4pi*sin(2pi*t)), 0 ≤ t <1

                1/t, t ≥ 1

    When dx/dt = 0, we have:

    t*(-4pi*sin(2pi*t)) = 0 

    Note that if we need to have finite values of t when the particle is at rest, we cannot set 1/t = 0 since t will be infinite.

    So we have  sin(2pi*t) = 0 (dividing throughout by -4pi*t)

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