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1 July, 03:54

A particle moving in the x-y plane has a position vector given by r = 1.89t2i + 1.17t3j, where r is in inches and t is in seconds. Calculate the radius of curvature? of the path for the position of the particle when t = 2.1 sec.

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  1. 1 July, 04:42
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    r=89.970 m

    Explanation:

    We need to calculate radius of curvature, using first and second derivatives of the functions x (t) and y (t).

    From the given equation, we can get, that:

    x (t) = 1.89t^2 and y (t) = 1.17t^3

    Then, the first derivative:

    x' (t) = 2*1.89*t=3.78t and y' (t) = 3*1.17*t^2=3.51t^2

    The second derivatives are:

    x'' (t) = 3.78 and y'' (t) = 3.51*2*t=7.02t

    Equation for the radius of curvature, can be found as:

    r = (x'^2+y'^2) ^ (3/2) / (x''y'-x'y'')

    Note, that the denominator should be taken by its absolute value.

    For the given time, we should calculate numerical values for the derivatives. For t=2.1 s:

    x' (2.1) = 7.938 m/s; x'' (2.1) = 3.78 m/s^2; y' (2.1) = 15.4791 m/s and y'' (2.1) = 14.742 m/s^2

    Using equation of the curvature radius and the values, we can get:

    r=89.970 m
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