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    Finding velocity and position vectors

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    Given that the acceleration vector is a(t) = (-9cos(-3t)) i + (-9sin(-3t)) j + (-2t) k , the initial velocity is v(0) = i + k , and the initial position vector is r(0) = i+j+k , compute:

    A. The velocity vector

    B. The position vector

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    https://brainmass.com/math/vector-calculus/finding-velocity-and-position-vectors-33741

    Solution Preview

    a(t) = (-9cos(-3t)) i + (-9sin(-3t)) j + (-2t) k

    initial velocity is v(0) = i + k , and the initial position vector is r(0) = i+j+k

    We know that acceleration = dV/dt, the Rate of change of velocity wrt Time

    So V(t) = integral [a*dt] = integral[ (-9cos(-3t)) i + (-9sin(-3t)) j + (-2t) k]*dt

    = -3Sin(3t) i - 3Cos(3t) j - t^2 k + C --------A

    When t= 0, We get V(0) = -3Sin(0) ...

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