Byrd Gross
03/08/2023 · High School
Given that the acceleration vector is \( \mathbf{a}(t)=(-16 \cos (4 t)) \mathbf{i}+(-16 \sin (4 t)) \mathbf{j}+(-2 t) \mathbf{k} \), the initial velocity is \( \mathbf{v}(0)=\mathbf{i}+\mathbf{k} \), and the initial position vector is \( \mathbf{r}(0)=\mathbf{i}+\mathbf{j}+\mathbf{k} \), compute: A. The velocity vector \( \mathbf{v}(t)= \) B. The position vector \( \mathbf{r}(t)= \)
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A. The velocity vector is:
\[
\mathbf{v}(t) = (-4 \sin(4t) + 1) \mathbf{i} + (4 \cos(4t) - 4) \mathbf{j} + (-t^2 + 1) \mathbf{k}
\]
B. The position vector is:
\[
\mathbf{r}(t) = \cos(4t) + t \mathbf{i} + (\sin(4t) - 4t + 1) \mathbf{j} + \left(-\frac{t^3}{3} + t + 1\right) \mathbf{k}
\]
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