If the position vector of a particle is vec{r | vedclass

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(B) The velocity vector $\vec{v}$ is the time derivative of the position vector $\vec{r}$:$\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(-\cos t \hat{i

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(B) The velocity vector $\vec{v}$ is the time derivative of the position vector $\vec{r}$:$\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(-\cos t \hat{i} + \sin t \hat{j} - 18t \hat{k}) = \sin t \hat{i} + \cos t \hat{j} - 18 \hat{k}$The acceleration vector $\vec{a}$ is the time derivative of the velocity vector $\vec{v}$:$\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(\sin t \hat{i} + \cos t \hat{j} - 18 \hat{k}) = \cos t \hat{i} - \sin t \hat{j}$The magnitude of the acceleration is given by:$|\vec{a}| = \sqrt{(\cos t)^2 + (-\sin t)^2} = \sqrt{\cos^2 t + \sin^2 t}$Since $\sin^2 t + \cos^2 t = 1$,we get:$|\vec{a}| = \sqrt{1} = 1$

If the position vector of a particle is $\vec{r} = -\cos t \hat{i} + \sin t \hat{j} - 18t \hat{k}$,then what is the magnitude of its acceleration?

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