Given the position function, find the acceleration function. $$ \vec{r}(t)=3 t \vec{\imath}+t \vec{\jmath}+\frac{1}{4} t^{2} \vec{k} $$ $$ \vec{a}(t)=\frac{[?]}{[\square]} \vec{k} $$

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To find the acceleration function $$\vec{a}(t)$$ from the given position function $$\vec{r}(t)$$, we need to compute the second derivative of $$\vec{r}(t)$$ with respect to time $$t$$.

Given:
$$
\vec{r}(t) = 3t \, \vec{\imath} + t \, \vec{\jmath} + \frac{1}{4} t^{2} \, \vec{k}
$$

**Step 1: Find the Velocity Function $$\vec{v}(t)$$**

The velocity is the first derivative of the position function:
$$
\vec{v}(t) = \frac{d\vec{r}(t)}{dt} = \frac{d}{dt}\left(3t \, \vec{\imath} + t \, \vec{\jmath} + \frac{1}{4} t^{2} \, \vec{k}\right) = 3 \, \vec{\imath} + 1 \, \vec{\jmath} + \frac{1}{2} t \, \vec{k}
$$

**Step 2: Find the Acceleration Function $$\vec{a}(t)$$**

The acceleration is the second derivative of the position function, or the first derivative of the velocity function:
$$
\vec{a}(t) = \frac{d\vec{v}(t)}{dt} = \frac{d}{dt}\left(3 \, \vec{\imath} + 1 \, \vec{\jmath} + \frac{1}{2} t \, \vec{k}\right) = 0 \, \vec{\imath} + 0 \, \vec{\jmath} + \frac{1}{2} \, \vec{k}
$$

So, the acceleration vector is:
$$
\vec{a}(t) = \frac{\frac{1}{2}}{1} \, \vec{k} = \frac{1}{2} \, \vec{k}
$$

**Final Answer:**
$$
\vec{a}(t) = \frac{\frac{1}{2}}{1} \, \vec{k}
$$
The acceleration function is $$ \vec{a}(t) = \frac{1}{2} \, \vec{k} $$.

Given the position function, find
the acceleration function.

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