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

To find the acceleration function \( \vec{a}(t) \), we first need to differentiate the position function \( \vec{r}(t) \) twice with respect to time \( t \). The position function given is: \[ \vec{r}(t) = 3t \vec{\imath} + t \vec{\jmath} + \frac{1}{4} t^2 \vec{k} \] The first derivative, which gives us the velocity function \( \vec{v}(t) \), is: \[ \vec{v}(t) = \frac{d}{dt} \vec{r}(t) = 3 \vec{\imath} + \vec{\jmath} + \frac{1}{2} t \vec{k} \] Now, we differentiate \( \vec{v}(t) \) to find the acceleration function \( \vec{a}(t) \): \[ \vec{a}(t) = \frac{d}{dt} \vec{v}(t) = 0 \vec{\imath} + 0 \vec{\jmath} + \frac{1}{2} \vec{k} \] So, in terms of the desired format, we have: \[ \vec{a}(t) = \frac{1}{2} \vec{k} \] This means that the acceleration is constant and points in the positive \( z \)-direction. Therefore, we can fill in the placeholders: \(\vec{a}(t)=\frac{1}{2}{[\square]}\vec{k}\).