Calc 3 - Vector Valued Function Initial Value Problem? The starting position of a particle is given by
$\mathbf p(0)=\langle 5,−2\rangle$
Suppose the initial velocity is given by $\mathbf v(0)=\langle 1,2\rangle$ and the acceleration is given by $\mathbf a(t)=\langle 2,3\rangle$. Find: the velocity function, $\mathbf f(t)$ (a vector), the speed function $s(t)$, and the position function $\mathbf p(t)$ (a vector). 
I'm not really sure how to use the information given in this problem! Thanks in advance. 
 A: I'd put it like this: $$ a(t) = \begin{bmatrix}a_1(t)\\a_2(t)\end{bmatrix}= \begin{bmatrix}2\\3\end{bmatrix} = \begin{bmatrix}v_1'(t)\\v_2'(t)\end{bmatrix} $$
So we integrate each component to get: $$ v(t)=\int a(t)\,dt = \int \begin{bmatrix} 2 \\ 3 \end{bmatrix} dt = \begin{bmatrix} 2t +c_1 \\ 3t+c_2 \end{bmatrix} $$
The initial conditions then say $$ v(0) = \begin{bmatrix} 1 \\ 2 \end{bmatrix}  = \begin{bmatrix} 2(0) +c_1 \\ 3(0)+c_2 \end{bmatrix} =  \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}\;\;\;\implies\;\;\; v(t)=\begin{bmatrix} 2t +1 \\ 3t+2 \end{bmatrix}$$ So the speed function is then: $$ s(t) = ||v(t)||_2 = \sqrt{(2t+1)^2+(3t+2)^2} = \sqrt{13t^2+16t+5} $$
The position function is basically the same:
$$ p(t) = \int v(t)\,dt = \int \begin{bmatrix} 2t +1 \\ 3t+2 \end{bmatrix}dt = \begin{bmatrix} t^2 +t+k_1 \\ 3t^2/2+2t+k_2 \end{bmatrix} $$
And again, the initial conditions: $$ p(0) = \begin{bmatrix} 5 \\ -2 \end{bmatrix} = \begin{bmatrix} k_1 \\ k_2 \end{bmatrix}\;\;\;\implies\;\;\; p(t)= \begin{bmatrix} t^2 +t+5 \\ 3t^2/2+2t-2 \end{bmatrix} $$
