c0639a…: “To the windy bay”

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The starting position of a particle is given by $\vec {p}(0) =\vector {5,-2}$ Suppose the initial velocity is given by $\vec {v}(0) =\vector {1,2}$ and the acceleration is given by $\vec {a}(t) =\vector {2,3}$ . Find:

The velocity function: $\vec {v}(t) =\vector {\answer {2t+1},\answer {3t+2}}$
The speed function: $\vec {s}(t) = \answer {\sqrt {(1+2t)^2+(2+3t)^2}}$
The position function: $\vec {p}(t) = \vector {\answer {t^2+t+5},\answer {3t^2/2+2t-2}}$