Answer is: option2
\( (\pi^2,\;3) \)Solution:
\[ \mathbf{v}(t)=\langle 2t,\sin t \rangle \quad \Rightarrow \quad \mathbf{s}(t)=\langle t^2+C,\,-\cos(t)+D \rangle \]
\[ \mathbf{s}(0)=\langle C,\,-1+D \rangle \]
Since we are given that \[ \mathbf{s}(0)=\langle 0,1 \rangle, \] we determine that \( C=0 \) and \( D=2 \).
Then \[ \mathbf{s}(t)=\langle t^2,\;2-\cos(t) \rangle \]
We then compute \[ \mathbf{s}(\pi)=\langle \pi^2,\;2-\cos(\pi) \rangle =\langle \pi^2,\;3 \rangle \]
