Answer is: option4
\( \langle 7,\; 4 + e^{-2} \rangle \)Solution:
With the acceleration vector \( \mathbf{a}=\langle 2,\;e^{-t}\rangle \), we antidifferentiate to obtain the velocity vector
\[ \mathbf{v}=\langle 2t+A,\;-e^{-t}+B\rangle \]
The particle is at rest when \( t=0 \), so \( \mathbf{v}(0)=\langle 0,0\rangle \).
Since \( \mathbf{v}(0)=\langle 0+A,\;-e^{0}+B\rangle \), we find that \( A=0 \) and \( B=1 \).
Hence \[ \mathbf{v}=\langle 2t,\;1-e^{-t}\rangle \]
We antidifferentiate again to obtain the position vector
\[ \mathbf{s}=\langle t^2+C,\;t+e^{-t}+D\rangle \]
At \( t=0 \), \( \mathbf{s}(0)=\langle 3,3\rangle \), so \( \langle C,\;1+D\rangle=\langle 3,3\rangle \), and therefore \( C=3 \), \( D=2 \).
\[ \mathbf{s}=\langle t^2+3,\;t+e^{-t}+2\rangle \]
When \( t=2 \), \[ \mathbf{s}=\langle 7,\;4+e^{-2}\rangle \]
