Answer is: option4
\( \langle 9,4 \rangle \)Solution:
\[ \mathbf{v}(t)=\langle 1+t,\; t^3 \rangle \]
\[ \mathbf{R}(t)=\left\langle t+\frac{t^2}{2}+C,\; \frac{t^4}{4}+D \right\rangle \]
When \( t=0 \), \[ \mathbf{R}(0)=\langle C,\;D \rangle \]
Since we are given that \[ \mathbf{R}(0)=\langle 5,\;0 \rangle, \] we know \( C=5 \) and \( D=0 \).
Hence \[ \mathbf{R}(t)=\left\langle t+\frac{t^2}{2}+5,\; \frac{t^4}{4} \right\rangle \]
From this we obtain \[ \mathbf{R}(2)=\langle 9,\;4 \rangle \]
