Answer is: option1
\( 2\sqrt{10} \)Solution:
The velocity vector is given by
\[ \begin{cases} \frac{dx}{dt} = 2t + 2 \\ \frac{dy}{dt} = 4t - 6 \end{cases} \]
When \( t = 2 \), the velocity vector is \( \mathbf{v}(2) = \langle 6, 2 \rangle \).
The speed of the particle is the length of this vector, \[ |\mathbf{v}| = \sqrt{6^2 + 2^2} = 2\sqrt{10}. \]
