49. The position vector of a particle moving in the xy-plane at time \( t \) is given by \[ \mathbf{p}=(3t^2-4t)\mathbf{i}+(t^2+2t)\mathbf{j}. \] The speed of the particle at \( t=2 \) is






Answer is: option3

\( 10 \)

Solution:

The velocity vector is \[ \mathbf{v}(t)=(6t-4)\mathbf{i}+(2t+2)\mathbf{j}. \]

At the time \( t=2 \), the velocity vector is \[ \mathbf{v}(2)=8\mathbf{i}+6\mathbf{j}=\langle 8,6 \rangle. \]

The speed of the particle is the magnitude (length) of this velocity vector.

\[ \text{Speed}=\sqrt{8^2+6^2}=10 \]

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