Answer
$v = -(2.70~m/s)\hat{i}-(1.80~m/s)\hat{j}$
Work Step by Step
We can find the radius of the circular path:
$r = \sqrt{(2.00~m)^2+(-3.00~m)^2} = 3.61~m$
We can find the speed of the particle:
$v = \frac{2\pi~r}{T}$
$v = \frac{(2\pi)~(3.61~m)}{7.00~s}$
$v = 3.24~m/s$
We can find the angle $\theta$ of the position vector below the positive direction of the x axis:
$tan~\theta = \frac{3.00~m}{2.00~m}$
$\theta = tan^{-1}~(\frac{3.00~m}{2.00~m})$
$\theta = 56.3^{\circ}$
The velocity vector is perpendicular to the position vector.
We can find the angle $\phi$ of the velocity vector below the negative direction of the x axis:
$\phi = 180^{\circ} - 90^{\circ} - 56.3^{\circ} = 33.7^{\circ}$
We can find the components of the velocity vector:
$v_x = -(3.24~m/s)~cos~33.7^{\circ} = -2.70~m/s$
$v_y = -(3.24~m/s)~sin~33.7^{\circ} = -1.80~m/s$
We can express the velocity in unit-vector notation:
$v = -(2.70~m/s)\hat{i}-(1.80~m/s)\hat{j}$
