Solution:

The particle moves from \((0,0)\) to \((\sqrt{3}, 3)\). The slope of this straight line path is \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 0}{\sqrt{3} - 0} = \frac{3}{\sqrt{3}} = \sqrt{3}\). The angle \(theta\) with the x-axis is given by \(tan\theta = m\). So, \(tan\theta = \sqrt{3}\) which implies \(\theta = 60^\circ\).