A particle is rotating in a circular path and at any instant its motion can be described as $$\theta = \frac{5t^4}{40} - \frac{t^3}{3}$$. The angular acceleration of the particle after 10 seconds is __________ rad/s$$^2$$.

The angular position as a function of time is given by $$\theta = \frac{5t^{4}}{40} - \frac{t^{3}}{3}$$.

Simplify the coefficients:
$$\theta = \frac{5}{40}\,t^{4} - \frac{1}{3}\,t^{3} = \frac{1}{8}\,t^{4} - \frac{1}{3}\,t^{3}$$

Angular velocity is the first time-derivative of angular position:
$$\omega = \frac{d\theta}{dt}$$

Differentiating term by term:
$$\omega = \frac{d}{dt}\left(\frac{1}{8}\,t^{4}\right) - \frac{d}{dt}\left(\frac{1}{3}\,t^{3}\right) = \frac{1}{8}\,(4t^{3}) - \frac{1}{3}\,(3t^{2}) = \frac{4}{8}\,t^{3} - t^{2} = \frac{1}{2}\,t^{3} - t^{2}$$

Angular acceleration is the second time-derivative of angular position, i.e. the first derivative of angular velocity:
$$\alpha = \frac{d\omega}{dt}$$

Differentiating $$\omega$$:
$$\alpha = \frac{d}{dt}\left(\frac{1}{2}\,t^{3}\right) - \frac{d}{dt}\left(t^{2}\right) = \frac{1}{2}\,(3t^{2}) - 2t = \frac{3}{2}\,t^{2} - 2t$$

Substitute $$t = 10 \text{ s}$$:
$$\alpha = \frac{3}{2}\,(10)^{2} - 2(10) = \frac{3}{2}\,(100) - 20 = 150 - 20 = 130 \text{ rad/s}^2$$

Therefore, the angular acceleration after 10 seconds is $$130 \text{ rad/s}^2$$.

Option C which is: $$130$$