A home-built model rocket is launched straight up into the air. At time \(\displaystyle t = 0 \) the rocket is at rest, about to be launched. The position of the rocket as a function of time is given by:

\(\displaystyle y(t) = \frac{1}{2}(a_0-g)t^2 - \frac{1}{30}\frac{a_0 }{t_{0}^{4}}t^{6} \) for \(\displaystyle 0 \lt t \lt t_0 \)

where \(\displaystyle a_0 \) is a positive constant, \(\displaystyle g\) is the acceleration of gravity and \(\displaystyle a_0 > g \). The contant \(\displaystyle t_0\) is the amount of time that the fuel takes to burn out. Express your answer in terms of g, t,\(\displaystyle a_0\), and \(\displaystyle t_0\).

Find \(\displaystyle a\), the \(\displaystyle y \)-component of the acceleration as a function of time.