The mechanical work can be defined as
\Delta E_k = F \cdot s
where E_k is the kinetic energy, F is the force (the thrust of the rocket which is considered constant), and s is the distance. Differentiating with respect to time, we obtain
\frac{\operatorname{d}E_k}{\operatorname{d}t} = F \cdot \frac{\operatorname{d}s}{\operatorname{d}t}
or
\frac{\operatorname{d}E_k}{\operatorname{d}t} = F \cdot v
where v is the velocity. Dividing by the instantaneous mass m to express this in terms of specific energy (e_k), we get
\frac{\operatorname{d}e_k}{\operatorname{d}t} = \frac F m \cdot v = a \cdot v
where a is the acceleration vector.
No not really.