For simplicity, Let: y = the kinetic energy
The kinetic energy changes by
\Delta y= (135-75)KJ=60KJ
over a time period of 9 seconds.
The rate of change of the kinetic energy, which I'm calling y, is
{dy}/dt={60KJ}/{9s}=6.7{KJ}/s
Therefore the kinetic energy as a function of time is
y_{(t)}=(6.7{KJ}/s)t+75KJ
This follows from y_{(t)}=y_{(t=0)}+\int_0^t{dy}/{dt}dt,
where I used that at t=0, \quad y_{(t=0)}=75KJ.
We know that Kinetic Energy =y=1/2mv^2, Therefore
v_{(t)}=\sqrt{{2y}/m}.
The average speed is
v_{ave}=1/9\int_0^9 v_{(t)}dt=1/{9s}\sqrt{2/m}\int_0^9\sqrt{y\quad}dt
We could substitute y and integrate in terms of t. But it looks like it would be easier to integrate over y. The endpoints of the integral correspond to the initial and final kinetic energies and we have that dt=(1/6.7 s/{KJ})dy
v_{ave}=1/{9s}\sqrt{2/m}(1/6.7 s/{KJ})\int_{y_i}^{y_f}y^{1/2}dy
Note that (9s)(6.7{KJ}/s)=y_f-y_i, because that's how we found the rate of change of the kinetic energy.
v_{ave}=1/{(y_f-y_i)}\sqrt{2/m} 2/3 [y_f^{3/2}-y_i^{3/2}]
Be sure to use y_i and y_f in Joules, the answer is
v_{ave}=237m/s
