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#