Given acceleration #a(t)=t-3#. We know that #a=x''#, therefore, to find distance moved during the period of interest #x=int int a(t).dt.dt# over the period of interest

(First integral will give us the velocity in terms of time.) Inserting given values.

#x=int int(t-3).dt.dt#

#=int_3^6(t^2/2-3t+C).dt# .....(1), where #C# is constant of integration.

We need to evaluate #C# from initial conditions

Velocity #v(t)=t^2/2-3t+C#

Now #v(0)=0^2/2-3xx0+C=0#, initial condition

or #C=0#

Inserting in (1)

#x=int_3^6(t^2/2-3t).dt#

#x=|t^3/6-3 t^2/2|_3^6#, constant of integration ignored due to definite integral

#=(6^3/6-3xx 6^2/2)-(3^3/6-3 xx3^2/2)#

#=36-54-9/2+27/2#

#=-9#,

We know that #"Average speed"=("Total Displacement")/"Time Taken"#

#"Average speed"=-9/(6-3)#

#"Average speed"=-3#