The speed of the particle is #dot s(t) = omega r = sqrt2 omega(t)#
For angular speed #omega#, we note that:
#dot omega(t) = alpha(t) = pi/4#
#implies omega(t) = pi/4t + C# and #omega(0) = 0 implies C = 0#
We also note that:
#dot theta(t) = omega(t) = pi/4 t#
#implies theta(t) = pi/8t^2 + C# and #theta(0) = 0 implies C = 0#
From that, for a quarter turn in time #t = tau#, we have:
#pi/2 = pi/8 tau^2 implies tau = 2#
Returning to speed #dot s#, we now have: #dot s(t) = sqrt2 pi/4t#
The average speed #dot s_(ave)# over the time period #tau# is:
#dot s_(ave) = (int_0^tau dot s dt)/(tau)#
The numerator is just the distance travelled, ie a quarter turn along the circumference:
# = (pi/2 sqrt2 )/(2) = ( pi)/(2sqrt2) #
