Question

a point moves along a circle of radius 1m with speed v=at .The tangential acceleration of the point at a time when it has completed 1/8 of the revolution from start is ?

Answer 1

tangential acceleration is asked... It is =  the derivative of the tangential velocity.
               acceleration = dv / dt = d (a t ) / dt = a

this is always constant.
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Completion of 1/8 of revolution means:  Ф = angle turned = π/4 radians,  or the distance traveled is = 2πR / 8 = π R/4

t = time
v = instantaneous linear speed along the circular arc = a * t
radius = R
ω = instantaneous angular velocity = v / r  = a t / R

linear displacement s = integral v * dt
   So    s = a t² / 2,    we choose  s = 0  at  t = 0
               = π R / 4
   =>  t² = (πR) / (2a)
         t = √ [ πR / (2a) ]
This is the time taken to reach the point where the body covered 1/8 of the revolution.

tangential or Linear acceleration =  dv/dt = a, it is always a constant. 
centripetal acceleration =  v² / R = a² t² / R
resultant acceleration will be the net result of the two components.