The velocity will be maximum to y = ½.

U_max  = - 1/2μ ∂p / ∂x [ t x t / 2 - (t / 2)²] = - 1/2μ ∂p / ∂x [ t² / 2 - t² / 4 ]

U_max  =  - 1/ 8μ ∂p / ∂x t²

Average velocity      u  =  Q/A

dQ  = Velocity at a distance y x Area of strip

= - 1/2μ ∂p / ∂x  [ ty - y² ] x dy x 1

Integrating from 0 to t

       t          t
Q = ∫ dQ = ∫ - 1/2μ ∂p / ∂x [ ty - y² ]  dy x 1
      0         0

Q  = - 1/2μ ∂p / ∂x [ t ( y² / 2)₀^t - (y³ / 3)₀¹ ] = - 1/2μ ∂p / ∂x  [ t³ / 2 - t³ / 3 ]

Q = - 1/2μ ∂p / ∂x t³ / 6

Q = - 1/ 12μ ∂p / ∂x t³

u  = Q / A = - 1/12μ . ∂p / ∂x . t3 / t x 1

u = - 1/ 12μ ∂p / ∂x t²

Ratio Umax to u

U_max  = - 1 / 8μ (∂p / ∂x) t² / 1/ 12μ ∂p / ∂x t²
-
u

U_max  = 3 / 2
-
u

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