If θ=0rad at t=0s, what is the blade's angular position at t=20s

The attached figure represents the relation between ω (rpm) and t (seconds)
To find the blade's angular position in radians ⇒ ω will be converted from (rpm) to (rad/s)
              ω = 250 (rpm) = 250 * (2π/60) = (25/3)π    rad/s
              ω = 100 (rpm) = 100 * (2π/60) = (10/3)π    rad/s

and from the figure it is clear that the operation is at constant speed but with variable levels
            ⇒   ω = dθ/dt   ⇒   dθ = ω dt

            ∴    θ = ∫₀²⁰  ω dt  
 
while ω is not fixed from (t = 0) to (t =20)
the integral will divided to 3 integrals as follow;
       ω = 0                                          from t = 0  to t = 5
       ω = 250 (rpm) = (25/3)π            from t = 5   to t = 15
       ω = 100 (rpm) = (10/3)π            from t = 15 to t = 20

∴ θ = ∫₀⁵  (0) dt   + ∫₅¹⁵  (25/3)π dt + ∫₁₅²⁰  (10/3)π dt
     
the first integral = 0
the second integral = (25/3)π t = (25/3)π (15-5) = (250/3)π
the third integral = (10/3)π t = (25/3)π (20-15) = (50/3)π

∴ θ = 0 + (250/3)π + (50/3)π = 100 π

while the complete revolution = 2π
so instantaneously at t = 20
∴ θ = 100 π - 50 * 2 π = 0 rad

Which mean:
the blade will be at zero position making no of revolution = (100π)/(2π) = 50

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