The water in the river flows at a rate that varies with time. This is given by the expression 30 + 2t m^3/s (it has been assumed that t represents time in terms of seconds). At 10 AM, t = 0. The volume of water that flows in time t is given by the definite integral `int_0^t 30 + 2t dt` .

This is equal to 4000, T seconds after 10 AM where `int_0^T 30 + 2t dt = 4000`

=> `(30t + t^2)_0^T = 4000`

=> `30T + T^2 = 4000`

=> `T^2 + 30T - 4000 = 0`

=> `T^2 + 80T - 50T - 4000 = 0`

=> `T(T + 80) - 50(T + 80) = 0`

=> `(T - 50)(T + 80) = 0`

=> T = 50 and T = 80

The positive solution of this equation is T = 50

**The time at which 4000 m^3 of water has flowed is 10:00:50 AM.**

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