Water in a river flows at a rate 30 + 2t m^3/s with 10:00 AM taken as t = 0. What is the time when has 4000 m^3 of water has flowed.
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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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