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    Topics || Problems

    It takes a motorboat \(1\frac{1}{3}\) hours to go 20 miles downstream and \(2\frac{2}{9}\) hours to return. Find the rate of the current and rate of the boat in still water. Solution:

    \(V_d = V_b + V_c\)

    \(\frac{distance}{time} = V_b + V_c\)

    \(\frac{20}{1\frac{1}{3}} = V_b + V_c\)

    \(15 = V_b + V_c\)

    \( V_b = 15 - V_c\)

    \(V_u = V_b - V_c\)

    \(\frac{distance}{time} = V_b - V_c\)

    \(\frac{20}{2\frac{2}{9}} = V_b - V_c\)

    \(V_b = 9 + V_c\)

    \(15-V_c = 9 +V_c\)

    \(V_c = 3\)


    \(V_b = 9 + 3 = 12\)

    Thus the rate of the boat on still water is 12 mph and the current is 3 mph