1.1. Um barco a motor movendo-se rio abaixo ultrapassou uma balsa no ponto $A$. Após um tempo $t=60$ minutos ele voltou e, depois de algum tempo, passou pela balsa a uma distância $L=6,0 \mathrm{~km}$ do ponto $A$. Determine a velocidade da correnteza, considerando que o motor trabalhe num ritmo constante.

Step 1 ：Let the speed of the boat in still water be Vb and the speed of the river current be Vr.

Step 2 ：When the boat is moving downstream, its effective speed is \((Vb + Vr)\), and when it's moving upstream, its effective speed is \((Vb - Vr)\).

Step 3 ：Let D1 be the distance traveled downstream and D2 be the distance traveled upstream. We have: \[D1 = (Vb + Vr) * 1\] and \[D2 = (Vb - Vr) * t2\]

Step 4 ：We also know that \(D1 + D2 = 6\) km. We can substitute D1 and D2 from the above equations and solve for Vr: \[(Vb + Vr) * 1 + (Vb - Vr) * t2 = 6\]

Step 5 ：We don't know the value of t2, but we can express it in terms of Vb and Vr using the fact that the boat traveled the same distance downstream and upstream: \[D1 = D2\] and \[(Vb + Vr) * 1 = (Vb - Vr) * t2\]

Step 6 ：Now we can solve for t2 in terms of Vb and Vr: \[t2 = \frac{(Vb + Vr)}{(Vb - Vr)}\]

Step 7 ：Substitute this expression for t2 back into the equation for the total distance: \[(Vb + Vr) * 1 + (Vb - Vr) * \frac{(Vb + Vr)}{(Vb - Vr)} = 6\]

Step 8 ：Now we can solve for Vr. Since we have only one equation and two unknowns (Vb and Vr), we cannot find a unique solution for Vr. However, we can express Vr in terms of Vb: \[Vr = \frac{(6 - 2 * Vb)}{2}\]

Step 9 ：\(\boxed{Vr = \frac{(6 - 2 * Vb)}{2}}\) is the formula to calculate the speed of the river current for a given speed of the boat in still water (Vb). There is no unique solution for Vr without knowing the value of Vb.