The cruise ship

The cruise ship has a speed of 12 km / h at a calm surface. When we sail 45 km along the river and 45 km back, it took us exactly 8 hours. Which (constant) speed of flow of the river?

Result

v =  3 km/h

Solution:

45=t1(12+v) 45=(8t1) (12v) t1=45/(12+v) 45=(845/(12+v)) (12v)   45 (12+v)=(8 (12+v)45) (12v) 8v272=0  a=8;b=0;c=72 D=b24ac=0248(72)=2304 D>0  v1,2=b±D2a=±230416 v1,2=±4816 v1,2=±3 v1=3 v2=3   Factored form of the equation:  8(v3)(v+3)=0  v>0 v=v1=3 km/h t1=45/(12+v)=45/(12+3)=3 h t2=45/(12v)=45/(123)=5 h45=t_{1}(12+v) \ \\ 45=(8-t_{1}) \cdot \ (12-v) \ \\ t_{1}=45 / (12+v) \ \\ 45=(8-45/(12+v)) \cdot \ (12-v) \ \\ \ \\ \ \\ 45 \cdot \ (12+v)=(8 \cdot \ (12+v)-45) \cdot \ (12-v) \ \\ 8v^2 -72=0 \ \\ \ \\ a=8; b=0; c=-72 \ \\ D=b^2 - 4ac=0^2 - 4\cdot 8 \cdot (-72)=2304 \ \\ D>0 \ \\ \ \\ v_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ \pm \sqrt{ 2304 } }{ 16 } \ \\ v_{1,2}=\dfrac{ \pm 48 }{ 16 } \ \\ v_{1,2}=\pm 3 \ \\ v_{1}=3 \ \\ v_{2}=-3 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 8 (v -3) (v +3)=0 \ \\ \ \\ v>0 \ \\ v=v_{1}=3 \ \text{km/h} \ \\ t_{1}=45 / (12+v)=45 / (12+3)=3 \ \text{h} \ \\ t_{2}=45 / (12-v)=45 / (12-3)=5 \ \text{h}

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