Water current

John is swimming upstream. After a while, he passes a floating bottle. From that moment, he continues swimming upstream for 20 minutes. He then turns around and swims downstream. From the point where he first passed the bottle, he swims 2 km before meeting the bottle again. What is the speed of the current? John swims at the same constant speed relative to the water in both directions.

Final Answer:

f =  3 km/h

Step-by-step explanation:

$$\begin{gathered} t_1=20\text{min}\to \text{h}=20:60\text{h}=0.33333\text{h} \\ s=2\text{km} \\ {s}_1+s_2=s \\ {s}_1=f\cdot t_1 \\ {s}_2=f\cdot t_2 \\ s+(j-f)\cdot t_1=(j+f)\cdot t_2 \\ s+j\cdot t_1-f\cdot t_1=j\cdot t_2+f\cdot t_2 \\ s=f\cdot t_1+f\cdot t_2 \\ s+j\cdot t_1-(s-f\cdot t_2)=j\cdot t_2+f\cdot t_2 \\ s+j\cdot t_1-s+f\cdot t_2=j\cdot t_2+f\cdot t_2 \\ j\cdot t_1=j\cdot t_2 \\ {t}_2=t_1=\frac{1}{3}\doteq 0.3333\text{h} \\ f\cdot (t_1+t_2)=s \\ f=\frac{s}{t_1+t_2}=\frac{2}{0.3333+0.3333}=3=3\text{km/h} \\ \text{ Verifying Solution: } \\ {s}_1=f\cdot t_1=3\cdot \frac{1}{3}=\frac{3\cdot 1}{3}=\frac{3}{3}=1\text{km} \\ {s}_2=f\cdot t_2=3\cdot 0.3333=1\text{km} \\ {j}_1=8\text{km/h} \\ {s}_3=(j_1-f)\cdot t_1=(8-3)\cdot \frac{1}{3}=\frac{5}{3}\doteq 1.6667\text{km} \\ {t}_{22}=\frac{s+s_3}{j_1+f}=\frac{2+\frac{5}{3}}{8+3}=\frac{1}{3}\doteq 0.3333\text{h} \\ {t}_{22}=t_2 \\ {j}_2=2\text{km/h} \\ {s}_4=(j_2-f)\cdot t_1=(2-3)\cdot \frac{1}{3}=-\frac{1}{3}\doteq -0.3333\text{km} \\ {t}_{33}=\frac{s+s_4}{j_2+f}=\frac{2+\frac{-1}{3}}{2+3}=\frac{1}{3}\doteq 0.3333\text{h} \\ {t}_{33}=t_2 \end{gathered}$$

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