Water current

John swims upstream. After a while, he passes the bottle, and from that moment, he floats for 20 minutes in the same direction. He then turns around and swims back, and from the first meeting with the bottle, he sails 2 kilometers before he reaches the bottle. What is the current speed? John is still swimming at the same speed regardless of the current.

Correct answer:

f =  3 km/h

Step-by-step explanation:

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

Try another example