Given:
The distance covered while travelling with the current is, d(1) = 60 miles.
The distance covered while travelling against the current is, d(2) = 10 miles.
The speed of the current is, s(w) = 5 mph.
The objective is to find the rate of team s(t) in still water.
Explanation:
It is given that, the time taken during the direction of current and against the current is same.
[tex]\begin{gathered} t(1)=t(2) \\ \frac{d(1)}{s(1)}=\frac{d(2)}{s(2)} \\ \frac{d(1)}{s(t)+s(w)}=\frac{d(2)}{s(t)-s(w)}\text{ . . . . . (1)} \end{gathered}[/tex]On plugging the given values in equation (1),
[tex]\begin{gathered} \frac{60}{s(t)+5}=\frac{10}{s(t)-5} \\ 60\lbrack s(t)-5\rbrack=10\lbrack s(t)+5\rbrack \\ 60s(t)-300=10s(t)+50 \\ 60s(t)-10s(t)=300+50 \\ 50s(t)=350 \\ s(t)=\frac{350}{50} \\ s(t)=7 \end{gathered}[/tex]Hence, the rate of team still in the water = 7 mph.
