Question

A motor boat traveled 18 miles down a river in 2 hours but took 4.5 hours to return upstream. Find the rate of the motor boat in still water and the rate of the current.

Answer 1

b = speed of the boat in still water

c = speed of the current

when going Upstream, the boat is not really going "b" fast, is really going slower, is going "b - c", because the current is subtracting speed from it, likewise, when going Downstream the boat is not going "b" fast, is really going faster, is going "b + c", because the current is adding its speed to it.

now, going either way the boat travelled the same 18 miles.

`\begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Upstream&18&b-c&4.5\\ Downstream&18&b+c&2 \end{array}\hspace{5em} \begin{cases} 18=(b-c)4.5\\\\ 18=(b+c)2 \end{cases} \\\\[-0.35em] ~\dotfill`

`\stackrel{\textit{using the 1st equation}}{18=(b-c)4.5}\implies \cfrac{18}{4.5}=b-c\implies 4=b-c\implies \boxed{4+c=b} \\\\\\ \stackrel{\textit{substituting on the 2nd equation}}{18=[~(4+c)+c~]2}\implies \cfrac{18}{2}=4+2c\implies 9=4+2c \\\\\\ 5=2c\implies \blacksquare~~ \cfrac{5}{2}=c ~~\blacksquare\hspace{5em}4+\left( \cfrac{5}{2} \right)=b\implies \blacksquare~~ \cfrac{13}{2}=b ~~\blacksquare`