Mathematics: Atour boat left the dock and traveled west at an average speed of 10 mph. a smaller fishing boat left some time later traveling in the same direction but at an average speed of 12 mph. after traveling for 10 hours, the fishing boat caught up to the tour boat. the tour boat left hours earlier than the fishing boat.

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Atour boat left the dock and traveled west at

Ответ:

agabu38

12.03.2022

recall your d = rt, distance = rate * time.

so the tour boat left at a speed of 10 mph, say the fishing boat left "h" hours later, traveling faster at 12 mph in the same direction.

By the time the fishing boat caught up with the tour boat, the distances travelled by both must be the same, say "d" miles, thus the idea of catching up, the distance of the fishing boat is the same as the distance of the tour boat.

Since the fishing boat caught up with the tour boat after going for 10 hours, and had left "h" hours later, then when that happened, the tour boat has already been travelling for 10 + h hours.

$\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ \textit{Tour boat}&d&10&10+h\\ \textit{Fishing boat}&d&12&10 \end{array}~\hfill \begin{cases} d=(10)(10+h)\\ d=(12)(10) \end{cases} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{doing substitution on the 2nd equation}~\hfill }{(10)(10+h)=(12)(10)\implies \cfrac{~~\begin{matrix} (10) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~(10+h)}{~~\begin{matrix} 10 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}=12} \\\\\\ 10+h=12\implies h=2$

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