A motorboat takes 4 hours to travel 128km going upstream. The return trip takes 2 hours going downstream. What is the rate of the boat in still water? What is the rate of the current?

Question

anonymous · Answer

I was able to figure that one out. Now I've moved on to the following: two cyclists leave town 168 miles apart at the same time and travel toward each other. One cyclist travels 4 mi/h slower than the other. If they meet in 4 hours what is the rate of each cyclist?

michele_laino · Answer

If I call with v the speed of the motorboat with respect to the earth and with V the speed of the current, then the time nedded to go upstream is:
$$\Large {t_a} = \frac{L}{{v - V}}$$
whereas the time needed to got to downstream is:
$$\Large  {t_d} = \frac{L}{{v + V}}$$

michele_laino · Answer

where L= 128 Km

michele_laino · Answer

and ta= 4 hours, td= 2 hours

anonymous · Answer

Okay - I understand that...

michele_laino · Answer

so we can write this algebraic system:
$$\Large  \left\{ \begin{gathered}
  v - V = 32 \hfill \
   \hfill \
  v + V = 64 \hfill \ 
\end{gathered}  \right.$$

anonymous · Answer

Okay

michele_laino · Answer

if we subtract the first equation from the second one, we get:
$$\Large  2V = 32$$
what is V?

michele_laino · Answer

@Juliette2120

michele_laino · Answer

here I call with v_A the speed of the first cyclist and with v_B the speed of the second cyclist, so I can write this:
$$\Large {v_B} = {v_A} - 4$$

michele_laino · Answer

here is the situation described in your problem:
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