## anonymous one year ago 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?

1. anonymous

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?

2. Michele_Laino

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: $\Large {t_d} = \frac{L}{{v + V}}$

3. Michele_Laino

where L= 128 Km

4. Michele_Laino

and ta= 4 hours, td= 2 hours

5. anonymous

Okay - I understand that...

6. Michele_Laino

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

7. anonymous

Okay

8. Michele_Laino

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

9. Michele_Laino

@Juliette2120

10. Michele_Laino

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: $\Large {v_B} = {v_A} - 4$

11. Michele_Laino

here is the situation described in your problem: |dw:1435553779177:dw|

12. Michele_Laino

we can write: $\Large {v_A}t + {v_B}t = L$ where L=168 miles, and t=4 hours

13. Michele_Laino

after a substitution, we get: $\large \begin{gathered} {v_A}t + \left( {{v_A} - 4} \right)t = L \hfill \\ \hfill \\ 2{v_A}t = L + 4t \hfill \\ \hfill \\ {v_A} = \frac{{L + 4t}}{{2t}} = \frac{{168 + 4 \times 4}}{{2 \times 4}} = \frac{{184}}{8} = 23\;miles/hour \hfill \\ \hfill \\ {v_B} = {v_A} - 4 = 23 - 4 = 17\;miles/hour \hfill \\ \end{gathered}$

14. Michele_Laino

oops.. $\Large {v_B} = {v_A} - 4 = 23 - 4 = 19\;miles/hour$

15. Michele_Laino

sorry, for my error!

16. Michele_Laino

reassuming, the requested rates or speed, are: $\Large \begin{gathered} {v_A} = 23\;miles/hour \hfill \\ {v_B} = {v_A} - 4 = 23 - 4 = 19\;miles/hour \hfill \\ \end{gathered}$

17. Michele_Laino

speeds*

18. Abhisar

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