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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?

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  1. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    where L= 128 Km

  4. Michele_Laino
    • one year ago
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    and ta= 4 hours, td= 2 hours

  5. anonymous
    • one year ago
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    Okay - I understand that...

  6. Michele_Laino
    • one year ago
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    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
    • one year ago
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    Okay

  8. Michele_Laino
    • one year ago
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    if we subtract the first equation from the second one, we get: \[\Large 2V = 32\] what is V?

  9. Michele_Laino
    • one year ago
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    @Juliette2120

  10. Michele_Laino
    • one year ago
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    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
    • one year ago
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    here is the situation described in your problem: |dw:1435553779177:dw|

  12. Michele_Laino
    • one year ago
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    we can write: \[\Large {v_A}t + {v_B}t = L\] where L=168 miles, and t=4 hours

  13. Michele_Laino
    • one year ago
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    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
    • one year ago
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    oops.. \[\Large {v_B} = {v_A} - 4 = 23 - 4 = 19\;miles/hour\]

  15. Michele_Laino
    • one year ago
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    sorry, for my error!

  16. Michele_Laino
    • one year ago
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    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
    • one year ago
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    speeds*

  18. Abhisar
    • one year ago
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    Hello @Juliette2120 ! Are you satisfied with the response? please don't forget to rate the user. Thanks.

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