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Rate problem

  1. Jul 17, 2008 #1
    1. The problem statement, all variables and given/known data
    A boat takes 2.0 hr to travel 31 km down a river, then 6.0 hr to return. How fast is the river flowing?

    2. Relevant equations

    3. The attempt at a solution

    first started out by finding the rate the boat traveled in each trip.

    31/2=15.5 km/hr

    31/6=5.2 km/hr

    Now this is where im not sure if im solving this correctly. I then find the difference between the two rates and the time taken to travel both ways.

    10.3 (difference of 15.5-5.2)/4(difference of 6-2)=2.6 km/hr

    Is this the correct way of solving this problem? Thanks!
  2. jcsd
  3. Jul 17, 2008 #2


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    Hi osker246,

    No, I don't think that's right. Notice is you keep units on your calculation you'll get km/hr^2.

    So work with the total rate on each trip that you found. The boat has a speed, and the river has a speed. How do those combine to give 15.5 km/hr on the one trip, and how do they combine to give 5.2 km/hr going the other way?
  4. Jul 17, 2008 #3
    hey alphycist,

    Im not sure if I follow what your trying to say. Obviously when the boat travels 15.5 km/hr it is traveling with the current and when it travels 5.2 hm/hr its traveling against the current. I'm at a loss how you find the rate the current moves though.
  5. Jul 17, 2008 #4


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    The boat has a speed [itex]v_b[/itex] (measured in still water), and the water has a speed [itex]v_w[/itex]. How do you combine those to get 15.5 km/h? How do you combine those to get 5.2 km/h? That will give you two equations and then you can solve for both of the unknown speeds. Does that make sense?
  6. Jul 17, 2008 #5
    Ok I think I do understand. Tell me if this is correct.

    Vb + Vw = 15.5


    Vb-Vw = 5.2

    Ok so I solved Vb + Vw = 15.5 for Vb. Giving me Vb=15.5 - Vw.

    I then pluged in Vb=15.5 - Vw, into the equation Vb-Vw = 5.2 then solving for Vw.





    Is this correct?
  7. Jul 17, 2008 #6


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    That sounds right to me.
  8. Jul 17, 2008 #7
    Alright! The answer was correct! Thank you very much alphysicist. I appreciate it.
  9. Jul 17, 2008 #8


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    Sure, glad to help!
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