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Time of boat completing round trip

  1. Mar 13, 2005 #1
    A boat made a round trip of 3km on a river in 2h. The speed of the river's flow was 1m/s. In what time will the boat complete the same round trip, if the speed of the flow is 2 times smaller?

    I suck at these kinds of questions, so would anybody please help me out here?

    - Kamataat
     
  2. jcsd
  3. Mar 13, 2005 #2
    Assume that the boat would move at a constant speed [tex]v[/tex] if there were no current.

    What is the speed of the boat on each leg of the trip?
     
  4. Mar 13, 2005 #3
    W/o current, it's 1,5km/h. With current it's [itex]v + v_{water} = 1,5 km/h[/itex]????????

    - Kamataat
     
    Last edited: Mar 13, 2005
  5. Mar 13, 2005 #4
    You need to be a little bit more careful. I am asking, in the situation where, when there is a current of 1 m/s, it takes the boat 2h to complete the trip, what speed [tex]v[/tex] must the boat be capable of travelling at neglecting current?

    Hint: On one leg of the trip, the speed of the boat is [tex]v + 1\frac{\mbox{m}}{\mbox{s}}[/tex], and on the other, [tex]v - 1\frac{\mbox{m}}{\mbox{s}}[/tex].
     
  6. Mar 13, 2005 #5
    I don't get this. It must be able to travel at 1,5km/h???? Then it's 1,5km/h-1m/s on one leg and 1,5km/h+1m/s on the other leg. That's what I said in my last post (sort of).

    EDIT: I already got as far as your hint on my own, but replacing 1m/s with 0,5m/s in these formulas still gives me 2h for the round trip.

    EDIT2: Aaaargh. The 1,4 and 1,6 km/h figures were wrong. I realize this.

    - Kamataat
     
    Last edited: Mar 13, 2005
  7. Mar 13, 2005 #6
    1m/s = 3,6km/h

    So I have to figure out [itex]v[/itex] and then replace 3,6km/h with 1,8km/h in your hint forulas??? And then for each leg of 1,5km I have to do [itex]1,5/v_{hint}[/itex] to get the time it took to complete that leg????

    - Kamataat
     
    Last edited: Mar 13, 2005
  8. Mar 14, 2005 #7
    Anybody??? :cry:
     
  9. Mar 14, 2005 #8

    HallsofIvy

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    You keep trying to do each leg separately and you can't: you don't know what time each leg took.

    If v is the speed of the boat in still water, then going upstream, its speed is v-1 so it would take 1500/(v-1) s to go 1500 m upstream. Going downstream, its speed is v+1 so it would take 1500/(v+1) s to go 1500 m downstream. The total round trip would take 1500/(v+1)+ 1500/(v-1) seconds. We are told that it actually takes 2 hours= 7200 seconds to do that: solve 1500/(v+1)+ 1500/(v-1)= 7200 to find the speed in still water.

    Now I have a question: what, exactly, is meant by "2 times smaller"? That's peculiar wording. Two times anything makes it larger, not smaller! I could see that being interpreted as either "1/2 as fast" or "1/3 as fast".
     
  10. Mar 14, 2005 #9
    You can probably guess that the original excercise wasn't written in English. "Two times smaller" in my language means 1/2 times the original size, "three times smaller" would be 1/3 times the original size and so on...

    I solved for "v" in the equation you gave and then plugged it in and by replacing 1m/s with 0,5m/s got an answer that is not right according to the excercise book. The correct answer should be 1,25h.

    Since solving for "v" resulted in a quadratic equation, I got two values for "v". One "v" gave 48min as the final answer and the other 98min.

    - Kamataat
     
  11. Mar 14, 2005 #10
    I found out at what speed the river must flow for the book's answer to be correct, and it isn't 0,5m/s. Can we conclude from this that the book has the wrong answer?

    I'm just asking this because I want to be sure that I've done everything correctly.

    - Kamataat
     
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