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Thread: How long does the trip take

  1. #1
    Junior Member
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    How long does the trip take

    question:
    A river has a steday speed of 0.345 m/s. A studen swims upstream a distance of 0.543 km and returns (Still swimming)to the starting point.
    If the studen can swim at a speed of 1.12m/s in still water, how long does the trip take. Answer in units of s.

    Any suggestions on this one. I can't find an example in my book to help me out.
    Stacey Rhone

  2. #2
    Elite Member
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    Re: How long does the trip take

    Hint:
    swims upstream at speed 1.12 - .345 (against current)
    swims coming back at speed 1.12 + .345
    I'm just an imagination of your figment !

  3. #3
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    I'm still having trouble figuring out how to solve this.
    Stacey Rhone

  4. #4
    Elite Member
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    Re: How long does the trip take

    Hello, Stacey!

    This problem doesn't require any Algebra,
    . . but you have to do some Thinking.

    Since they're talking about distance, speed, and time,
    . . it is reasonable that we need: [tex]\,\text{Distance}\:=\:\text{Speed }\times\text{ Time}[/tex]

    Since they ask for the Time ("How long does the trip take?")
    . . we'll use this variation: [tex]\:\text{Time}\;=\;\frac{\text{Distance}}{\text{Spe ed}}[/tex]
    Still with me?


    A river has a steady speed of 0.345 m/s.
    A student swims upstream a distance of 0.543 km and returns to the starting point.
    If the student can swim at a speed of 1.12m/s in still water,
    how long does the trip take? .Answer in units of seconds.

    The student swims at 1.12 m/s in still water.

    Swimming against the current, his speed is: [tex]1.12\,-\,0.345\:=\:0.775[/tex] m/s.
    . . To swim 0.543 km = 543 m, it will take: [tex]\frac{543}{0.775}\:\approx\:700.645[/tex] seconds.

    Swimming with the current, his speed is: [tex]1.12\,+\,0.345\:=\:1.465[/tex] m/s.
    . . To swim 543 m, it will take: [tex]\,\frac{543}{1.465} \:\approx\:370.648[/tex] seconds.


    Therefore, the entire trip takes: [tex]\,700.645\,+\,370.648\:=\:1071.293\:\approx\:1071. 3[/tex] seconds.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    And stop looking in the book for an identical problem.
    . . What kind of learning is that?

    "If I give you a map, can you find your way to City Hall?"
    "Sure, but ... um ... could you drive me there first?"

    I'm the other of the two guys who "do" homework.

  5. #5
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    I agree with you 100% here. I'm just finding that with physics there are so many rules and laws and formulas that i'm getting all mixed up on what applies to what and when to use what. Am I ever going to understand this! I've always been bad at word problems to make matters worse. I really appreciate your help.
    Stacey Rhone

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