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Thread: Current's Speed

  1. #1
    Member harpazo's Avatar
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    Current's Speed

    A motorboat maintained a constant speed of 15 miles per hour relative to the water in going 10 miles upstream and then returning. The total time for the trip was 1.5 hours. Find the speed of the current.

    Let c = current's speed

    Distance is given to be 10 miles.

    D = rt is manipulated to be t = D/r.

    [10/(c - 15)] + [10/(c - 15) = 1.5

    Is this correct?
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  2. #2
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    Re: Current's Speed

    Quote Originally Posted by harpazo View Post
    [10/(c - 15)] + [10/(c - 15) = 1.5
    Is this correct?
    Not quite; should be:
    [10/(15 - c)] + [10/(15 + c)] = 1.5

    You won't get anywhere upstream if current is more than speed
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  3. #3
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    Re: Current's Speed

    Not quite. In order that the boat be able to go upstream, its speed, 15 mph relative to the water, must be greater than the current speed. That is, its speed upstream 15- c, not c- 15. Since the boat maintains a speed of 15 mph relative to the water, relative to the river bank, going upstream its speed is 15- c mph and going downstream its speed is 15+ c mph. The time it takes to go 10 m at 15- c mph is 10/(15- c) hours. The time it takes to go 10 m at 15+ c mph is 10/(15+ c) hours. So the total time to go up and down stream is 10/(15- c)+ 10/(15+ c)= 1.5.
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  4. #4
    Member harpazo's Avatar
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    Re: Current's Speed

    I meant to place the total miles going up and down stream as 10. It was a typo at my end.
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  5. #5
    Member harpazo's Avatar
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    Re: Current's Speed

    Multiply both sides by (15 + c)(15 - c).

    When I do that, I get the following several steps later:

    300 = 337.5 - 1.5c^2

    300 - 337.5 = -1.5c^2

    -37.5 = -1.5c^2

    -37.5/-1.5 = c^2

    25 = c^2

    sqrt{25} = sqrt{c^2}

    5 = c

    -5 = c

    We reject -5 mph.

    The current's speed is 5 pmh.
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  6. #6
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    Re: Current's Speed

    Correct!
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  7. #7
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    Re: Current's Speed

    Wonderful.
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