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31 January, 03:10

A boat traveled downstream a distance of 80 mi and then came right back. If the speed of the current was 8 mph and the total trip took 5 hours and 20 minutes, find the average speed of the boat relative to the water.

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  1. 31 January, 05:21
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    Velocity of the boat relative to the stream = b

    Velocity ot the boat downstream = b + 8

    Velocity of the boat upstream = b - 8

    v = d / t = > t = d / v

    Time downstream = 80 / [b + 8]

    Time upstream = 80 / [b - 8]

    Total time 80 / [b + 8] + 80 / [b - 8] = 17/3 ... [ I converted 5 h + 20 m to 1/

    hours]

    Now you have to solve

    80 / [b + 8] + 80 / [b - 8] = 17/3

    80[b - 8] + 80[b + 8] = [17/3] (b+8) (b-8)

    Divide by 80

    b - 8 + b + 8 = [17 / (80*3) ] (b^2 - 64)

    2b = [17/240]b^2 - 64*17/240

    0.0708b^2 - 2b - 4.53333 = 0

    Use the quadratic formula to solve. You will obtain b = - 2.11 and b = 30.36

    Only the positve result make sense. Then the answer is b = 30.36 mph.

    Let's verify that that result is coherent:

    Downstream velocity = 30.36 + 8 = 38.36 mi / h

    Downstream time = 80mi / 38.36 mi/h = 2.085 h

    Upstream velocity = 30.36 - 8 = 22.36 mi / h

    Upstream time = 80 mi / 22.36 mi/h = 3.578 h

    Total time = 2.085 h + 3.578 h = 5.663 h, which is 5 h + 20 min; then, the result is correct.
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