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17 March, 22:33

Orson rides his power boat up and down a canal. The water in the canal flows at 6 miles per hour. Orson takes 5 hours longer to travel 360 miles against the current than he does to travel 360 miles with the current. What is the speed of orson's boat?

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  1. 18 March, 02:28
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    To solve this problem, let us say that:

    v1 = the speed or velocity of Orson in travelling against the current

    t1 = the time taken in travelling against the current

    v2 = the speed or velocity of Orson in travelling with the current

    t2 = the time taken in travelling with the current

    It was stated in the problem that the time difference in travelling against the current and travelling with the current is 5 hrs, therefore:

    t1 - t2 = 5

    We know that:

    t = d / v

    Therefore:

    (360 / v1) - (360 / v2) = 5

    However we also know that:

    v1 = v - 6

    v2 = v + 6

    where v is the velocity of the boat alone

    [360 / (v - 6) ] - [360 / (v + 6) ] = 5

    Multiplying everything by (v - 6) * (v + 6):

    360 (v + 6) - 360 (v - 6) = 5 (v - 6) (v + 6)

    360 v + 2160 - 360 v + 2160 = 5 v^2 - 180

    4320 = 5 v^2 - 180

    5 v^2 = 4500

    v^2 = 900

    v = 30

    Therefore the speed of Orson’s boat is 30 miles per hour
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