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12 April, 03:35

A powerboat heads due northwest at 13 m/s relative to the water across a river that flows due north at 5.0 m/s. what is the velocity (both magnitude and direction) of the motorboat relative to the shore?

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  1. 12 April, 03:42
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    The boat is travelling at 16.9 m/s at 57 degrees North of West.

    This is a vector problem.

    The '^' symbol denotes a power, ('4^2' is '4 squared')

    The first step is to break down the components of the NorthWest motion (13 m/s). The phrase Northwest means it has an angle of 45 degrees north of west. So both the north and west components will be equal.

    Pythagorean theorem: A^2 + B^2 = C^2

    We have C^2, it is 13^2 which is 169.

    Divide that by two to get the squares of one of the components: 84.5,.

    The square root of 84.5 is 9.19.

    So the components of the Northwest motion at 13 m/s is

    9.19 m/s north and ...

    9.19 m/s west.

    Add the north motion of the boat relative to the water to the motion of the water (north at 5 m/s)

    9.19+5 = 14.19 m/s North.

    Our new vector components are

    14.19 m/s North and

    9.19 m/s West.

    Pthagorean theorem again tells us the magnitude of the motion:

    14.19^2 + 9.19^2 = C^2 = 285.81

    Take the square root and get 16.9 m/s

    Now we have a vector triangle.

    Opposite wall: 14.19

    Adjacent Wall: 9.19

    Hypotenuse: 16.9

    Now we need the angle in the bottom right corner:

    Sin (angle) = opposite / hypotenuse = 14.19/16.9

    Sin (angle) =.8396449704

    ArcSin (.8396449704) = 57.1 degrees

    The magnitude and direction:

    16.9 m/s at 57 degrees North of West.

    ArcSin is a reverse trigonometry operation available on most graphing calculators, it appears as "Sin^-1"
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