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?

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"