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"
