The boat is travelling at 16.9 m/s at 57 degrees North of West.Β
This is a vector problem.Β
The '^' symbol denotes a power, ('4^2' is '4 squared')Β
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.Β
Pythagorean theorem: A^2 + B^2 = C^2Β
We have C^2, it is 13^2 which is 169.Β
Divide that by two to get the squares of one of the components: 84.5,.Β
The square root of 84.5 is 9.19.Β
So the components of the Northwest motion at 13 m/s isΒ
9.19 m/s north and..Β
9.19 m/s west.Β
Add the north motion of the boat relative to the water to the motion of the water (north at 5 m/s)Β
9.19+5= 14.19 m/s North.Β
Our new vector components areΒ
14.19 m/s North andΒ
9.19 m/s West.Β
Pthagorean theorem again tells us the magnitude of the motion:Β
14.19^2 + 9.19^2 = C^2 = 285.81Β
Take the square root and get 16.9 m/sΒ
Now we have a vector triangle.Β
Opposite wall: 14.19Β
Adjacent Wall: 9.19Β
Hypotenuse: 16.9Β
Now we need the angle in the bottom right corner:Β
Sin(angle)= opposite / hypotenuse = 14.19/16.9Β
Sin(angle)= .8396449704Β
ArcSin(.8396449704) = 57.1 degreesΒ
The magnitude and direction:Β
16.9 m/s at 57 degrees North of West.Β
ArcSin is a reverse trigonometry operation available on most graphing calculators, it appears as "Sin^-1"
