Asailboat is heading directly north at a speed of 20 knots (1 knot 50.514 m/s). the wind is blowing toward the east with a speed of 17 knots. (a) determine the magnitude and direction of the wind velocity as measured on the boat. (b) what is the component of the wind velocity in the direction parallel to the motion of the boat?

(a) The magnitude of the wind as it is measured on the boat will be the result of the two vectors. Since they are at 90°, the resultant can be determined by the Pythagorean theorem.
 
                   R = sqrt ((20 knots)² + (17 knots)²)
                    R = sqrt (400 + 289)
                      R = 26.24 knots

The direction of the wind will have to be angle between the boat and the resultant.
                   cos θ = (20 knots)/(26.24 knots)
                        θ = 40.36°

Hence, the direction is 40.36° east of north.

(b) As stated, the wind is blowing in the direction that is to the east. This means that it only has one direction. Parallel to the motion of the boat, the magnitude of the wind velocity will have to be zero. 

It equals just about 6,300 km per hour each day an week