Answer:
The resultant velocity of the jet as a vector in component form 426.87 mi/hr 5.36 degrees North.
Explanation:
Vectors are quantities that have their magnitude and direction .
Sketching out the problem given, by using straight lines to represent each of the vectors, we will have a right angled triangle as shown below.
The solution can be obtained by applying Pythagoras theorem to
resolve the vectors.
Velocity of jet plane = 425 mi/hr
velocity of air = 40 mi/hr
Resultant of the vectors =[tex]\sqrt[]{425^{2}+40^{2}}=426.87[/tex] mi/hr
Vector direction =[tex]tan^{-1}(\frac{40}{425})= 5.36 degrees[/tex]
hence the velocity is 426.87 mi/hr in a direction 5.36 degrees inclined Northward
