Question

A plane, initially traveling 150 mi/hr due

Answer 1

By working with the velocity vectors, we will find that the direction of motion of the plane is 28.7 south of east.
Working with the velocity vectors:
First, we need to define our coordinate axis, I will define the x-axis as the east and the y-axis as the north.
We know that the plane travels at 150 mi/hr at 100° east of north, notice that if we measure from the positive x-axis, this is equivalent to an angle of -10°.
Then the x-component of the velocity is: 150mi/hr*cos(-10°) = 147.7 mi/hr
The y-component of the velocity is: 150mi/hr*sin(-10°) = -26 mi/hr.
So the vector is:
V = < 147.7 mi/hr, -26 mi/hr >
Now we know that the wind blowing at 50mi/hr at southwest, exactly southwest would be at an angle of 225°, so the components of the vector are:
x-component: 50mi/hr*cos(225°) = -35.4 mi/hr
y-component: 50mi/hr*sin(225°) = -35.4 mi/hr.
So the vector is:
W = < -35.4 mi/hr, -35.4 mi/hr>
The sum of the vectors gives the total velocity of the plane in the wind, we will get:
V + W = < 147.7 mi/hr -35.4 mi/hr , -26 mi/hr -35.4 mi/hr >
V + W = <112.3 mi/hr, -61.4 mi/hr>
The direction of a vector is given by:
θ = Atan(y/x)
Then in this case the direction of the plane is:
θ = Atan(-61.4 mi/hr/112.3 mi/hr) = -28.7°
So the direction of the plane is 28.7° south of east.