Need help solving a Vector problem, a step by step solution would be great!

Question

A jet is flying through a wind that is blowing with a speed of 55 mph in the direction of N 30°E. The jet has a speed of 765 mph in the direction of N 45°E.

a. Find the true velocity of the jet as a vector.

b. Find the true speed (ground speed) and direction (true course) of the jet.

Answer 1

I am interpreting the notation N X^o E as meaning the direction is X^o East of North

So, the x coordinate is x = v sin X^o and the y coordinate is y = v cos X^o, where we are assuming the y-axis is positive toward the north and the positive x-axis is to the east.

Step 1 (for both parts): break the wind and jet velocities into their components.

The velocity of the wind is 55 mph at 30 degrees east of north, so

xw = (55 mph) sin 30 = 27.5 mph

yw = (55 mph) cos 30 = 47.6 mph

The velocity of the jet is 765 mph at 45 degrees east of north, so

xj = (765 mph) sin 45 = 541 mph

yj = (765 mph) cos 45 = 541 mph

(a) the true velocity of the plane (in vector form) has the components:

xT = xw + xj = 27.5 + 541 = 569 mph (round the decimal up)

yT = yw + yj = 47.6 + 541 = 589 mph

V = 569 i + 589 j

(b) The magnitude (speed) and direction are found using the Pythagorean Theorem and inverse tangent function:

speed: s = √[(xT)² + (yT)²] = √[(569)² + (589)²] = 819 mph (or 818 mph, depending on where rounded the values)

direction: θ = tan^-1 (xT/yT) = tan^-1 (569/589) = 44 The true direction of the plane is N 44^o E or 44 degrees east of north.

Since the wind is blowing in roughly the same direction as the plane's flight (slightly more north), it increases the ground speed of the plane by 54 mph and deflects the trajectory slightly (1 degree) to the north.

Answer 2

Assuming it is flying with the wind.

Draw and label a diagram. That includes angle measures.

The angle between the two vectors is 165°

Use Law of Cosines toe determine resultant vector magnitude.

Use Law of Sines to determine bearing.