Question

A small plane flies 37.0 km in a direction 45° north of east and then flies 28.0 km in a direction 25° north of east.

Answer 1

Answer:

d= 64.1 km θ = 36.4º

Explanation:

a) In order to find the plane's straight-line distance from the starting point, we need to know the coordinates of the final and initial position of the plane, so we can find the total displacement, as the difference between the final and initial position.

If we choose to put our origin at the initial point of trajectory, we have that (x₀, y₀) = (0, 0)

In order to find the position of the plane after finishing the flight, we need to find its final coordinates (x₁, y₁).

In order to get x₁, we need to add the x-coordinate after flying 45º north of east, and the Δx after  completing the flight in a direction 25º of east, that we can find applying trigonometry, as follows:

x₁ = 37.0 km * cos 45º + 28.0 km* cos 25º = 51.6 Km

Appying the same considerations for the y-coordinate, we have:

y₁ = 37.0 km * sin 45º + 28.0 km* sin 25º = 38.0 km

Now, as the initial position coincides with the origin, the distance in a straight line from this point to the origin, is just the hypotenuse of the triangle determined by the coordinates (x₁, y₁) and (0,0), as follows:

$d = \sqrt{x1^{2}+y1^{2}} =\sqrt{(51.6km)^{2}+(38km)^{2}} =64.1 km$

The geographic direction of the displacement vector (which coincides in magnitude with the distance we have just found), is just the angle that this distance forms with the east axis, that we can find getting the tangent of this angle as follows:

tg θ =$\frac{y1}{x1} = \frac{38km}{56.1km} =0.736$

⇒ θ = 36.4º North of East (counterclockwise from the east axis).