Answer to Question #171466 in Classical Mechanics for Jethro Torio

Question

A plane flies 20 km in a direction 60° north of east, then 30 km straight east, then 10 km straight north. How far and in what direction is the plane from the starting point? Use i and j unit vectors.

Answer 1

Let's find $x$- and $y$-components of the resultant displacement of the plane:

$$d_x=20\ km\cdot cos60^{\circ}+30\ km\cdot cos0^{\circ}+10\ km\cdot cos90^{\circ}=40\ km,$$ $$d_y=20\ km\cdot sin60^{\circ}+30\ km\cdot sin0^{\circ}+10\ km\cdot sin90^{\circ}=27.32\ km.$$

We can write the resultant displacement of the plane in unit vector notation as follows:

$$d=(40\ km)i+(27.32\ km)j.$$

We can find the magnitude of the resultant displacement from the Pythagorean theorem:

$$d=\sqrt{d_x^2+d_y^2}=\sqrt{(40\ km)^2+(27.32\ km)^2}=48.44\ km.$$

We can find the direction of the plane from the starting point from the geometry:

$$\theta=cos^{-1}(\dfrac{d_x}{d}),$$ $$\theta=cos^{-1}(\dfrac{40\ km}{48.44\ km})=34.3^{\circ}\ N\ of\ E.$$