Question

VECTORS:

A ship was seen on a radar screen to be 10 km east of a particular location. Sometime later, the ship is at 12 km north of west. What is the displacement of the ship?

Answer 1

Let's first find $x$- and $y$-components of the displacement of the ship:

$$d_x=10\ km+12\ km\cdot cos(90^{\circ}+45^{\circ})=1.51\ km,$$ $$d_y=12\ km\cdot sin(90^{\circ}+45^{\circ})=8.48\ km.$$

We can find the displacement of the ship from the Pythagorean theorem:

$$d=\sqrt{d_x^2+d_y^2}=\sqrt{(1.51\ km)^2+(8.48\ km)^2}=8.6\ km.$$

We can find the direction from the geometry:

$$\theta=cos^{-1}(\dfrac{d_x}{d}),$$ $$\theta=cos^{-1}(\dfrac{1.51\ km}{8.6\ km})=79.8^{\circ}\ N\ of\ E.$$