Question

A jet flies 700 km due north, then 300 km west, after which it moves 400 km due north and 800 km due east before it arrived at it's destination and landed.

(a) using vector notation, draw this movement.

(b) find the resultant displacement of the jet

Answer 1

(a) Let's draw this movement:

(b) Let's first find $x$- and $y$-components of resultant displacement of jet:

$R_{x}=700\ km\times cos90^{\circ}+300\ km\times cos180^{\circ}+400\ km\times cos90^{\circ}+800\ km\times cos0^{\circ}=500\ km,$

$R_{y}=700\ km\times sin90^{\circ}+300\ km\times sin180^{\circ}+400\ km\times sin90^{\circ}+800\ km\times sin0^{\circ}=1100\ km.$

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

$$R=\sqrt{R_x^2+R_y^2}=\sqrt{(500\ km)^2+(1100\ km)^2}=1208.3\ km.$$

We can find the direction of resultant displacement from the geometry:

$$\theta=tan^{-1}(\dfrac{R_y}{R_x})=tan^{-1}(\dfrac{1100\ km}{500\ km})=65.5^{\circ}\ N\ of\ E.$$