Question

A jet flies 700km due north, then 300km west, after which it moves 400km due north and 800km east before it arrived at its 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:

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