# An aeroplane flies 400 m due north and then 300 m due south and then flies 1200 m upwards, the net displacement is ?

Jun 21, 2015

The net displacement is 1204 m.

#### Explanation:

The net displacement of the plane will be the shortest distance between its starting point and its finish point.

Let's assume that your plane starts in point $A$. The first stage of its flight will take it 400 m north to point $B$. The second stage of its flight will take it 300 m south to point $C$.

Since north and south are opposite directions, the plane will end up very close to where it started, i.e. point $A$. More specifically, point $A C$ will be

underbrace(AC)_(color(blue)("horizontal displacement")) = AB - BC = 400 - 300 = "100 m"

So, horizontally, the plane only travelled 100 m from its starting point. At this point, its flight path turns upwards.

The plane flies 1200 m upwards to point $D$, which is its finish point. The distance between point $A$ and point $D$ can be seen as the hypothenuse of a right triangle with sides 100 and 1200.

This means that the net displacement, which is equal to $A D$, will be

${\left(A D\right)}^{2} = {\underbrace{{\left(A C\right)}^{2}}}_{\textcolor{b l u e}{\text{horiz displacement")) + overbrace((CD)^2)^(color(green)("vert displacement}}}$

${\left(A D\right)}^{2} = 100 {\text{^2 + 1200}}^{2}$

$A D = \sqrt{100 \text{^2 + 1200""^2) ~= color(green)("1204 m}}$