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

1 Answer
Jun 21, 2015

Answer:

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 #AC# 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 #AD#, will be

#(AD)^2 = underbrace((AC)^2)_(color(blue)("horiz displacement")) + overbrace((CD)^2)^(color(green)("vert displacement"))#

#(AD)^2 = 100""^2 + 1200""^2#

#AD = sqrt(100""^2 + 1200""^2) ~= color(green)("1204 m")#