A body travels a distance of 3 km towards East, then 4 km towards North and finally 9 km towards East.
(i) What is the total distance travelled?
(ii) What is the resultant displacement?

Solution
(a) Given: A body travels a distance of 3 km towards East, then 4km towards North and finally 9 km towards East.
So, the total distance traveled by the body is the length of the path traversed = AB + BC + CD = (3 + 4 + 9) km = 16 km

(b) Displacement is the shortest path between source and destination.
So, in this case, it is Path DA.

If we see the diagram,
AB + BE = 3+ 9 = 12 Km
DE = 4 Km
Consider right angle triangle DEA,
By Pythagoras theorem,
$$\begin{gathered} A{D}^2 =A{E}^2 +D{E}^2 \\ AD =\sqrt{A{E}^2 +D{E}^2} =\sqrt{{12}^2+4^2} =\sqrt{144+16} =\sqrt{160} \\ AD = 12.65 Km \end{gathered}$$