Question

A particle moves 3m north, then 4m east and finally 6m south. Calculate the distance travelled and displacement.

Answer 1

Hint: Distance: Distance is the actual length of the path that is travelled by any object. It is the scalar quantity.
Displacement: The shortest distance between the initial point and final point. It is the vector quantity.

Complete step by step answer:
First of all we have to make a diagram according to the question.

Distance covered by the particle is AD
Here
\[ \Rightarrow AD = AB + BC + CD\]
\[ \Rightarrow AD = 3 + 4 + 6\]
\[ \Rightarrow AD = 13m\]
So Distance covered by a particle is 13m.
Now Displacement covered by the particles is AD
Here
\[ \Rightarrow ED = CD - CE\]
\[ \Rightarrow ED = 6 - 3\]
\[ \Rightarrow ED = 3m\]
Now Displacement,
$ \Rightarrow AD = \sqrt {A{E^2} + E{D^2}} $
$ \Rightarrow AD = \sqrt {{4^2} + {3^2}} $
$ \Rightarrow AD = \sqrt {16 + 9} $
$ \Rightarrow AD = \sqrt {25} $
\[ \Rightarrow AD = 5m\]
So the displacement covered by the particle is 5m.

Note: Distance is always greater than displacement. Displacement of the circular path is always zero because the starting and ending points lie on a single point.