Question

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

Answer 1

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.