Question

A man walks $30m$ towards north, then $20m$, towards east and in the last $30\sqrt{2}m$ towards south west. The displacement from origin is

• A. $10m$ towards west
• B. $10m$ towards east
• C. $60\sqrt{2}m$, towards north west
• D. $60\sqrt{2}m$, towards east north

Answer 1

The correct option is A $10m$ towards west

Displacement is the shortest distance between the initial and final position of the particle.

From the above figure,

$\overrightarrow{AB}=30m$

$\overrightarrow{BC}=20m$

$\overrightarrow{CD}=30\sqrt{2}m$

$\Delta \left(OCD\right)$ is right angle triangle,

By Pythagoras theorem,

$(DC{)}^2=(OC{)}^2+(DO{)}^2$

$(30\sqrt{2}{)}^2=(OC{)}^2+(30{)}^2$

$OC=30m$. . . . . .(1)

$OD=DA+OA=30m$

$DA=30m-20m=10m$

$DA=10m$

The displacement from the origin is $10m$ towards the west.

The correct option is A.