Question

A person moves $30m$ north, then $20m$ east, then $30\sqrt{2}m$ south west. His displacement from the original position is:
A. Zero
B. 28 m towards south
C. 10 m towards west
D. 15 m towards east

Answer 1

We have to find the value of displacement of the person from his original position. The person is moving in a different direction. His displacement in north, east and south west is given. We are going to use vector methods to find the displacement of the person.

Complete step by step answer:
Given,
Displacement in north $= 30m$
Displacement in east $= 20m$
Displacement in south west = $30\sqrt 2$ m
Now, we will draw a diagram of movement of the person.

The movement along north direction is $OY = 30\widehat j$
The movement along east direction is $YX = 20\widehat i$
The movement along south west direction is $XZ = 30\sqrt 2 (\cos \theta \widehat i + \sin \theta \widehat j)$

As we can see in the diagram below that the value of $\theta$ is$45^\circ$, after putting these values we get the following.
$XZ = 30\sqrt 2 (\cos 45^\circ \widehat i + \sin 45^\circ \widehat j)$
$XZ = 30\widehat i + 30\widehat j$
Now, using triangle law, we get following;
OZ=OX-XZ
$OZ = 20\widehat i + 30\widehat j - 30\widehat i - 30\widehat j$
$OZ = - 10\widehat i$
$\therefore$ The displacement of the person and the value of displacement is $OZ = - 10\widehat i$. It means that the person moves 10 m towards the west direction
Hence, the correct answer is option (C).

Note: In such types of questions, it is important to draw the diagram of movement because it gives a clear idea in which direction the object is moving. It is important to write the vectors carefully otherwise we will get the wrong answer. The angle between the vectors should be measured carefully. In case of moving in two directions at the same time; like south west, north east etc. we should use the vector form in terms of cosine and sine.