Question

A person moves thirty meter north, then twenty meter east and then $30\sqrt 2$ meter south-west. Evaluate its displacement from the original position.

Answer 1

Hint: Displacement is a vector quantity and it is the shortest distance between the initial and final position of an object during motion.

Complete step-by-step answer:
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Given that,
Person moves thirty meter north i.e. $AB = 30$ meter
And twenty meter east i.e. $BC = 20$ meter
Then $30\sqrt 2$meter south west i.e. $CD = 30\sqrt 2$ meter
Since displacement is the distance between initial and final position of object during motion
Therefore displacement is, $AD = CD - CA$
Here $CD = 30\sqrt 2$ meter given and in order to calculate $CA$ apply Pythagoras theorem in right angle triangle $ABC$
$\therefore C{A^2} = A{B^2} + B{C^2} \\\ C{A^2} = {30^2} + {20^2} \\\ C{A^2} = 900 + 400 \\\ C{A^2} = 1300 \\\ CA = 36.05 \\\$
As, $AD = CD - CA$
Putting the values we get:
$AD = 30\sqrt 2 - 36.05 \\\ AD = 42.42 - 36.05 \\\ AD = 6.37 \\\$
Hence the displacement from the original position is $6.37$ meter.

Note: Here, from the figure displacement is represented by $AD$ which is equal to $CD - CA$, the value of $CD$ is $30\sqrt 2$meter which is given and in order to calculate $CA$ we applied the Pythagoras theorem in right angle triangle $ABC$and calculated the value of $CA$ as $36.05$ meter, hence the value of $AD$is calculated as $6.37$ meter which represents the displacement.