Question

# A person moves $30m$ north and then $20m$ east and then $30\sqrt{2}m$southwest. His displacement from the origin position will be?

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Solution

## Step 1: Given dataFirst, the person moves in the north direction for $30m$Then, the person moves in the east direction for $20m$Then, the person moves in the southeast direction for $30\sqrt{2}m$Step 2: Formula usedDisplacement is the change in positions and can be written as $\stackrel{\to }{OC}=\stackrel{\to }{OA}+\stackrel{\to }{AB}+\stackrel{\to }{BC}$Step 3: SolutionFirst, the person moves toward $30m$north.$\stackrel{\to }{OA}=0\stackrel{^}{\mathbb{i}}+30\stackrel{^}{\mathbb{j}}+0\stackrel{^}{\mathbb{k}}$Now, the person moves toward $20m$east. $\stackrel{\to }{AB}=20\stackrel{^}{\mathbb{i}}+0\stackrel{^}{\mathbb{j}}+0\stackrel{^}{\mathbb{k}}$Now, the person moving toward $30\sqrt{2}m$the southwest$\stackrel{\to }{BC}=-30\sqrt{2}\mathrm{cos}45°\stackrel{^}{\mathbb{i}}+-30\sqrt{2}\mathrm{sin}45°\stackrel{^}{\mathbb{j}}+0\stackrel{^}{\mathbb{k}}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{BC}=-30\stackrel{^}{\mathbb{i}}-30\stackrel{^}{\mathbb{j}}+0\stackrel{^}{\mathbb{k}}$ [Here, the horizontal component of $30\sqrt{2}m$southwest $=-30\sqrt{2}\mathrm{cos}45°$, the vertical component of $30\sqrt{2}m$southwest $=-30\sqrt{2}\mathrm{sin}45°$ ]Now, displacement $\stackrel{\to }{OC}$ is$\stackrel{\to }{OC}=\stackrel{\to }{OA}+\stackrel{\to }{AB}+\stackrel{\to }{BC}=-10\stackrel{^}{\mathbb{i}}$ [Here, displacement is found by vector sum because it is a vector quantity]Hence, the displacement is of magnitude $10m$ in the west direction.

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