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?

Answer 1

Step 1: Given data

First, 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 used

Displacement is the change in positions and can be written as $\overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{AB}+\overrightarrow{BC}$

Step 3: Solution

First, the person moves toward $30m$north.

$\overrightarrow{OA}=0\hat{\mathbb{i}}+30\hat{\mathbb{j}}+0\hat{\mathbb{k}}$

Now, the person moves toward $20m$east.

$\overrightarrow{AB}=20\hat{\mathbb{i}}+0\hat{\mathbb{j}}+0\hat{\mathbb{k}}$

Now, the person moving toward $30\sqrt{2}m$the southwest

$$\begin{gathered} \overrightarrow{BC}=-30\sqrt{2}\cos 45°\hat{\mathbb{i}}+-30\sqrt{2}\sin 45°\hat{\mathbb{j}}+0\hat{\mathbb{k}} \\ \overrightarrow{BC}=-30\hat{\mathbb{i}}-30\hat{\mathbb{j}}+0\hat{\mathbb{k}} \end{gathered}$$

[Here, the horizontal component of $30\sqrt{2}m$southwest $=-30\sqrt{2}\cos 45°$, the vertical component of $30\sqrt{2}m$southwest $=-30\sqrt{2}\sin 45°$ ]

Now, displacement $\overrightarrow{OC}$ is

$\overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{AB}+\overrightarrow{BC}=-10\hat{\mathbb{i}}$

[Here, displacement is found by vector sum because it is a vector quantity]

Hence, the displacement is of magnitude $10m$ in the west direction.