Question

A particle is displaced by 4 m in the south west direction and then by 5m EasT and lastly through 6m in a direction 60 degree north of East calculate the total displacement from the starting point

Answer 1

Given:

A particle is displaced by 4 m in the south west direction and then by 5m EasT and lastly through 6m in a direction 60 degree north of East.

To find:

Net Displacement from starting point ?

Calculation:

We will apply vectors to solve these kind of questions:

When particle moves 4 m south:

$\therefore \: \vec{r1} = 4( - \hat{j})$

When particle moves 5 m east:

$\therefore \: \vec{r2} = 5 \hat{i}$

When particle moves 6m at 60° north of East:

$\therefore \: \vec{r3} = 6 \cos( {60}^{ \circ} ) \hat{i} + 6 \sin( {60}^{ \circ} ) \hat{j}$

$\implies\: \vec{r3} = 3\hat{i} + 3 \sqrt{3} \hat{j}$

Now , net Displacement:

$\therefore \: \vec{r} = \vec{r1} + \vec{r2} + \vec{r3}$

$\implies\: \vec{r} = 4( - \hat{j}) + 5 \hat{i} + 3 \hat{i} + 3 \sqrt{3} \hat{j}$

$\implies\: \vec{r} = 8 \hat{i} + (3 \sqrt{3} - 4) \hat{j}$

$\implies\: \vec{r} = 8 \hat{i} +(1.19)\hat{j}$

$\implies\: | \vec{r}| = \sqrt{ {8}^{2} + {(1.19)}^{2} }$

$\implies\: | \vec{r}| = \sqrt{65.41}$

$\implies\: | \vec{r}| = 8.08 \: m$

So, net Displacement is 8.08 metres.