Question

the position co-ordinate of particle moving in x - y plane changes from (1m,1m) to (3m, -5m ) its displacement vector is represented as options : 1. (i^ + j^)m 2. (i^ + 3j^)m 3.( 3i^ - 5j^)m. 4. (2i^ - 6j^) ​

Answer 1

Given:

The position co-ordinate of particle moving in x - y plane changes from (1m,1m) to (3m, -5m ).

To find:

Displacement vector .

Calculation:

The initial position of object is (1,1) , hence its position vector w.r.t origin :

$\therefore \: \vec{r1} = \hat{i} + \hat{j}$

The final position of object is (3,-5) , hence its position vector w.r.t origin :

$\therefore \: \vec{r2} = 3 \hat{i} + 5( - \hat{j} )$

$= > \: \vec{r2} = 3 \hat{i} - 5 \hat{j}$

Now , displacement vector is defined as a change in position vector of an object :

$\therefore \: \vec{d} = \vec{r2} - \vec{r1}$

$= > \: \vec{d} = ( 3 \hat{i} - 5 \hat{j}) - ( \hat{i} + \hat{j})$

$= > \: \vec{d} = 3 \hat{i} - 5 \hat{j}- \hat{i} - \hat{j}$

$= > \: \vec{d} = 2 \hat{i} - 6 \hat{j}$

So, final answer is:

$\boxed{ \large{ \red{ \bold{\: \vec{d} = 2 \hat{i} - 6 \hat{j}}}}}$