Question

calculate the average velocity of the particle whose position vector changes from r1=5î+3j r2= t2î-3j in a time 5 sec

Answer 1

Average Velocity is defined as:

$v_{avg} = \frac{Displacement}{Time}$

We need to find the displacement.

$\text{Initial Position}=\vec{r_1}=5\hat{\imath}+6\hat{\jmath} \\ \\ \text{Final Position}=\vec{r_2}=2\hat{\imath}+3\hat{\jmath} \\ \\ \text{Time}=t=5 \, \, s$

Displacement is given by

$\text{Displacement = Final Position - Initial Position}$

Here, the displacement vector becomes:

$\vec{r} = \vec{r_2}-\vec{r_1} \\ \\ \implies \vec{r} = (2\hat{\imath}+3\hat{\jmath})-(5\hat{\imath}+6\hat{\jmath})\\ \\ \implies \vec{r}=-3\hat{\imath}-3\hat{\jmath} \\ \\ \implies \vec{r} = -3(\hat{\imath}+\hat{\jmath})$

The magnitude of displacement is:

$|\,\vec{r}\, |=3 \sqrt{1^2+1^2} \\ \\ \implies r = 3\sqrt{2}$

So, now average velocity simply becomes:

$v_{avg}=\frac{r}{t} \\ \\ \\ \implies \boxed{\bold{v_{avg} = \frac{3\sqrt{2}}{5} \, \, m/s}}$

Answer 2

yes that is correct answer