A particle is moving such that its position coordinates (x,y) are (2m, 3m) at time t = 2s and (13 m, 14m) at time t = 5sAverage velocity vector v→av from t = 0 to t = 5 s is

A
$\frac{1}{5} (13 \hat{i} + 14 \hat{j})$
B
$\frac{7}{3} (\hat{i} + \hat{j})$
C
$2 (\hat{i} + \hat{j})$
D
$\frac{11}{5} (\hat{i} + \hat{j})$

Solution:

At time t = 0, the position vector of the particle is ${\vec{r}}_{1} = 2 \hat{i} + 3 \hat{j}$

At time t = 5s, the position vector of the particle is ${\vec{r}}_{2} = 13 \hat{i} + 14 \hat{j}$

Displacement from ${\vec{r}}_{1} to {\vec{r}}_{2}$ is $Δ \vec{r} = {\vec{r}}_{2} - {\vec{r}}_{1} = (13 \hat{i} + 14 \hat{j}) - (2 \hat{i} + 3 \hat{j}) = 113 \hat{i} + 11 \hat{j}$

$∴$ Average velocity,

${\vec{v}}_{av} = \frac{Δ \vec{r}}{Δ t} = \frac{11 \hat{i} + 11 \hat{j}}{5 - 0} = \frac{11}{5} (\hat{i} + \hat{j})$

A particle is moving such that its position coordinates (x,y) are (2m, 3m) at time t = 2s and (13 m, 14m) at time t = 5sAverage velocity vector $({\vec{v}}_{av})$ from t = 0 to t = 5 s is

Solution:

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