First slide

Average Velocity

Question

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 = 5s
Average velocity vector $({\vec{v}}_{av})$ from t = 0 to t = 5 s is

Easy

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

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