Question

A particle is moving such that its position coordinates (x, y) are (2 m, 3 m) at time t = 0, (6 m, 7 m) at time t = 2 s and (13 m, 14 m) at time t = 5 s. Average velocity vector $ ( \vec{ v_{av}})$ from t = 0 to t = 5 s is .

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

Answer 1

The Correct Option is D

At time t = 0, the position vector of the particle is
$ \vec{r_1} = 2 \widehat{i} + 3 \widehat{j}$
At time t = 5 s, the position vector of the particle is
$ \vec{r_2} = 13 \widehat{i} + 14 \widehat{j}$
Displacement from $ \vec{r_1} $ to $\vec{r_2} $ is
$\triangle \vec{r} = \vec{r_2} - \vec{r_1} = (13 \widehat{i} + 14 \widehat{j}) - (2 \widehat{i} + 3 \widehat{j})$
$ =11 \widehat{i} + 11 \widehat{j}$
$\therefore$ Average velocity,
$\vec{v_{ av}} = \frac{ \triangle \vec{r}}{ \triangle t} = \frac{ 11 \widehat{i} + 11 \widehat{j}}{ 5 - 0 } = \frac{ 11}{ 5} ( \widehat{i} + \widehat{j})$