Question

A car is moving with a velocity of $4~\text{m/s}$ towards east. After a time of $4~\text{s}$, it is heading north-east with a velocity of $4\sqrt{2}~\text{m/s}$. Then the average velocity of the car is

• A. $2\sqrt{5}~\text{m/s}$
• B. $3\sqrt{5}~\text{m/s}$
• C. $4\sqrt{3}~\text{m/s}$
• D. $5\sqrt{3}~\text{m/s}$

Hint:

Use vector addition and displacement over time to calculate average velocity when direction changes.

Answer 1

The Correct Option is A

Initial velocity vector: $\vec{u} = 4\hat{i}$
Final velocity vector: $\vec{v} = 4\sqrt{2} \cdot \dfrac{1}{\sqrt{2}}(\hat{i} + \hat{j}) = 4\hat{i} + 4\hat{j}$
Displacement vector (assuming uniform change):
$\vec{r} = \dfrac{\vec{u} + \vec{v}}{2} \cdot t = \dfrac{(4\hat{i} + 4\hat{i} + 4\hat{j})}{2} \cdot 4 = (4\hat{i} + 2\hat{j}) \cdot 4 = 16\hat{i} + 8\hat{j}$
Average velocity $= \dfrac{16\hat{i} + 8\hat{j}}{4} = 4\hat{i} + 2\hat{j}$
Magnitude $= \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}~\text{m/s}$