Question

A particle has initial velocity \( 2\hat{i} + 3\hat{j} \) \( ms^{-1} \) and acceleration \( 0.8\hat{i} + 0.6\hat{j} \) \( ms^{-2} \). Its velocity after 5 sec is (in \( ms^{-1} \)):

• A. \( 6\sqrt{2} \)
• B. \( 6 \)
• C. \( 7\sqrt{2} \)
• D. \( 7 \)

Hint:

When solving vector kinematics, always decompose the vectors into their Cartesian components and operate on them separately.

Answer 1

The Correct Option is A

Step 1: Pick the right equation.
With constant acceleration, velocity follows $\vec{v} = \vec{u} + \vec{a}\,t$, applied separately to each component.
Step 2: List what is given.
$\vec{u} = 2\hat{i} + 3\hat{j}$, $\vec{a} = 0.8\hat{i} + 0.6\hat{j}$, and $t = 5$ s.
Step 3: Find the change in velocity.
$\vec{a}\,t = (0.8\hat{i} + 0.6\hat{j})\times 5 = 4\hat{i} + 3\hat{j}$.
Step 4: Add to the initial velocity.
$\vec{v} = (2 + 4)\hat{i} + (3 + 3)\hat{j} = 6\hat{i} + 6\hat{j}$.
Step 5: Take the magnitude.
The question asks for speed, so $|\vec{v}| = \sqrt{6^2 + 6^2} = \sqrt{72}$.
Step 6: Simplify the surd.
$\sqrt{72} = \sqrt{36\times 2} = 6\sqrt{2}$.
\[ \boxed{6\sqrt{2}\ \text{ms}^{-1}} \]