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 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 \( ms^{-1} \)):
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When solving vector kinematics, always decompose the vectors into their Cartesian components and operate on them separately.
Updated On: Jun 9, 2026
- \( 6\sqrt{2} \)
- \( 6 \)
- \( 7\sqrt{2} \)
- \( 7 \)
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The Correct Option is A
Solution and Explanation
Concept:
We utilize the kinematic equation for constant acceleration in vector form: \(\vec{v} = \vec{u} + \vec{a}t\), where \(\vec{u}\) is the initial velocity, \(\vec{a}\) is acceleration, and \(t\) is time.
Step 1: List the given vectors and time.
$$ \vec{u} = 2\hat{i} + 3\hat{j} $$ $$ \vec{a} = 0.8\hat{i} + 0.6\hat{j} $$ $$ t = 5 \text{ s} $$
Step 2: Calculate final velocity components.
$$ \vec{v} = (2\hat{i} + 3\hat{j}) + (0.8\hat{i} + 0.6\hat{j}) \times 5 $$ $$ \vec{v} = 2\hat{i} + 3\hat{j} + 4\hat{i} + 3\hat{j} $$ $$ \vec{v} = (2+4)\hat{i} + (3+3)\hat{j} = 6\hat{i} + 6\hat{j} $$
Step 3: Determine the magnitude of the velocity vector.
The magnitude is given by \( |\vec{v}| = \sqrt{v_x^2 + v_y^2} \): $$ |\vec{v}| = \sqrt{6^2 + 6^2} $$ $$ |\vec{v}| = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} $$ $$\boxed{6\sqrt{2}}$$
Step 1: List the given vectors and time.
$$ \vec{u} = 2\hat{i} + 3\hat{j} $$ $$ \vec{a} = 0.8\hat{i} + 0.6\hat{j} $$ $$ t = 5 \text{ s} $$
Step 2: Calculate final velocity components.
$$ \vec{v} = (2\hat{i} + 3\hat{j}) + (0.8\hat{i} + 0.6\hat{j}) \times 5 $$ $$ \vec{v} = 2\hat{i} + 3\hat{j} + 4\hat{i} + 3\hat{j} $$ $$ \vec{v} = (2+4)\hat{i} + (3+3)\hat{j} = 6\hat{i} + 6\hat{j} $$
Step 3: Determine the magnitude of the velocity vector.
The magnitude is given by \( |\vec{v}| = \sqrt{v_x^2 + v_y^2} \): $$ |\vec{v}| = \sqrt{6^2 + 6^2} $$ $$ |\vec{v}| = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} $$ $$\boxed{6\sqrt{2}}$$
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