Question:
If the velocity (in \( \text{m s}^{-1} \)) of a particle at any instant \( t \) is given by \( 2.0\hat{i} + 3.0t\hat{j} \), then the magnitude of its acceleration (in \( \text{m s}^{-2} \)) is
If the velocity (in \( \text{m s}^{-1} \)) of a particle at any instant \( t \) is given by \( 2.0\hat{i} + 3.0t\hat{j} \), then the magnitude of its acceleration (in \( \text{m s}^{-2} \)) is
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Constant components vanish on differentiation.
Updated On: Apr 21, 2026
- \(5 \)
- \(3 \)
- \(2 \)
- \(4 \)
- \(6 \)
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The Correct Option is B
Solution and Explanation
Concept:
Acceleration = derivative of velocity.
Step 1: Differentiate.
\[ \vec{a} = \frac{d}{dt}(2\hat{i} + 3t\hat{j}) = 0\hat{i} + 3\hat{j} \]
Step 2: Magnitude.
\[ |\vec{a}| = \sqrt{0^2 + 3^2} = 3 \]
Step 1: Differentiate.
\[ \vec{a} = \frac{d}{dt}(2\hat{i} + 3t\hat{j}) = 0\hat{i} + 3\hat{j} \]
Step 2: Magnitude.
\[ |\vec{a}| = \sqrt{0^2 + 3^2} = 3 \]
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