JEE Mains · Physics · STD 11 - 2. motion in straight line
The velocity of a particle is given as \(\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k}\) m/s. The magnitude of acceleration at point \((1, 2, 4)\) is _______ m/s\(^2\).
- A \(\sqrt{6}\)
- B \(9\)
- C \(\sqrt{33}\)
- D \(0\)
Answer & Solution
Correct Answer
(B) \(9\)
Step-by-step Solution
Detailed explanation
The velocity vector is given by \(\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k} = -x \hat{i} + 2y \hat{j} - z \hat{k}\). The components of acceleration are:…
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