The velocity of a particle is given as vec v = - x hat i +2 y hat j - z hat k m / s The magnitude of acceleration at point (1, 2, 4) is m/s2.

Physics · Motion in one Dimension

JEE Main 2026 — 2 April, Morning Shift — Question 2

The velocity of a particle is given as $\vec{\mathrm{v}} = -\mathrm{x}\hat{\mathrm{i}} +2\mathrm{y}\hat{\mathrm{j}} -\mathrm{z}\hat{\mathrm{k}}\mathrm{m} / \mathrm{s}$ The magnitude of acceleration at point (1, 2, 4) is m/s2.

Option A:
$\sqrt{6}$
Option B:
9
Correct
Option C:
$\sqrt{33}$
Option D:
0

Answer: B

Step-by-step solution

$\vec{\mathrm{a}} = \frac{\mathrm{d}\vec{\mathrm{v}}}{\mathrm{d}t} = -\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}t}\hat{\mathrm{i}} +2\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}t}\hat{\mathrm{j}} -\frac{\mathrm{d}\mathrm{z}}{\mathrm{d}t}\hat{\mathrm{k}} = \mathrm{x}\hat{\mathrm{i}} +4\mathrm{y}\hat{\mathrm{j}} +\mathrm{z}\hat{\mathrm{k}}$ $\therefore \vec{\mathrm{a}}\mathrm{at}(1,2,4) = \hat{\mathrm{i}} +8\hat{\mathrm{j}} +4\hat{\mathrm{k}}$ $\therefore |\vec{\mathrm{a}} | = 9\mathrm{m} / \mathrm{s}^{2}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 2 April, Morning Shift
Subject: Physics
Chapter: Motion in one Dimension
Topic: Uniformly Accelerated Motion