Physics · 2026
JEE Main · 2 April 2026, Shift 1 · Q28
The velocity of a particle is given as v =-x i +2 y j -z k m / s. The magnitude of acceleration at point (1, 2, 4) is ____ m / s^2.
The velocity of a particle is given as $\displaystyle \vec{v}=-x \hat{i}+2 y \hat{j}-z \hat{k} \mathrm{~m} / \mathrm{s}$. The magnitude of acceleration at point ($\displaystyle 1$, $\displaystyle 2$, $\displaystyle 4$) is $\displaystyle \_\_\_\_$ $\displaystyle \mathrm{m} / \mathrm{s}^2$.
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle 9$
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JEE Main 2026 Physics question, with the answer from NTA’s final answer key. Where our answers come from.