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

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Kinematics

The velocity of a particle is given as

v = - x \hat{i} + 2 y \hat{j} - z \hat{k} m / s

. The magnitude of acceleration at point

(1, 2, 4)

is ____

m / s^{2}

(A) $6$
(B)9✓ Correct
(C) $33$
(D)0

Answer: (B)

Solution

Each velocity component depends only on its own coordinate, so by the chain rule

a_{x} = \frac{d}{d t} (- x) = - \overset{x}{˙} = - v_{x} = x

a_{y} = \frac{d}{d t} (2 y) = 2 \overset{y}{˙} = 2 v_{y} = 4 y

, and

a_{z} = \frac{d}{d t} (- z) = - \overset{z}{˙} = - v_{z} = z

. Hence

a = x \hat{i} + 4 y \hat{j} + z \hat{k}

. At the point

(1, 2, 4)

a = \hat{i} + 8 \hat{j} + 4 \hat{k}

, whose magnitude is

1^{2} + 8^{2} + 4^{2} = 81 = 9 m/ s^{2}

. The magnitude of the acceleration is

9 m/ s^{2}

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Single correct option

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