JEE Main 2026 — Apr 2, Shift 1 — Physics solution

JEE Main2026Apr 2, Shift 1Physics

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
$9$
C
$33$
D
$0$

Solution

Differentiate each component of $v$ with respect to time, recognising that $\overset{x}{˙} = v_{x} = - x$ , $\overset{y}{˙} = v_{y} = 2 y$ , $\overset{z}{˙} = v_{z} = - z$ :

$a = - \overset{x}{˙} \hat{i} + 2 \overset{y}{˙} \hat{j} - \overset{z}{˙} \hat{k} = x \hat{i} + 4 y \hat{j} + z \hat{k}$

At $(1, 2, 4)$ : $a = \hat{i} + 8 \hat{j} + 4 \hat{k}$

$∣ a ∣ = 1 + 64 + 16 = 81 = 9$ m/s $^{2}$

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