Q53 / 75 · JEE Main 2026 (02 April Shift 1) · Physics

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
    $\sqrt{6}$
  • Correct answer
    B
    $9$
  • C
    $\sqrt{33}$
  • D
    $0$

Solution

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:

$a_x = v_x \dfrac{\partial v_x}{\partial x} + v_y \dfrac{\partial v_x}{\partial y} + v_z \dfrac{\partial v_x}{\partial z} = (-x)(-1) + 0 + 0 = x$

$a_y = v_x \dfrac{\partial v_y}{\partial x} + v_y \dfrac{\partial v_y}{\partial y} + v_z \dfrac{\partial v_y}{\partial z} = 0 + (2y)(2) + 0 = 4y$

$a_z = v_x \dfrac{\partial v_z}{\partial x} + v_y \dfrac{\partial v_z}{\partial y} + v_z \dfrac{\partial v_z}{\partial z} = 0 + 0 + (-z)(-1) = z$

The acceleration vector is $\vec{a} = x \hat{i} + 4y \hat{j} + z \hat{k}$.

At the point $(1, 2, 4)$, substituting $x = 1$, $y = 2$, and $z = 4$:

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

The magnitude of acceleration is:

$|\vec{a}| = \sqrt{1^2 + 8^2 + 4^2} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9$ m/s$^2$

Answer: $9$

Hence, the correct answer is option: 'B'.

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