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JEE Main 20262 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
2 April
Shift
Morning Shift
Subject
Physics
Chapter
Motion in One Dimension

JEE Main 2026 · 2 April · Morning Shift · Physics · Motion in One Dimension

The velocity of a particle is given as v = -x i + 2y j - z k m/s....

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

Options

  1. A6\sqrt{6}
  2. B99
  3. C33\sqrt{33}
  4. D00

Correct answer

B. 99

Step-by-step solution

The velocity vector is given by v⃗=vxi^+vyj^+vzk^=−xi^+2yj^−zk^\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:

ax=vx∂vx∂x+vy∂vx∂y+vz∂vx∂z=(−x)(−1)+0+0=xa_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

ay=vx∂vy∂x+vy∂vy∂y+vz∂vy∂z=0+(2y)(2)+0=4ya_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

az=vx∂vz∂x+vy∂vz∂y+vz∂vz∂z=0+0+(−z)(−1)=za_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 a⃗=xi^+4yj^+zk^\vec{a} = x \hat{i} + 4y \hat{j} + z \hat{k}.

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

a⃗=1i^+4(2)j^+4k^=i^+8j^+4k^\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:

∣a⃗∣=12+82+42=1+64+16=81=9|\vec{a}| = \sqrt{1^2 + 8^2 + 4^2} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9 m/s2^2

Answer: 99
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