JEE PYQ: Vector Algebra - Question ID 24f2d8b8395f (JEE Main 2026)

In kinematics, acceleration is the time derivative of velocity. When the velocity vector $\vec{v}$ is expressed as a function of position coordinates $(x, y, z)$, the acceleration $\vec{a}$ at any point is obtained by computing the material derivative of $\vec{v}$: $\vec{a} = \frac{d\vec{v}}{dt} = \frac{\partial \vec{v}}{\partial x}\frac{dx}{dt} + \frac{\partial \vec{v}}{\partial y}\frac{dy}{dt} + \frac{\partial \vec{v}}{\partial z}\frac{dz}{dt}.$ Here $\frac{dx}{dt} = v_x$, $\frac{dy}{dt} = v_y$, and $\frac{dz}{dt} = v_z$ are the components of the velocity itself. Thus the acceleration vector becomes $\vec{a} = \bigl(v_x \partial_x + v_y \partial_y + v_z \partial_z\bigr)\vec{v}.$ Once $\vec{a}$ is found, its magnitude is simply $|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}.$

1. Write the given velocity field:
$\displaystyle \vec{v} = -x\,\hat{i} + 2y\,\hat{j} - z\,\hat{k}\quad\text{(in m/s).}$

2. Compute the partial derivatives of each component:
$$\begin{aligned} \frac{\partial v_x}{\partial x} &= \frac{\partial (-x)}{\partial x} = -1,\quad \frac{\partial v_x}{\partial y} = 0,\quad \frac{\partial v_x}{\partial z} = 0,\\[6pt] \frac{\partial v_y}{\partial x} &= 0,\quad \frac{\partial v_y}{\partial y} = \frac{\partial (2y)}{\partial y} = 2,\quad \frac{\partial v_y}{\partial z} = 0,\\[6pt] \frac{\partial v_z}{\partial x} &= 0,\quad \frac{\partial v_z}{\partial y} = 0,\quad \frac{\partial v_z}{\partial z} = \frac{\partial (-z)}{\partial z} = -1. \end{aligned}$$

3. Evaluate the velocity components at the point $(1,2,4)$:
$$\begin{aligned} v_x &= -x = -1,\\ v_y &= 2y = 4,\\ v_z &= -z = -4. \end{aligned}$$

4. Assemble the acceleration vector $\vec{a}$:
$$\begin{aligned} a_x &= v_x\,\frac{\partial v_x}{\partial x} + v_y\,\frac{\partial v_x}{\partial y} + v_z\,\frac{\partial v_x}{\partial z} = (-1)(-1) + 4\cdot0 + (-4)\cdot0 = 1,\\[6pt] a_y &= v_x\,\frac{\partial v_y}{\partial x} + v_y\,\frac{\partial v_y}{\partial y} + v_z\,\frac{\partial v_y}{\partial z} = (-1)\cdot0 + 4\cdot2 + (-4)\cdot0 = 8,\\[6pt] a_z &= v_x\,\frac{\partial v_z}{\partial x} + v_y\,\frac{\partial v_z}{\partial y} + v_z\,\frac{\partial v_z}{\partial z} = (-1)\cdot0 + 4\cdot0 + (-4)(-1) = 4. \end{aligned}$$ Hence $\vec{a} = 1\,\hat{i} + 8\,\hat{j} + 4\,\hat{k}\quad\text{(in m/s²).}$

5. Compute the magnitude of $\vec{a}$:
$|\vec{a}| = \sqrt{1^2 + 8^2 + 4^2} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9.$

6. Match with the given options:
The magnitude $9$ m/s² corresponds to option B.

• Many students forget that acceleration is the material derivative of velocity and mistakenly compute only the partial derivatives $\partial \vec{v}/\partial t$ (which is zero here). • A frequent arithmetic error is squaring the components incorrectly, e.g. $8^2 = 64$ is sometimes miscalculated as $16$ or $32$. • Always double-check the signs of each velocity component at the given point before substituting into the acceleration formula.