The velocity of a particle is given as v = -xi + 2yj - zk m/s. The magnitude of acceleration at point (1, 2, 4) is ________ m/s2. (a) √(6) (b) 9 (c) √(33) (d) 0

Answer: (b). Velocity depends on position, so ax = vx(∂ vx)/(∂ x), and similarly for y and z. ax = (-x)(-1) = x, ay = (2y)(2) = 4y, az = (-z)(-1) = z At (1, 2, 4): a = i + 8j + 4k |a| = √(1 + 64 + 16) = 9 m/s² So the answer is (b) 9.

The velocity of a particle is given as v = -xi + 2yj - zk m/s. The… | JEE Main 2026 (2 Apr, Shift 1)

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Vector Basics, Unit Vectors & Components

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/s2.

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Answer: (b)

Velocity depends on position, so $a_x = v_x\frac{\partial v_x}{\partial x}$ , and similarly for $y$ and $z$ .
$a_x = (-x)(-1) = x$ , $a_y = (2y)(2) = 4y$ , $a_z = (-z)(-1) = z$
At $(1, 2, 4)$ : $\vec{a} = \hat{i} + 8\hat{j} + 4\hat{k}$
$|\vec{a}| = \sqrt{1 + 64 + 16} = 9\ \text{m/s}^2$
So the answer is (b) 9.

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