The velocity of a particle is given as v = -x i + 2y j - z k m/s. The magnitude of acceleration at point (1, 2, 4) is _______ m/s^2.

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q28 – JEE-MAIN 2026

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Home Physics Motion in a Straight LineQuestion #318

The velocity of a particle is given as

\vec{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}

Clue 1 of 3: Understanding Acceleration

Recall that acceleration is the time derivative of velocity, $\vec{a} = \frac{d \vec{v}}{d t}$ .

🧠 Full Solution Path

● Video Walkthrough

Step 1: Determine the components of velocity✦ Active

The given velocity vector is $\vec{v} = - x \hat{i} + 2 y \hat{j} - z \hat{k}$ . From this, we can identify the components of velocity:

v_{x} = - x

v_{y} = 2 y

v_{z} = - z

Step 2: Calculate the components of acceleration○ Expand

Acceleration is the time derivative of velocity, $\vec{a} = \frac{d \vec{v}}{d t}$ . We apply the chain rule for each component, noting that $\frac{d x}{d t} = v_{x}$ , $\frac{d y}{d t} = v_{y}$ , and $\frac{d z}{d t} = v_{z}$ :

a_{x} = \frac{d v_{x}}{d t} = \frac{d (- x)}{d t} = - \frac{d x}{d t} = - v_{x} = - (- x) = x

a_{y} = \frac{d v_{y}}{d t} = \frac{d (2 y)}{d t} = 2 \frac{d y}{d t} = 2 v_{y} = 2 (2 y) = 4 y

a_{z} = \frac{d v_{z}}{d t} = \frac{d (- z)}{d t} = - \frac{d z}{d t} = - v_{z} = - (- z) = z

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

💡 Teacher's Secret Hint

Remember to use the chain rule correctly when differentiating position-dependent velocity components with respect to time.

Step 3: Evaluate acceleration at the given point and find its magnitude○ Expand

We need to find the acceleration at point $(1, 2, 4)$ . Substitute $x = 1$ , $y = 2$ , and $z = 4$ into the acceleration vector:

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

The magnitude of the acceleration vector is given by $| \vec{a} | = \sqrt{a_{x}^{2} + a_{y}^{2} + a_{z}^{2}}$ :

| \vec{a} | = \sqrt{(1)^{2} + (8)^{2} + (4)^{2}} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9

The magnitude of acceleration at point $(1, 2, 4)$ is $9$ m/s $^{2}$ .

💡 Teacher's Secret Hint

Ensure correct substitution of coordinates and calculation of the magnitude.

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Physics Question 28 – JEE-MAIN 2026

Clue 1 of 3: Understanding Acceleration

🧠 Full Solution Path