Q.28. The velocity of a particle is given as . The magnitude of acceleration at point is ____ .

Q: What is the correct answer for JEE Main Physics Q28?

The correct answer is B. Sol. Given velocity $\vec{v}=-x\hat{i}+2y\hat{j}-z\hat{k}$. Acceleration $\vec{a} = \frac{d\vec{v}}{dt} = \frac{dx}{dt} \frac{\partial \vec{v}}{\partial x} + \frac{dy}{dt} \frac{\partial \vec{v}}{\partial y} + \frac{dz}{dt} \frac{\partial \vec{v}}{\partial z}$ We know $\frac{dx}{dt} = v_x = -x$, $\frac{dy}{dt} = v_y = 2y$, $\frac{dz}{dt} = v_z = -z$. $\frac{\partial \vec{v}}{\partial x} = -\hat{i

Q: Q.28. The velocity of a particle is given as . The magnitude of acceleration at point is ____ .

The correct answer is B. Sol. Given velocity $\vec{v}=-x\hat{i}+2y\hat{j}-z\hat{k}$. Acceleration $\vec{a} = \frac{d\vec{v}}{dt} = \frac{dx}{dt} \frac{\partial \vec{v}}{\partial x} + \frac{dy}{dt} \frac{\partial \vec{v}}{\partial y} + \frac{dz}{dt} \frac{\partial \vec{v}}{\partial z}$ We know $\frac{dx}{dt} = v_x = -x$, $\frac{dy}{dt} = v_y = 2y$, $\frac{dz}{dt} = v_z = -z$. $\frac{\partial \vec{v}}{\partial x} = -\hat{i

Q: What are the options for this JEE Main Physics question?

A. $\sqrt{6}$ · B. 9 · C. $\sqrt{33}$ · D. 0

Question

Q.28. The velocity of a particle is given as . The magnitude of acceleration at point  is ____ .

FirstInTest · Accepted Answer

Sol. Given velocity $\vec{v}=-x\hat{i}+2y\hat{j}-z\hat{k}$.
Acceleration $\vec{a} = \frac{d\vec{v}}{dt} = \frac{dx}{dt} \frac{\partial \vec{v}}{\partial x} + \frac{dy}{dt} \frac{\partial \vec{v}}{\partial y} + \frac{dz}{dt} \frac{\partial \vec{v}}{\partial z}$
We know $\frac{dx}{dt} = v_x = -x$, $\frac{dy}{dt} = v_y = 2y$, $\frac{dz}{dt} = v_z = -z$.

$\frac{\partial \vec{v}}{\partial x} = -\hat{i}$
$\frac{\partial \vec{v}}{\partial y} = 2\hat{j}$
$\frac{\partial \vec{v}}{\partial z} = -\hat{k}$

So, $\vec{a} = (-x)(-\hat{i}) + (2y)(2\hat{j}) + (-z)(-\hat{k})$
$\vec{a} = x\hat{i} + 4y\hat{j} + z\hat{k}$

At point $(1, 2, 4)$, we have $x=1, y=2, z=4$.
$\vec{a} = (1)\hat{i} + 4(2)\hat{j} + (4)\hat{k} = \hat{i} + 8\hat{j} + 4\hat{k}$

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

$\sqrt{6}$

Answer

9

Answer

$\sqrt{33}$

Answer

0

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