The velocity of a particle is given as vec{v} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The acceleration vector is given by the material derivative of velocity: $\vec{a} = \frac{d\vec{v}}{dt} = (\vec{v} \cdot 
abla) \vec{v}$.Given $\

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) The acceleration vector is given by the material derivative of velocity: $\vec{a} = \frac{d\vec{v}}{dt} = (\vec{v} \cdot 
abla) \vec{v}$.Given $\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k}$, we calculate the components:$a_x = v_x \frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} + v_z \frac{\partial v_x}{\partial z} = (-x)(-1) + (2y)(0) + (-z)(0) = x$.$a_y = v_x \frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} + v_z \frac{\partial v_y}{\partial z} = (-x)(0) + (2y)(2) + (-z)(0) = 4y$.$a_z = v_x \frac{\partial v_z}{\partial x} + v_y \frac{\partial v_z}{\partial y} + v_z \frac{\partial v_z}{\partial z} = (-x)(0) + (2y)(0) + (-z)(-1) = z$.Thus, $\vec{a} = x\hat{i} + 4y\hat{j} + z\hat{k}$.At point $(1, 2, 4)$, $\vec{a} = 1\hat{i} + 4(2)\hat{j} + 4\hat{k} = 1\hat{i} + 8\hat{j} + 4\hat{k}$.The magnitude is $|\vec{a}| = \sqrt{1^2 + 8^2 + 4^2} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9 	ext{ m/s}^2$.

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

Similar Questions

If velocity and acceleration are in opposite directions in one-dimensional motion,what happens to the magnitude of velocity?

The velocity of a particle is given by the equation $v(x) = 3x^2 - 4x$,where $x$ is the distance covered by the particle. The expression for its acceleration is

$A$ body is moving with variable acceleration $a$ along a straight line. The average acceleration of the body in the time interval $t_1$ to $t_2$ is

$A$ particle moves along a straight line such that its displacement at any time $t$ is given by $s = (t^3 - 3t^2 + 2) \, m$. The displacement when the acceleration becomes zero is ........ $m$.

In what different ways can velocity be changed?

Vedclass Test Series

Exam Paper Generator

Online Exam Module