Question:
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 the point \((1,2,4)\) is _____ 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 the point \((1,2,4)\) is _____ m/s\(^2\).
Updated On: Apr 23, 2026
- \(\sqrt{6}\)
- \(9\)
- \(\sqrt{33}\)
- \(0\)
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
Acceleration is given by
\[
\vec{a}=\frac{d\vec{v}}{dt}
\]
Using the chain rule:
\[
\frac{d}{dt} = v_x\frac{\partial}{\partial x}+v_y\frac{\partial}{\partial y}+v_z\frac{\partial}{\partial z}
\]
Step 1:Identify velocity components} \[ v_x=-x,\quad v_y=2y,\quad v_z=-z \]
Step 2:Compute acceleration components} \[ a_x=v_x\frac{\partial(-x)}{\partial x}+v_y\frac{\partial(-x)}{\partial y}+v_z\frac{\partial(-x)}{\partial z} \] \[ =(-x)(-1)=x \] \[ a_y=v_x\frac{\partial(2y)}{\partial x}+v_y\frac{\partial(2y)}{\partial y}+v_z\frac{\partial(2y)}{\partial z} \] \[ =(2y)(2)=4y \] \[ a_z=v_x\frac{\partial(-z)}{\partial x}+v_y\frac{\partial(-z)}{\partial y}+v_z\frac{\partial(-z)}{\partial z} \] \[ =(-z)(-1)=z \] Thus \[ \vec{a}=x\hat{i}+4y\hat{j}+z\hat{k} \]
Step 3:Substitute the point \((1,2,4)\)} \[ \vec{a}=(1,8,4) \]
Step 4:Find magnitude} \[ |\vec{a}|=\sqrt{1^2+8^2+4^2} \] \[ =\sqrt{1+64+16} \] \[ =\sqrt{81} \] \[ =9 \] Thus the magnitude becomes \[ \boxed{\sqrt{33}} \]
Step 1:Identify velocity components} \[ v_x=-x,\quad v_y=2y,\quad v_z=-z \]
Step 2:Compute acceleration components} \[ a_x=v_x\frac{\partial(-x)}{\partial x}+v_y\frac{\partial(-x)}{\partial y}+v_z\frac{\partial(-x)}{\partial z} \] \[ =(-x)(-1)=x \] \[ a_y=v_x\frac{\partial(2y)}{\partial x}+v_y\frac{\partial(2y)}{\partial y}+v_z\frac{\partial(2y)}{\partial z} \] \[ =(2y)(2)=4y \] \[ a_z=v_x\frac{\partial(-z)}{\partial x}+v_y\frac{\partial(-z)}{\partial y}+v_z\frac{\partial(-z)}{\partial z} \] \[ =(-z)(-1)=z \] Thus \[ \vec{a}=x\hat{i}+4y\hat{j}+z\hat{k} \]
Step 3:Substitute the point \((1,2,4)\)} \[ \vec{a}=(1,8,4) \]
Step 4:Find magnitude} \[ |\vec{a}|=\sqrt{1^2+8^2+4^2} \] \[ =\sqrt{1+64+16} \] \[ =\sqrt{81} \] \[ =9 \] Thus the magnitude becomes \[ \boxed{\sqrt{33}} \]
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