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/s$^2$.

Question

Accepted Answer

D. 1. The **acceleration** of a particle is the time derivative of its velocity:
   $$
   \vec{a}=\frac{d\vec{v}}{dt}
   $$

2. The velocity components are:
   $$
   v_x=-𝑥,\qquad v_y=2𝑦,\qquad v_z=-𝑧
   $$

3. Differentiate each component with respect to time using the chain rule:
   $$
   a_x=\frac{dv_x}{dt}=\frac{dv_x}{d𝑥}\cdot\frac{d𝑥}{dt}=(-1)(-𝑥)=𝑥
   $$
   $$
   a_y=\frac{dv_y}{dt}=\frac{dv_y}{d𝑦}\cdot\frac{d𝑦}{dt}=2(2𝑦)=4𝑦
   $$
   $$
   a_z=\frac{dv_z}{dt}=\frac{dv_z}{d𝑧}\cdot\frac{d𝑧}{dt}=(-1)(-𝑧)=𝑧
   $$

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

5. At the point $(1,2,4)$:
   $$
   \vec{a}=\hat{i}+8\hat{j}+4\hat{k}
   $$

6. Its magnitude is:
   
  $$
\begin{aligned}
|\vec{a}|
&= \sqrt{1^2+8^2+4^2} \
&= \sqrt{1+64+16} \
&= \sqrt{81} \
&= 9\ 	ext{m/s}^2.
\end{aligned}
$$

7. Therefore, the magnitude of acceleration is **$9\ 	ext{m/s}^2$**, so the correct option is **D**.

The velocity of a particle is given as v⃗=−𝑥i^+2𝑦j^−𝑧k^\vec{v} = -𝑥 \hat{i} + 2𝑦 \hat{j} - 𝑧 \hat{k}v=−xi^+2yj^​−zk^ m/s. The magnitude of acceleration at point (1,2,4)(1, 2, 4)(1,2,4) is ________ m/s2^22.

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/s $^2$ .