Question

a slow neutron produced in a nuclear reactor travels from location <0.4, -0.01, 0.6> m to location <-0.211, 0.053, 0.093> m in 2 microseconds (1μs = 1×10⁻⁶ s). (a) what is the average velocity of the neutron? (express your answer in vector form.)

Explanation:

Step1: Find displacement vector

The initial position vector $\vec{r}_1 = \langle 0.4, -0.01, 0.6
angle$ m and the final position vector $\vec{r}_2 = \langle -0.211, 0.053, 0.093
angle$ m. The displacement vector $\Delta\vec{r} = \vec{r}_2 - \vec{r}_1$.
Calculating each component:

• x - component: $-0.211 - 0.4 = -0.611$
• y - component: $0.053 - (-0.01) = 0.063$
• z - component: $0.093 - 0.6 = -0.507$

So, $\Delta\vec{r} = \langle -0.611, 0.063, -0.507
angle$ m.

Step2: Convert time to seconds

The time taken $t = 2$ microseconds. Since $1\ \mu\text{s} = 1\times 10^{-6}\ \text{s}$, then $t = 2\times 10^{-6}\ \text{s}$.

Step3: Calculate average velocity

The formula for average velocity $\vec{v}_{\text{avg}} = \frac{\Delta\vec{r}}{t}$.
Dividing each component of $\Delta\vec{r}$ by $t$:

• x - component: $\frac{-0.611}{2\times 10^{-6}} = -3.055\times 10^{5}$
• y - component: $\frac{0.063}{2\times 10^{-6}} = 3.15\times 10^{4}$
• z - component: $\frac{-0.507}{2\times 10^{-6}} = -2.535\times 10^{5}$

Answer:

$\vec{v} = \langle -3.055\times 10^{5}, 3.15\times 10^{4}, -2.535\times 10^{5}
angle\ \text{m/s}$