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). part 1 (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: $-0.211 - 0.4 = -0.611$ m
• y - component: $0.053 - (-0.01) = 0.063$ m
• z - component: $0.093 - 0.6 = -0.507$ m

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

Step2: Find time interval

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

Step3: Calculate average velocity

The formula for average velocity $\vec{v}_{\text{avg}} = \frac{\Delta\vec{r}}{t}$.

Substituting the values:

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

Answer:

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