Question: An electron passes location $\left(0.02,0.04,-0.06\right)\mathit{m}$, and $2\mu s$later is detected at location $\left(0.02,1.84,-0.86\right)\mathbf{}\mathit{m}$, (1 microsecond is$\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathit{s}$). a) What is the average velocity of the electron? b) If the electron continues to travel at this average velocity, where will it be in another $5\mu s$?

The given data can be listed below as:

• The electron passes the location$\left(0.02,0.04,-0.06\right)\mathbf{}\mathit{m}$.

• The electron has been detected in the location$\left(0.02,1.84,-0.86\right)\mathbf{}\mathit{m}$.

The electron is detected after$2\mu s$.

This law states that an object will continue to move in a uniform motion unless it is resisted by an external object.

a) From Newton’s first law, the formula for the average velocity of the electron can be expressed as:

${\overrightarrow{v}}_{average}=\left(\frac{\left(x_B-x_A\right)}{\Delta t_{AB}},\frac{\left(y_B-y_A\right)}{\Delta t_{AB}},\frac{\left(z_B-z_A\right)}{\Delta t_{AB}}\right)$

Here, $t_{AB}$The time difference of a particle from location A to location B

$x_A,y_A,z_A=$The positions of the ball at A position in the x, y, and z coordinates

$x_B,y_{B,}{z}_B=$The positions of the ball at B position in the x, y, and z coordinate

Substituting the values localid="1662456388346" $t_{AB}=2\mu s,x_A,y_A,z_A=(0.02,0.04,-0.06)\text{m}$$and{x}_B,y_{B,}{z}_B=(0.02,1.84,-0.86)\text{m}$in the above equation, we get:

$$\begin{gathered} {\overrightarrow{v}}_{average}=\left(\frac{\left(0.02-0.02\right)\text{m}}{2\times {10}^{-6}\text{s}},\frac{\left(0.04-1.84\right)\text{m}}{2\times {10}^{-6}\text{s}},\frac{\left(-0.06+0.86\right)\text{m}}{2\times {10}^{-6}\text{s}}\right) \\ {\overrightarrow{v}}_{average}=\left(0,-900000,400000\right)\text{m/s} \end{gathered}$$

Thus, the average velocity of the ball is $\left(0,-900000,400000\right)\text{m/s}$.

b) From Newton’s first law, the equation of displacement of the ball in the x-coordinate can be expressed as:

$x_C=x_B+v_{average,x}·\Delta t_{BC}$

Here,$x_C=$The displacement of the particle in the C point in the x-quadrant,

$t_{BC}=$The displacement of the particle in the B point in the x-qua

$x_B=$The time difference of a particle from location B to location C

$v_{average,x}=$The average velocity of the particle in the localid="1662455496234" $x$quadrant

Substituting the value , localid="1662456406627" $t_{Bc}=3\mu s,v_{average,x}=0\text{m/s}and{x}_B=0.02\text{m}$n the above equation, we get:

$$\begin{gathered} x_C=0.02\text{m + 0}·\Delta {\text{t}}_{BC} \\ {x}_C=0.02\text{m} \end{gathered}$$

Similarly, the equation of displacement of the ball in the y coordinate can be expressed as:

$y_C=y_B+v_{average,y}·\Delta t_{BC}$

$y_C=$The displacement of the particle in the C point in the y -quadrant,

$y_B=$The displacement of the particle in the B point in the y -quadrant,

$t_{BC}=$The time difference of a particle from location B to location C

$v_{average,y}·$The average velocity of the particle in the y -quadrant

Substituting the value$t_{BC}=3\mu c,v_{average,y}=-900000\text{m/sand}{y}_B=1.84\text{m}$, in the above equation, we get:

$$\begin{gathered} {\text{y}}_{\text{C}}\text{=}\left(\text{1.84}\text{m}\right)\text{+}\left(\text{900000}\text{m/s}\right)\times \left(\text{3μ}\text{s}\right) \\ \text{= 1.84}\text{m - 2.7}\text{m} \\ \text{= - 0.86}\text{m} \end{gathered}$$

Similarly, the equation of displacement of the ball in the z coordinate can be expressed as:

$z_C=z_B+v_{average,z}·\Delta t_{BC}$

$z_C=$The displacement of the particle in the C point in the z-quadrant,

$z_B=$The displacement of the particle in the B point in the z -quadrant,

$t_{BC}=$The time difference of a particle from location B to location C

$v_{average,z}=$The average velocity of the particle in the z quadrant

Substituting the values $t_{BC}=3\mu s,v_{average,z}=400000\text{m/sand}{z}_B=-0.86\text{m}$in the above equation, we get:

$$\begin{gathered} z_C=\left(-0.86\text{m}\right)\text{+}\left(400000\text{m/s}\right)\times \left(\text{3s}\right) \\ =-0.86\text{m}+1.2\text{m} \\ =0.34\text{m} \end{gathered}$$

Thus, the electron will be at the position$(0.02,-0.86,0.34)\text{m}$.