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Expert-verified Found in: Page 45 ### Matter & Interactions

Book edition 4th edition
Author(s) Ruth W. Chabay, Bruce A. Sherwood
Pages 1135 pages
ISBN 9781118875865 # Question: An electron passes location $\left(0.02,0.04,-0.06\right){\mathbit{m}}$, and $2\mu s$ later is detected at location $\left(0.02,1.84,-0.86\right){\mathbf{}}{\mathbit{m}}$, (1 microsecond is${\mathbf{1}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{6}}{\mathbf{}}{\mathbit{s}}$). 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 $5\mu s$?

a) The average velocity of the ball is $\left(0,-900000,400000\right)m/s$

b) The electron will be at the position$\left(0.02,-0.86,0.34\right)m$.

See the step by step solution

## Step 1: Identification of the given data

The given data can be listed below as:

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

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

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

## Step 2: Significance of Newton’s first law to calculate the average velocity and position

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

## Step 3: Determination of the average velocity and distance

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

${\stackrel{\to }{v}}_{average}=\left(\frac{\left({x}_{B}-{x}_{A}\right)}{\Delta {t}_{AB}},\frac{\left({y}_{B}-{y}_{A}\right)}{\Delta {t}_{AB}},\frac{\left({z}_{B}-{z}_{A}\right)}{\Delta {t}_{AB}}\right)$

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}=\left(0.02,0.04,-0.06\right)\text{m}$$and{x}_{B},{y}_{B,}{z}_{B}=\left(0.02,1.84,-0.86\right)\text{m}$in the above equation, we get:

${\stackrel{\to }{v}}_{average}=\left(\frac{\left(0.02-0.02\right)\text{m}}{2×{10}^{-6}\text{s}},\frac{\left(0.04-1.84\right)\text{m}}{2×{10}^{-6}\text{s}},\frac{\left(-0.06+0.86\right)\text{m}}{2×{10}^{-6}\text{s}}\right)\phantom{\rule{0ex}{0ex}}{\stackrel{\to }{v}}_{average}=\left(0,-900000,400000\right)\text{m/s}\phantom{\rule{0ex}{0ex}}$

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:

${x}_{C}=0.02\text{m + 0}·\Delta {\text{t}}_{BC}\phantom{\rule{0ex}{0ex}}{x}_{C}=0.02\text{m}$

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${t}_{BC}=3\mu c,{v}_{average,y}=-900000\text{m/s and}{y}_{B}=1.84\text{m}$, in the above equation, we get:

${\text{y}}_{\text{C}}\text{=}\left(\text{1.84}\text{m}\right)\text{+}\left(\text{900000}\text{m/s}\right)×\left(\text{3μ}\text{s}\right)\phantom{\rule{0ex}{0ex}}\text{= 1.84}\text{m - 2.7}\text{m}\phantom{\rule{0ex}{0ex}}\text{= - 0.86}\text{m}\phantom{\rule{0ex}{0ex}}$

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 ${t}_{BC}=3\mu s,{v}_{average,z}=400000\text{m/s and}{z}_{B}=-0.86\text{m}$in the above equation, we get:

${z}_{C}=\left(-0.86\text{m}\right)\text{+}\left(400000\text{m/s}\right)×\left(\text{3s}\right)\phantom{\rule{0ex}{0ex}}=-0.86\text{m}+1.2\text{m}\phantom{\rule{0ex}{0ex}}=0.34\text{m}\phantom{\rule{0ex}{0ex}}$

Thus, the electron will be at the position$\left(0.02,-0.86,0.34\right)\text{m}$. ### Want to see more solutions like these? 