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Expert-verified Found in: Page 84 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # An electron’s position is given by $\stackrel{\mathbf{\to }}{\mathbf{r}}{\mathbf{=}}{\mathbf{3}}{\mathbf{.}}{\mathbf{00}}{\mathbit{t}}{\mathbf{}}\stackrel{\mathbf{^}}{\mathbf{i}}{\mathbf{-}}{\mathbf{4}}{\mathbf{.}}{\mathbf{00}}{{\mathbf{t}}}^{{\mathbf{2}}}{\mathbf{}}\stackrel{\mathbf{^}}{\mathbf{j}}{\mathbf{+}}{\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}\stackrel{\mathbf{^}}{\mathbf{k}}$, with${\mathbit{t}}$ in seconds and $\stackrel{\mathbf{\to }}{\mathbf{r}}$ in meters. (a)In unit-vector notation, what is the electron’s velocity$\stackrel{\mathbf{^}}{\mathbf{v}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$? At ${\mathbit{t}}{\mathbf{=}}{\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{s}}$, what is $\stackrel{\mathbf{\to }}{\mathbf{v}}$(b) in unit-vector notation and as (c) a magnitude and (d) an angle relative to the positive direction of the ${\mathbit{x}}$ axis?

1. The velocity of electron is $\left(3.0\stackrel{^}{\mathrm{i}}-8.0t\stackrel{^}{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}$ .
2. The velocity of electron at t=2.0 s is $\left(3.0\stackrel{^}{\mathrm{i}}-16.0t\stackrel{^}{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}$ .
3. The magnitude of velocity of electron is 16.3 m/s.
4. The angle of velocity relative to the positive direction of x-axis is $-79.4°$.
See the step by step solution

## Step 1: Given data

The position vector of electron, $\stackrel{\to }{r}=\left(3.0t\right)\stackrel{^}{i}-\left(4.0{t}^{2}\right)\stackrel{^}{j}+\left(2.0\right)\stackrel{^}{k}$

## Step 2: Understanding the velocity

The velocity is the rate of change of position of an object with respect to time. It is a vector quantity, which has magnitude as well as direction.

The expression for velocity in general can be written as:

$\stackrel{\to }{v}=\frac{d\stackrel{\to }{r}}{dt}$ ...(i)

Here, $\stackrel{\to }{r}$ is the position vector.

The magnitude of is given as:

$v=\sqrt{{{v}^{2}}_{x}+{{v}^{2}}_{y}}$ ....(ii)

Here, ${v}_{x}$ and ${v}_{y}$ are the x and y components of velocity.

The direction of $\stackrel{\to }{v}$ is given as:

$\theta ={\mathrm{tan}}^{-1}\left(\frac{{v}_{y}}{{v}_{x}}\right)$ … (iii)

## Step 3: (a) Determination of theelectron’s velocity

Using equation (i), the velocity of electron is calculated as:

$\stackrel{\to }{v}\left(t\right)=\frac{d\left[\left(3.0t\right)\stackrel{^}{i}-\left(4.0{t}^{2}\right)\stackrel{^}{j}+2.0\stackrel{^}{k}\right]}{dt}\phantom{\rule{0ex}{0ex}}=\left(3.0\stackrel{^}{\mathrm{i}}-8.0t\stackrel{^}{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}$

Thus, the velocity of electron is $\left(3.0\stackrel{^}{\mathrm{i}}-8.0t\stackrel{^}{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}$.

## Step 4: (b) Determination of theelectron’s velocity at

Substitute $t=2.0\mathrm{s}$ in the above expression to find the velocity.

$\stackrel{\to }{v}=\left(3.0\stackrel{^}{\mathrm{i}}-8.0×2.0\stackrel{^}{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}=\left(3.0\stackrel{^}{\mathrm{i}}-16.0t\mathit{}\stackrel{\mathit{^}}{\mathit{j}}\right)\mathrm{m}/\mathrm{s}$

Thus, the velocity of electron at t=2.0 s is $\left(3.0\stackrel{^}{\mathrm{i}}-16.0t\mathit{}\stackrel{\mathit{^}}{j}\right)\mathrm{m}/\mathrm{s}$ .

## Step 5: (c) Determination of themagnitude of velocity

Using equation (ii), the magnitude of velocity is calculated as:

$v=\sqrt{{\left(3.0\mathrm{m}/\mathrm{s}\right)}^{2}+{\left(16.0\mathrm{m}/\mathrm{s}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=16.3\mathrm{m}/\mathrm{s}$

Thus, the magnitude of velocity of electron is 16.3 m/s .

## Step 6: (d) Determination ofthedirection of velocity

Using equation (iii), the direction of electron’s velocity is calculated as:

$\theta ={\mathrm{tan}}^{-1}\left(\frac{-16.0\mathrm{m}/\mathrm{s}}{3.0\mathrm{m}/\mathrm{s}}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-}\left(-5.333\right)\phantom{\rule{0ex}{0ex}}=-79.4°$

Thus, the angle of velocity relative to the positive direction of x-axis is $-79.4°$. ### Want to see more solutions like these? 