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Q6P

Expert-verifiedFound in: Page 84

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**An electron’s position is given by $\overrightarrow{\mathbf{r}}{\mathbf{=}}{\mathbf{3}}{\mathbf{.}}{\mathbf{00}}{\mathit{t}}{\mathbf{}}\hat{\mathbf{i}}{\mathbf{-}}{\mathbf{4}}{\mathbf{.}}{\mathbf{00}}{{\mathbf{t}}}^{{\mathbf{2}}}{\mathbf{}}\hat{\mathbf{j}}{\mathbf{+}}{\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}\hat{\mathbf{k}}$, with${\mathit{t}}$ in seconds and $\overrightarrow{\mathbf{r}}$ in meters. (a)In unit-vector notation, what is the electron’s velocity$\hat{\mathbf{v}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$? At ${\mathit{t}}{\mathbf{=}}{\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{s}}$, what is $\overrightarrow{\mathbf{v}}$(b) in unit-vector notation and as (c) a magnitude and (d) an angle relative to the positive direction of the ${\mathit{x}}$ axis?**

- The velocity of electron is $\left(3.0\hat{\mathrm{i}}-8.0t\hat{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}$ .
- The velocity of electron at
*t=2.0 s*is $\left(3.0\hat{\mathrm{i}}-16.0t\hat{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}$ . - The magnitude of velocity of electron is 16.3 m/s.
- The angle of velocity relative to the positive direction of x-axis is $-79.4\xb0$.

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

**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:

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

Here, $\overrightarrow{r}$ 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 $\overrightarrow{v}$ is given as:

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

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

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

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

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

$\overrightarrow{v}=\left(3.0\hat{\mathrm{i}}-8.0\times 2.0\hat{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}=\left(3.0\hat{\mathrm{i}}-16.0t\mathit{}\hat{\mathit{j}}\right)\mathrm{m}/\mathrm{s}$

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

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 .

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

$\theta ={\mathrm{tan}}^{-1}\left(\frac{-16.0\mathrm{m}/\mathrm{s}}{3.0\mathrm{m}/\mathrm{s}}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-}\left(-5.333\right)\phantom{\rule{0ex}{0ex}}=-79.4\xb0$

Thus, the angle of velocity relative to the positive direction of x-axis is $-79.4\xb0$.

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