Question

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?

Short answer

1. The velocity of electron is $\left(3.0\hat{\mathrm{i}}-8.0t\hat{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}$.
2. The velocity of electron at t=2.0 sis $\left(3.0\hat{\mathrm{i}}-16.0t\hat{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}$.
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 $-79.4°$.

Step-by-step solution

Given data

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

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:

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

$\theta ={\tan }^{-1}\left(\frac{v_y}{v_x}\right)$ … (iii)

(a) Determination of theelectron’s velocity

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

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

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

(b) Determination of theelectron’s velocity at  

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

$$\begin{gathered} \overrightarrow{v}=\left(3.0\hat{\mathrm{i}}-8.0\times 2.0\hat{\mathrm{j}}\right)\mathrm{m}/\mathrm{s} \\ =\left(3.0\hat{\mathrm{i}}-16.0t\mathit{}\hat{\mathit{j}}\right)\mathrm{m}/\mathrm{s} \end{gathered}$$

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

(c) Determination of themagnitude of velocity

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

$$\begin{gathered} v=\sqrt{{\left(3.0\mathrm{m}/\mathrm{s}\right)}^2+{\left(16.0\mathrm{m}/\mathrm{s}\right)}^2} \\ =16.3\mathrm{m}/\mathrm{s} \end{gathered}$$

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

(d) Determination ofthedirection of velocity

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

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{-16.0\mathrm{m}/\mathrm{s}}{3.0\mathrm{m}/\mathrm{s}}\right) \\ ={\tan }^{-}\left(-5.333\right) \\ =-79.4° \end{gathered}$$

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