91Ó°ÊÓ

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

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:

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

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

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

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

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

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

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

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

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

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 .

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

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

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