91Ó°ÊÓ

$$\begin{gathered} \mathbf{v}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s} \\ \mathbf{a}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}} \\ \mathbf{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s} \end{gathered}$$

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. Using kinematic equations, the velocity of the electron at given instants if the acceleration and time are known can be determined. The equation used is given below.

Formula:

The final velocity in kinematic equation is given by,

${\mathit{V}}_{\mathit{t}}\mathit{=}{\mathit{V}}_{\mathit{0}\mathit{}}\mathit{+}\mathit{(}\mathit{a}\mathit{×}\mathit{t}\mathit{)}$

(i)

Where${}^{{\mathbf{V}}_{\mathbf{0}}}$is the initial velocity, a is an acceleration and t is time.

To find velocity${}^{\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}{\mathbf{sec}}_{\mathbf{}}}$earlier, consider${}^{\mathbf{t}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}{\mathbf{s}}_{\mathbf{}}}$.

Thus the velocity by equation (i) is given by,

$$\begin{gathered} {}^{{\mathbf{V}}_{\mathbf{f}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{6}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{2}\mathbf{×}\mathbf{2}\mathbf{.}\mathbf{5}}} \\ \mathbf{}\mathbf{}{\mathbf{}}^{\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}} \end{gathered}$$

The velocity of the electron${}^{\mathbf{2}\mathbf{.}{\mathbf{5}}_{\mathbf{}\mathbf{s}}}$earlier is${}^{\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$

To find velocity${}^{\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{sec}}$earlier, consider${}^{\mathbf{t}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{s}}$.

Thus the velocity by equation (i) is given by,

$$\begin{gathered} {}^{{\mathbf{V}}_{\mathbf{f}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{6}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{2}\mathbf{×}\mathbf{2}\mathbf{.}\mathbf{5}}} \\ \mathbf{}\mathbf{}{\mathbf{}}^{\mathbf{=}\mathbf{18}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}} \end{gathered}$$

The velocity of the electron${}^{\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{s}}$later is ${}^{\mathbf{18}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$.