Question

An electron has a constant acceleration of +${}^{\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}}$. At a certain instant, its velocity is +localid="1654238193944" ${}^{\mathbf{9}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$.What is its velocity (a) localid="1654238215362" ${}^{\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{s}}$earlier and (b)localid="1654238228852" ${}^{\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{s}}$later?

Short answer

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

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

Step-by-step solution

Given information

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

Concept and formula used in the given question

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,

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

(i)

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

(a) Calculation for the velocity2.5 s earlier

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,

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

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

(b) Calculation for the velocity2.5 s later

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,

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

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