# An electron has a constant acceleration of

An electron has a constant acceleration of $+3.2 m/s^2$. At a certain instant its velocity is 9.6 m/s. What is its velocity (a) 2.5 s earlier and (b) 2.5 s later?

#### Solution:

To solve this question, we will use the following equation:

$v=v_0+at$

(where v is final velocity, $v_0$ is initial velocity, $a$ is acceleration, and $t$ is time.)

a) To find the velocity before 2.5 seconds, we will set $t=-2.5 s$. Thus, we have:

$v=(9.6 m/s)+(3.2 m/s^2)(-2.5 s)$

$v=1.6 m/s$

b) To find the speed after 2.5 seconds, we will set $t=+2.5 s$. Thus, we have:

$v=(9.6 m/s)+(3.2 m/s^2)(2.5 s)=18 m/s$