Physics constant acceleration: special case problem?

Question

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?

Answer 1

Using v = a*t  -->  t = v/a, you can find the time of the measured instant.

Then, either add or subtract 2.5 to/from that time to find the desired times.

Finally, substitute these new times back in to v = a*t to find the velocities at their respective times.