An electron has a constant acceleration of + 2.9 m/s2. At a certain instant its velocity is + 8.6 m/s. What is its velocity (a) 2.1 s earlier and (b) 2.1 s later?

(a) = 2.51m/s

(b) = 14.69m/s

Explanation:

Given.

Velocity v = 8.6m/s

Acceleration a = 2.9m/s2

The final velocity of particle is given by

Vf = Vo+at

. (a) 8.6 = Vo + (2.9) (2.1)

VP = 8.6 - (2.9) (2.1) = 2.51m/s

(b) Vf = 8.6 + (2.9) (2.1) = 14.69m/s