#### Answer

At t = 0:
$v = 0$
At t = 2 s:
$v = 5~m/s$
At t = 4 s:
$v = 20~m/s$
At t = 6 s:
$v = 30~m/s$
At t = 8 s:
$v = 30~m/s$

#### Work Step by Step

The change in velocity is equal to the area under the acceleration versus time graph. Note that the initial velocity is zero.
At t = 0:
$v = 0$
From t = 0 to t = 2 s:
$\Delta v = \frac{1}{2}(5~m/s^2)(2~s)$
$\Delta v = 5~m/s$
At t = 2 s:
$v = 0+\Delta v = 5~m/s$
From t = 2 s to t = 4 s:
$\Delta v = 15~m/s$
At t = 4 s:
$v = 5~m/s+\Delta v$
$v = 5~m/s+15~m/s$
$v = 20~m/s$
From t = 4 s to t = 6 s:
$\Delta v = \frac{1}{2}(10~m/s^2)(2~s)$
$\Delta v= 10~m/s$
At t = 6 s:
$v = 20~m/s+\Delta v$
$v = 20~m/s+10~m/s$
$v = 30~m/s$
From t = 6 s to t = 8 s:
$\Delta v = 0$
At t = 8 s:
$v = 30~m/s + \Delta v$
$v = 30~m/s + 0$
$v = 30~m/s$