Answer
$v_{(2)}=24\frac{m}{s}$
Work Step by Step
For the maximum positive velocity, acceleration should be zero:
$a=24-12t=0$
$t=\frac{24}{12}=2s$
Thus, the maximum velocity occurs at $t=2 s$, and the velocity can be determined as:
$v_{(t)}=24t-6t^2$
Now, plug in the value $t=2$
$v_{(2)}=24(2)-6(2)^2$
$v_{(2)}=24\frac{m}{s}$
