A

Given the velocity function:
$$ v(t) = t^3 - 6t^2 + 4 $$
To find the maximum and minimum velocities, we need to take the derivative of the velocity function:
$$ v'(t) = 3t^2 - 12t $$
Setting the derivative equal to zero to find critical points:
$$ 3t^2 - 12t = 0 $$
Factoring gives:
$$ 3t(t - 4) = 0 $$
Thus, we have:
$$ t = 0 $$ or $$ t = 4 $$
Next, we evaluate the second derivative to determine the nature of these critical points:
$$ v''(t) = 6t - 12 $$
Evaluating at $$ t = 0 $$:
$$ v''(0) = 6(0) - 12 = -12 $$ (maximum)
Evaluating at $$ t = 4 $$:
$$ v''(4) = 6(4) - 12 = 12 $$ (minimum)
Now we calculate the velocities at these points:
For $$ v(0) $$:
$$ v(0) = 0^3 - 6(0^2) + 4 = 4 $$
For $$ v(4) $$:
$$ v(4) = 4^3 - 6(4^2) + 4 = 64 - 96 + 4 = -28 $$
Therefore, the maximum velocity is 4 m/s at time t = 0 seconds, and the minimum velocity is -28 m/s at time t = 4 seconds.