Answer:
To determine the rider's acceleration after two minutes, we need to find the second derivative of the distance function s(t) = 3t^3 + 2t^2 + 4 with respect to time. In other words, we need to find the rate of change of the rider's velocity.
The first derivative of s(t) with respect to time is v(t) = 9t^2 + 4t.
The second derivative of s(t) with respect to time is a(t) = 18t + 4.
Now that we have the second derivative, we can find the acceleration after two minutes.
a(2) = 18(2) + 4 = 40 miles/min^2
Therefore the rider's acceleration after two minutes is C. 40 miles/min^2