Solve the below:
Q: After t hours on a particular day, a freight train is s(t) = 20t^2 - (4/3)t^3 miles due east of its starting point (for 0 t 18).
a) Where is the train (relative to its starting point) after 13 hours?
b) What is the velocity of the train after 13 hours?
c) What is the acceleration of the train after 13 hours?
d) Is the train speeding up or slowing down after 13 hours? How can you tell?