After t hours on a particular day a freight train is st20t24

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) What is the acceleration of the train after 13 hours?

a(t)= v \'(t)

So -64 miles/hour^2
Correct?

s(t)=20t^2-(4/3)t^3 v(t) = 40t - 4t^2 a(t) = 40 - 8t a(13) = 40 - 8*13 = -64 miles/hour^2 So it correct.