Question 989333
A body subject to air resistance is allowed to fall from the top of a 36 m tower. Given that the distance s metres, through which it falls in t seconds, is given by 
s = 5 t^2 - t^3 / 3,       0 < t < 3

what is the acceleration after 3 seconds, in m/s2. Please give an exact, possibly fractional, answer (no decimals).

Do I differentiate twice?
---------------
Acceleration is the 2nd derivative of displacement.
---
So, (10t - 2t^2) / 3  ***** = (10t - 3t^2)/3
Then (10 - 4t) / 3 ***** = (10 - 6t)/3
At t = 3: s" = (10 - 18)/3 = -8/3
------------
Seems odd that the acceleration is negative.
--
Maybe it's:
s = 5t^2 - t^3/3
s' = 10t - t^2
s" = 10 - 2t
s"(3) = 4 m/sec/sec
================
I have doubts about the given equation.