A moving train moves Y meters in t seconds. Find its acceleration.

I. $$Y = t^3 - 4t^2 + 16t - 2$$

II. Velocity at that moment was 20 m/sec.

We have been given an expression of displacement in terms of time. 
Thus, differentiating this equation with respect to time once will give us an expression for velocity, and differentiating once again with respect to time will give us an expression for acceleration. 

Hence,
V = $$3t^2-8t+16$$ ..... (A) and 
A = $$6t-8$$ ............. (B)

While we have obtained the expressions for velocity and acceleration, we cannot determine their absolute values at a particular time simply from the given statement 1. However, if we take into account statement 2, we have:
$$20=3t^2-8t+16$$ which when solved gives t = 4+2(sqrt 7) 

When this value of t is put in (B), we get value of acceleration and hence, the answer can be found by using both the statements together but not by using either of them alone.