Calculate Velocity of Particle at t=0.657s

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1. A particle moves in such a way that its position z, in meters, is given as a function of t, in seconds, by the equation z = 2.12m/s^2t^2 - 2.96m/s^3t^3

What is the velocity (m/s) of the particle at t = 0.657s

The answer is -1.05, which is not what I got.

2. First, I used the equation Xf = Xi + 1/2 (Vi + Vf) (t)

Plugging in .0757 for Xf

My answer was V=.231

Then I tried V= Displacement/time

.0757/.657 = .115

What am I doing wrong here?
 
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You are using the wrong equation.

Xf = Xi + 1/2 (Vi + Vf) (t) is valid only when the acceleration is constant. Here it is not. How do you find the velocity as a function of time if you know the position as a function of time? Hint: Velocity is the rate of change of position with respect to time.
 
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I already tried the equation

V=Displacement/time

.0757/.657 = .115

Do I now have to substitute that answer into the equation for position as a function of time?
 
or...do I have to take the derivative of v=displacement/time