# Chapter 2 - Motion Along a Straight Line - Problems: 18c

$-12 m/s^2$

#### Work Step by Step

The position is given by $x = 12t^2 - 2t^3$ Since the velocity equation is the derivative of the position equation we need to find $\frac{d}{dt} (12t^2 - 2t^3)$ Since $\frac{d}{dt} t^n = n*t^(n-1)$ then $\frac{d}{dt} (12t^2 - 2t^3) = 12(2)t - 2(3)t^2 = 24t - 6t^2$ Therefore $v = 24t - 6t^2$ The acceleration is given by the derivative of the velocity equation This is $\frac{d}{dt} (24t - 6t^2) = 24 - 6(2)t = 24 - 12t$ Therefore, to find acceleration at t =3, we plug in t = 3 into the equation This gives us $24 -12(3) = 24-36 = -12$ Therefore the acceleration is $-12m/s^2$

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