## Engineering Mechanics: Statics & Dynamics (14th Edition)

The acceleration of the particle after 2 seconds is equal to 2 m/$s^2$
1. Using equation 12-2: $$a = \frac {dv}{dt}$$ 2. Substitute the expression and derivate: $$a = \frac{d(0.5t^3 - 8t)}{dt} = 1.5t^2 - 8$$ 3. At t =2 s: $$a =1.5(2)^2 - 8 = -2 \space m/s^2$$