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

Published by Pearson

# Chapter 12 - Kinematics of a Particle - Section 12.2 - Rectilinear Kinematics: Continuous Motion - Fundamental Problems - Page 16: 4

#### Answer

The acceleration of the particle after 2 seconds is equal to 2 m/$s^2$

#### Work Step by Step

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: $$a =1.5(2)^2 - 8 = -2 \space m/s^2$$

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