## 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: 1

#### Answer

$$deceleration = 1.67 \space m/s^2$$

#### Work Step by Step

Using equation 12-4: $$v = v_0 + a_ct$$ 1. Solve for the acceleration and substitute the values: $$v - v_0 = a_ct$$ $$a_c = \frac{v - v_0}{t} = \frac{10 -35}{15} = -1.67 \space m/s^2$$ 2. The deceleration is equal to the negative of the acceleration: $$deleceration = -(-1.67 \space m/s^2) = 1.67 \space m/s^2$$

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