# Chapter 2 - Kinematics in One Dimension - Problems - Page 49: 32

-22 $m/s^{2}$

#### Work Step by Step

We set up a system of linear equations: $x = ut + \frac{1}{2}at^{2}$ $0 = (5800)t + \frac{1}{2}(-15)t^{2}$ $t(5800-\frac{15}{2}t)=0$ $t=0, t=\frac{2320}{3}$ $0 = (8600)t + \frac{1}{2}(a)t^{2}$ $t(8600+\frac{1}{2}at)=0$ $t=0, t=\frac{-17200}{a}$ $\frac{2320}{3}=\frac{-17200}{a}$ $2320a=-51600$ $a\approx-22\; m/s^{2}$

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